/sci/ - Science & Math

Seems interesting. Glad to see some well though out material being posted.

>>7496579
I read through your first thread OP. I'm excited to hear your updates. You mind going over everything you covered last time with a quick summary? It was a lot to follow, especially without any visuals.

I'm really glad this is back OP, I studied this in high school and missed the opportunity to post some stuff I wanted to. I'll be back soon with a conjecture, hopefully /sci/ can help figure out how to prove it (or find a counter example).

It is easily shown that for an Euler brick, it is impossible that any two of its edge lengths should be equal (a √2 term inevitably crops up which spoils the rational requirement of edge lengths). So where Euler bricks are concerned, we discuss integer edge-length triples (a,b,c) such that no two of a-c are equal, and the OP's equations work out. Take care not to confuse these "a-c" with the "a-c" of a standard Pythagorean theorem formula-all of our a-c are LEGS, c is NOT a hypotenuse here.

Now, if you imagine an Euler brick situated in R^3, such that it has a vertex (corner) at the origin, then the Euler brick could also be oriented in space so that its far corner is at the rectilinear coordinate (a,b,c). But since R^3 also involves negative coordinates, and we can permute (a,b,c) in six different ways, since no two are equal, it turns out that an Euler brick has (more generally) 48 different point-representations in R^3. It furthermore turns out that any six of these in a given octant define a plane (pick any three, and the other three are co-planar), and that these plane-segments describe a regular octahedron with vertices on the axes of the pleasing form +-(a+b+c).

Now, this remark, of itself, is not at all special to Euler brick-points. If you have some (x,y,z) in R^3 such that no two of x-z are equal, then it follows by the above that you can have 48 distinct, related points with variously permuted and signed coordinates (all one some sphere centered at the origin).

But it is in view of two observations about a FACE of the above described octahedron, that we develop the situation with respect to Euler bricks.

In the picture, "q" is a shorthand for a+b+c. Imagine the face of the regular octahedon in the strictly positive octant, and you're looking face-on toward the origin, with the positive x-axis going off to the left, y to the right.. and z up. But the lines on the triangle are not intended to suggest spatial depth, but regions on the face.

>>7496662

Especially notice here, that there are "MAJOR" diagonals, and "MINOR" diagonals (imagine planes slicing through the plane segment of this face of the octahedron). These describe general areas of possibility (or impossibility) for where the far vertex of an Euler brick may lie in R^3.

One given Euler brick endpoint (a,b,c) lies somewhere strictly inside the green region, for example. Its permutations which stay in the same octant, populate the other sectors.

In the overall scheme, the far point cannot possibly lie on the outside edge of our triangle, for then it should have an edge of zero, and be a filthy degenerate. Similarly, the far point cannot possibly lie on any of the MAJOR diagonals (let alone the center!), for the reason that this means that some two of its coordinates are equal, which is impossible. Imagine the planes x=y, x=z, y=z slicing through the principal planes at 45 degree angles (and thus also through our octahedral face).

But in the course of investigating Euler-brick point permutations and data (and in light of this geometric treatment), it became reasonable to consider these MINOR diagonals, which generally satisfy that three unequal terms are EVENLY SPACED- that is, that c-b = b-a, for example. Moreover, the author is not aware of any Euler brick which has evenly-spaced edge lengths, and he thinks he's close to dispensing with the idea (but we'll see). This leads us to the central reason for this thread's existence, a:

CONJECTURE: There does not exist an Euler brick with edge lengths a, b, c such that a<b<c, and c-b = b-a.

Regardless of what becomes of this conjecture, we are going to have a long and winding road which will take us into other parts of number theory, and the author is encouraged by the initial response.

Now it's time to present review details, previous results, and links.

Where can I find the first thread?

Hi OP. In the last thread I asked if anyone knew a derivation of the parameterization given in the Wikipedia article:

https://en.wikipedia.org/wiki/Euler_brick#Generating_formula

It's easy to check that it works, but is there a principled way to derive it? I don't think anyone answered before.

>>7496580
>>7496613
>>7496654

By way of previously covered material, the previous 4chan thread on same subject (where a few simple lemmas were developed) has happily been archived at

>>/sci/thread/7437394

Meanwhile, the general Wikipedia article is at

https://en.wikipedia.org/wiki/Euler_brick

This wiki link also has a link to the mathworld article, but I leave that one out for the simple reason that I dislike their notation. My notation generally conforms to wiki's.

Most importantly for development, the result about Almost-Isosceles Right Side(d triangles) by M.A. Nyblom

http://www.fq.math.ca/Scanned/36-4/nyblom.pdf

is /just/ the juice I needed to investigate my conjecture, one way or the other. Either way I will devote a significant chunk of the thread toward expositing this note of his.

Finally, in the following post, I will recap the big True Facts about Euler bricks that I presented in this post's first link.

>>7496692

The first thread has been archived at the first "warosu" link

>>/sci/thread/7437394

in this post >>7496710 . It has long since exited 4chan's archive.

>>7496705

I saw your post but as I didn't understand the generating formula at the time, I didn't bother to reply. I now think the generating formula in your link may be related to Euclid's formula for generating Pythagorean triples, which is here:

https://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple

In fact, I'd bet a dollar on it. The Euler brick involves three coupled equations for Pythagorean triples, and the above link is just too similar to your

https://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple

for them not to be related. Both men were without modern computers.

>>7496710

I lied! Now here are the BIG TRUE FACTS (lemmata) about Euler bricks, previously covered (and how to prove them):

1. No two edge lengths of an Euler brick are equal. Pf: Pythagorean thm and √2 is irrational, reductio ad absurdum (RAA).

2. An Euler brick does not have a unit (=1) edge. Pf. RAA, squares are also finite sums (series) of consecutive odd numbers.

3. The longest edge length c of an Euler brick may be shorter than its shortest face diagonal d, and vice verse. (inspect low known bricks at wiki for this: 240-252-275 & 44-117-240 for example)

4. No two face diagonals are equal. And, they "correspond" length-wise to a,b,c in the expected way (short-mid-long). Pf. Use 1. and 2, RAA, and manipulate various equalities and inequalities.

5. (INCOMPLETE!!!) "the expressions a-g^2 satisfy this big chard of inequalities". Partial proof: a long, ordered train of thought, in pic related. (help?)

6. The 48 versions of an EB in R^3 have endpoints on the faces of a regular octahedron with vertices "+- a,b,c", etc. Pf. describe planes in space, find vertices, etc.

(conjecture, I wrote: c-b = b-a)

>>7496613

Hello, this post >>7496745 covers the developments from last time.

Like last time, this feels like a multi-day thread, and I aim to keep it that way. I have two open issues that I haven't thought hard enough about, so I would REALLY ENJOY help:

--- Can anyone show the maximal volume of some cuboid whose opposite edge varies over some octahedral face as suggested in >>7496662 (I conjecture the center) ? This should be a fairly simple calc problem but I haven't been fussed to set up the integral(s).

--- Can anyone complete (counter-examples count) the table of inequalities in >>7496745 ? I can explain the image at will.


------------------------------------------------------------------------------------------------------------------------------------------------

Future posting (tonight-tomorrow) will develop an appreciation of Nyblom's great paper

http://www.fq.math.ca/Scanned/36-4/nyblom.pdf

As it relates to my conjecture at the end of >>7496684 .

I am also very interested to hear of >>7496654 's conjecture.

Great thread, I will study some of this and post some "thoughts" aswell

https://www.youtube.com/watch?v=J9FImc2LOr8
Thread theme

>>7496654
And I'm back. Here's what I've got.

I developed a fairly simple algorithm to search for Euler Bricks. Then idea works like this. Start with two primitive Pythagorean triples, which have legs given by <span class="math">(a_1, b_1)[/spoiler] and <span class="math">(a_2, b_2)[/spoiler].
Let <span class="math">l = lcm(a_1, a_2)[/spoiler].
Then we will scale our original Pythagorean triples, so that they both have a side equal to <span class="math">l[/spoiler]. That is:

<span class="math">a_1\prime = a_1 \cdot \frac{l}{a_1} = l[/spoiler]
<span class="math">b_1\prime = b_1 \cdot \frac{l}{a_1}[/spoiler]
<span class="math">a_2\prime = a_2 \cdot \frac{l}{a_2} = l[/spoiler]
<span class="math">b_2\prime = b_2 \cdot \frac{l}{a_2}[/spoiler]

Let <span class="math">a = l, b = b_1 \prime, c = b_2 \prime[/spoiler]. We know <span class="math">\sqrt{a^2 + b^2}[/spoiler] is an integer, as well as <span class="math">\sqrt{a^2 + c^2}[/spoiler], so if <span class="math">\sqrt{b^2 + c^2}[/spoiler] is an integer, then <span class="math">(a,b,c)[/spoiler] is an Euler Brick. Here is a short python script which generates Euler Bricks via this method: http://pastebin.com/qSM8ZPVs.. (cont.)

>>7496920
Also note that though the algorithm is given in terms of the lcm of <span class="math">a_1, a_2[/spoiler], there are 4 ways to potentially generate a brick using two triples. Specifically, using <span class="math">lcm(a_1, a_2)[/spoiler], <span class="math">lcm(b_1, a_2)[/spoiler], <span class="math">lcm(a_1, b_2)[/spoiler], <span class="math">lcm(b_1, b_2)[/spoiler]. The script given takes advantage of this.

The attached picture is the program's output using the settings in the pastebin. The first thing of note is that the bricks repeat. Specifically, they generated up to 3 times. This is not surprising, If <span class="math">(a,b,c)[/spoiler] is an Euler Brick, <span class="math">(a,b), (b,c), (a,c)[/spoiler] are the legs of Pythagorean triples, and have corresponding primitive triples <span class="math">(a_1,b_1), (a_2,b_2), (a_3,b_3)[/spoiler]. Then applying the algorithm to any two of these three triples should result in the original brick, and there are there are three ways to choose two out of three triples.

The next result is more subtle. There are an even number of results. I encourage you to play around with the number of bricks the program generates and see that this always seems to be the case, no matter how many it bricks it generates. I will begin to show why this is the case in the next post.

>>7496920
>>7496947

These are EXCELLENT contributions, thank you! One immediately recognizes multiple correct bricks in the output of >>7496947 !!

Please continue to develop your own train of thought in this thread as far as you can!

For my part (right now), I will go off into a separate post and re-establish a simple fact about Euler bricks which is mentioned on wiki.

>>7496993
Thank you :)

>>7496947
An obvious point of interest is how many Euler Bricks will a given triple generate if it is successful. Remember, given a pair of triples, the algorithm can be applied in 4 ways: <span class="math">lcm(a_1, a_2), lcm(a_1, b_2), lcm(b_1, a_2), lcm(b_1, b_2)[/spoiler]. So on the surface, a pair of triples may generate anywhere between 0 and 4 Euler Bricks. By investing a few specific examples, a very obvious pattern emerges. When a pair is successful, it will generate two different Euler Bricks, using opposing lcm's. That is, if <span class="math">lcm(a_1, a_2)[/spoiler] generates a brick, <span class="math">lcm(b_1, b_2)[/spoiler] will also generate a different brick. Further, <span class="math">lcm(a_1, b_2)[/spoiler] and <span class="math">lcm(b_1, a_2)[/spoiler] will not generate bricks. So, just by investigation of the data, a pair of triples will on generate either 0 bricks, or 2, and if it generates bricks, it will generate them using opposing lcms. This is my conjecture. Here is a modified version of the original script which tests this conjecture: http://pastebin.com/FxJCSv3z..

I am done for now, and I welcome anyone to attempt to see if they can prove (or disprove) my conjecture.

A simple result directly referred to in >>7496993 , which we will now prove (because it relates directly to the motivation >>7496662 >>7496684 ), is that scalings of primitive Euler bricks are also Euler bricks, AND vice verse. So this will be a biconditional strengthening of the first item given at https://en.wikipedia.org/wiki/Euler_brick#Properties . That is, in the other direction, if some triple's components give an Euler brick and are NOT all relatively prime (ka,kb,kc), then they must reduce to some relatively prime primitive (a,b,c).

It goes like this, LEMMA: Let a,b,c,k be positive natural numbers, and let "EB" denote the set of all natural edge triples (a,b,c) which satisfy the conditions of an Euler brick (and so we say that (a,b,c) is in EB), where gcd(a,b,c)=1. Then,

\displaystyle (a,b,c) \in EB \iff (ka,kb,kc) \in EB

The "gcd(x,y,z)" notation is NOT to be confused with other similar notations that we use, involving Pythagorean triples, points in space, and so on. We emphasize this, because in other treatments, (particularly Nyblom, mentioned already in the thread), the simple notation (a,b) is used to indicate the gcd of a and b.

In the "forward" direction, we will do one line and leave the others as exercises. If we begin with the assumption that (a,b,c) is an Euler brick (with OP's preferred (d,e,f) side diagonal notation), then we must consider (ka,kb,kc) for natural k. It must of course satisfy the form

<span class="math"> \displaystyle (ka)^2+(kb)^2=k^2(a^2+b^2)=k^2d^2=(kd)^2 [/spoiler]

And etc, from which the forward direction (P -> Q) is satisfied. Of course, its contrapositive (¬Q -> ¬P) also immediately follows from this result, but its CONVERSE (Q -> P) and its INVERSE (¬P -> ¬Q) do NOT, which is why we will take the trouble to do the reverse case below, to satisfy the biconditional! Our approach is a bit trickier and less straightforward, this time.

A simple result directly referred to in >>7496993, which we will now prove (because it relates directly to the motivation >>7496662 >>7496684 ), is that scalings of primitive Euler bricks are also Euler bricks, AND vice verse. So this will be a biconditional STRENGTHING of the first item given at https://en.wikipedia.org/wiki/Euler_brick#Properties . That is, in the other direction, if some triple's components give an Euler brick and are NOT all relatively prime (ka,kb,kc), then they must reduce to some relatively prime primitive (a,b,c).

It goes like this, LEMMA: Let a,b,c,k be positive natural numbers, and let "EB" denote the set of all natural edge triples (a,b,c) which satisfy the conditions of an Euler brick (and so we say that (a,b,c) is in EB), where gcd(a,b,c)=1. Then,

<span class="math"> \displaystyle (a,b,c) \in EB \iff (ka,kb,kc) \in EB [/spoiler]

The "gcd(x,y,z)" notation is NOT to be confused with other similar notations that we use, involving Pythagorean triples, points in space, and so on. We emphasize this, because in other treatments, (particularly Nyblom, mentioned already in the thread), the simple notation (a,b) is sometimes used to indicate the gcd of a and b (was confusing news to me!)

In the "forward" direction, we will do one line and leave the others as exercises. If we begin with the assumption that (a,b,c) is an Euler brick (with OP's preferred (d,e,f) side diagonal notation), then we must consider (ka,kb,kc) for natural k. It must of course satisfy the form

<span class="math"> (ka)^2+(kb)^2=k^2(a^2+b^2)=k^2d^2=(kd)^2 [/spoiler]

And etc, from which the forward direction (P -> Q) is satisfied. Of course, its contrapositive (¬Q -> ¬P) also immediately follows from this result, but its CONVERSE (Q -> P) and its INVERSE (¬P -> ¬Q) do NOT, which is why we will take the trouble to do the reverse case below, to satisfy the biconditional! Our approach is a bit trickier and less straightforward, this time.

>>7497051

So we did the "forward" half of the above biconditional, and to get at the other half, we consider first the INVERSE situation

<span class="math"> \displaystyle (a,b,c) \notin EB \rightarrow (ka,kb,kc) \notin EB [/spoiler]

and, in this rendering, what does it mean to say that some triple of naturals (a,b,c) is NOT "an Euler brick"? Quite simply, that ANY ONE (or two, or all three) of the involved equations >>7496579 don't give up some integer hypotenuse d,e or f.

So WLOG we can consider the first case, in the situation where d is NOT a natural number. This is enough to say that d^2 is not a perfect square, but we know something even better about perfect squares, and naturals that aren't: perfect squares by definition have natural square roots, and any natural number that ISN'T a perfect square, has for its positive square root an IRRATIONAL number (due to FTA, roots of primes etc), so the "not-natural-but-maybe-just-rational" business that a close reader could have picked up on as a possibility a clause or two ago, vanishes. Since non-zero d is not-natural by assumption, it is forced to be irrational due to the above. And, bringing natural k into the situation and considering the term kd (the crux here, again), we know from pic related that the product of a non-zero irrational and a non-zero rational number is ALWAYS irrational (to suppose otherwise is immediately absurd).

Thus etc, and we have an even stronger BICONDITIONAL about scalings of Euler bricks. As a reminder, the reason for being so careful about this result is that (questions about) scalings of Euler bricks are directly related to the conjecture early in the thread, conjecturing that edge lengths can't be evenly spaced!

>>7496996

This is a very interesting and mathematically data-based approach to exploring bricks, which is also along the lines of checks that I'm doing. I'm currently caught up in my train of thought but I mean to fully understand how this works at some point. Moreover, the focus on the "legs" of triples dovetails very nicely with the direction I'm moving in, which is why I want to especially encourage all development of this separate train of thought.

The direction I'm moving in is, to investigate and exposit properties of "almost-isoceles" Pythagorean triples, whose legs differ by a unit. I was surprised that (3,4,5) isn't the only one! There are (non-trivially) infinitely many of these triples, and per Nyblom, they have a one-to-one correspondence with the set of SQUARE TRIANGULAR NUMBERS: that is, the numbers which are BOTH square and triangular. These are spaced out very far apart and suck to calculate very quickly, but there /are/ infinitely many of them: 1,36,1225, 41616...

>>7497125
Square triangle numbers follow a recursive formula

http://www.johndcook.com/blog/2015/08/21/computing-square-triangular-numbers/

Not directly related to that, here is Matlab code to generate all bricks with a<b<c up to a given size of a. Remove the gcd() check if you want to print scaled bricks as well as primitive.

clear all

% look for all bricks a<b<c

for a=3:1000

bc=[a+1: floor((a^2-1)/2)]; % will be possible vals for b and c
q=sqrt(a^2+bc.^2); % this must be an integer
k=find(q==round(q));
bc=bc(k); % the possible b's and c's
bc2=repmat(bc.^2,[length(bc),1]);
q=sqrt(bc2+bc2');
[k,j]=find(q==round(q));
if length(k)>0
h=find(k<j);
abc=[ repmat(a,length(h),1) bc(k(h))' bc(j(h))'];
for j=1:length(h)
if gcd(gcd(abc(j,1),abc(j,2)),abc(j,3))==1
disp(abc(j,:))
end
end
end
end

>>7497125

Picking up here (and backing up a bit before continuing to Nyblom), let's brainstorm about scalings of natural triples (and candidate Euler bricks) a bit.

In the top case of the figure, if you have some (relatively prime) triple that you know for a fact is NOT an Euler brick, then it follows from >>7497051 , >>7497097 that none of its multiples are Euler bricks. But since these are always /natural/ scalings, they always produce pairwise strictly greater scalings of components - we don't consider zero, and we know that an Euler brick cannot have a unit edge length, but it's worth mentioning that not even 1 is safe from this "carrying-forward-to-the-right". A family of triples about which it is true that c-b = b-a are thus ruled out.

But of course, you could always make up some other such triple, centered around some other b, say 1,5,9, and inspect.

Or still further, you could observe that for some natural center b with its attendant a and c, there are finitely many triples (a,b,c) centered at b. Spacing the numbers eventually takes you back to 1 (or 2 if you like, since we've banished 1 where Euler bricks are concerned). Either way, we say that there are n-1 (or n-2) triples /centered/ on b, being various bases of other families of distinct, evenly spaced triples.

If the conjecture >>7496684 is true, it will eventually become necessary to completely classify evenly spaced triples along the lines developed in this post. But for now, we will really begin with a review of Nyblom and the AIRA triangles (3,4,5), (20,21,29) etc (not to be confused with the above candidate-Euler brick triples!!).

>>7497600

Thanks!

It's time to have a look at Nyblom, and his derivation of AIRA triangles and their one-to-one correspondence to square triangular numbers:

http://www.fq.math.ca/Scanned/36-4/nyblom.pdf

For the next few posts I'll be following along with Nyblom's argument. He starts out by mentioning Euclid's formula and a few other historical items for background. Note especially that when Nyblom writes something like "(x,y)=1" , this means the /GCD/ of x and y. The intro concludes, and then a

LEMMA: "there are infinitely many square triangular numbers."

How do we know this? the whole note is an invitation to the reader to follow along and work things out for one's self. But this treatment of the lemma is basically the same thing as

https://en.wikipedia.org/wiki/Square_triangular_number#Other_characterizations

So if you go from

<span class="math"> \displaystyle T_{n} \rightarrow T_{4n(n+1) [/spoiler]

step-by step, starting from T_1 takes you to T_8, for example. And, iterating with the same rule, T_8 takes you to T_288, and this continues. The algebra works out that the next thing up is necessarily a square number, as wiki shows. BUT!

But but but, the above is just a proof of infinitude, and NOT a complete classification, of itself, of the set! Because the above train of thought goes 1 -> 36 -> 41616, skipping 1225! You could of course just as easily start at 1225 and work up, but it's not the same thing. Nyblom eventually gives (the start of) a complete listing on the back end, by means of various fomulas. For the moment, all we have is "infinitude" plus whatever else we can search on the internet.

Thanks!

It's time to have a look at Nyblom, and his derivation of AIRA triangles and their one-to-one correspondence to square triangular numbers:

http://www.fq.math.ca/Scanned/36-4/nyblom.pdf

For the next few posts I'll be following along with Nyblom's argument. He starts out by mentioning Euclid's formula and a few other historical items for background. Note especially that when Nyblom writes something like "(x,y)=1" , this means the /GCD/ of x and y. The intro concludes, and then a

LEMMA: "there are infinitely many square triangular numbers."

How do we know this? the whole note is an invitation to the reader to follow along and work things out for one's self. But this treatment of the lemma is basically the same thing as

https://en.wikipedia.org/wiki/Square_triangular_number#Other_characterizations

So if you go from

<span class="math"> \displaystyle T_{n} \rightarrow T_{4n(n+1)} [/spoiler]

step-by step, starting from T_1 (or what Nyblom calls T(1) , this comes up later) takes you to T_8, for example. And, iterating with the same rule, T_8 takes you to T_288, and this continues. The algebra works out that the next thing up is necessarily a square number, as wiki shows. BUT!

But but but, the above is just a proof of infinitude, and NOT a complete specification, of itself, of the set! Because the above train of thought goes 1 -> 36 -> 41616, skipping 1225! You could of course just as easily start at 1225 and work up, but it's not the same thing. Nyblom eventually gives (the start of) a complete listing on the back end, by means of various fomulas. For the moment, all we have is "infinitude" plus whatever else we can search on the internet.

So, okay, we have infinitely many square triangular numbers, and we should be (are) able to write them all out via some other explicit formula given elsewhere (below). It's with reference to these numbers that Nyblom makes the theorem, which I will restate slightly:

"There are infinitely many non-trivial pairs of consecutive integers m, m+1, the sum of whose squares is itself a perfect square of some other number 2s+1, which three combine to give an almost-isoceles (AIRA) Pythagorean triple of the form (m,m+1,2s+1). The set of all AIRA triangles (or triples) has a one-to-one correspondence to the set of all square triangular numbers, the nth AIRA triple is given by a formula having the nth square triangular number as its argument, and a triple is an AIRA triple if and only if the below variable is a square triangular number. Symbolically,

<span class="math"> \displaystyle (m,m+1,2s+1) = \Bigg(\frac{4\sqrt{\diamondsuit_{n}}-1+\sqrt{8\diamondsuit_{n}+1}}{2},\frac{4\sqrt{\diamondsuit_{n}}+1+\sqrt{8\diamondsuit_{n}+1}}{2},2\sqrt{\diamondsuit_{n}}+\sqrt{8\diamondsuit_{n}+1}\Bigg) \iff \diamondsuit_{n} \in \diamondsuit [/spoiler]

where "diamondsuit_n" denotes the nth square triangular number, and "diamondsuit" denotes the set of all square triangular numbers."

This theorem has three important components:

- infinitude of AIRA triples,
- the AIRA triples are non-trivial ( gcd(m,m+1,2s+1)=1 ),
- and the AIRA triples correspond exactly to "diamond", iff, necessary and sufficient, etc. You can never ever come up with some other natural number which gives you an AIRA triple via the above.

So, okay, we have infinitely many square triangular numbers, and we should be (are) able to write them all out via some other explicit formula given elsewhere (below). It's with reference to these numbers that Nyblom makes the theorem, which I will restate slightly:

"There are infinitely many non-trivial pairs of consecutive integers m, m+1, the sum of whose squares is itself a perfect square of some other number 2s+1, which three combine to give an almost-isoceles (AIRA) Pythagorean triple of the form (m,m+1,2s+1). The set of all AIRA triangles (or triples) has a one-to-one correspondence with the set of all square triangular numbers, the nth AIRA triple is given by a formula having the nth square triangular number as its argument, and a triple is an AIRA triple if and only if the below argument is a square triangular number. Symbolically,

<span class="math"> \displaystyle (m,m+1,2s+1) = \Bigg(\frac{4 \sqrt{\diamond_{n}}-1+ \sqrt{8\diamond_{n}+1}}{2},\frac{4 \sqrt{\diamond_{n}}+1+ \sqrt{8\diamond_{n}+1}}{2},2 \sqrt{\diamond_{n}}+ \sqrt{8\diamond_{n}+1}\Bigg) \iff \diamond_{n} \in \diamond [/spoiler]

where "diamond_n" denotes the nth square triangular number, and "diamond" denotes the set of all square triangular numbers."

This theorem has three important components:

- infinitude of AIRA triples,
- the AIRA triples are non-trivial ( gcd(m,m+1,2s+1)=1 ),
- and the AIRA triples correspond exactly to "diamond", iff, necessary and sufficient, etc. You can never ever come up with some other number which gives you an AIRA triple via the above.

In the event that LaTeX doesn't behave, the theorem is also in pic related.

>>7497923

First of all, non-triviality of any of our triples is easily dispensed with in this case, since for any natural q, gcd(q,q+1) (or as Nyblom might write, simply (q,q+1) ) is equal to one (left as an exercise), and thus bringing any natural hypotenuse into the mix doesn't matter, a given AIRA triple has relatively prime components, and each one stands very much on its own (of course, the multiples wouldn't even be AIRA triples!). So what's left is to show infinitude, and the if-and-only-ifness of the thing. But first, we must relate square triangular numbers to AIRA triples in the first place, and Nyblom does this in the following way. The letters are (generally) natural numbers.

- m and its successor, squared, require some s (basis of some odd number).

- m and s necessarily satisfy certain inequalities, a series of expressions constrain m and s, requiring some r to exist

- a quadratic in s (taking r along for the ride) is solved, and the discriminant has to be such-and-such. The negative radical has to be thrown out (left as an exercise)

- The discriminant involving r requires still further that some n exist (basis of another odd number), phew, it's getting tedious by now, we're in the tall weeds. But this is the most delightful part of Nyblom's note --- :^) ---

- rearrangement of r and n is precisely A SQUARE TRIANGULAR NUMBER

<span class="math"> \displaystyle r^2 = \frac{n(n+1)}{2} [/spoiler]

in its two aspects, of which there are infinitely many per the lemma! Either side IS the square triangular number. So the whole thing gets walked back right up the track, establishing the formula, and infinitude of solutions!

>>7497923
>>7497938

To complete this train of thought, we should say a bit about the necessity and sufficiency of square triangular numbers to the above formula, and why I altered Nyblom's notation slightly. First, the regrettable choice of T_n to stand for the /square root/ of such-and-such a square triangular number (as opposed to the earlier T(n)!! ; although germane to the convenient recurrence relation given about a page down) doesn't quite convey the idea, the /feeling/ of the square triangular number in both of its aspects as we should appreciate it, to appreciate the above formula.

That larger radical form's "discriminant", repeated thrice, is necessarily an odd perfect square, so the whole thing /has/ to be and odd number (or natural number). But this happens exactly when "diamond_n" is a TRIANGULAR number, as was worked out here! https://boards.4chan.org/sci/thread/7494436/

Meanwhile, for /natural/ numbers, their square roots are integers precisely when they themselves are squares, that is PERFECT SQUARES. So "diamond_n" must pull double duty as both a square and as a triangular number in my long rephrased expression above. And there are infinitely many such, so we're good there!

So in the first two terms, you've got an even + odd + odd (=even), divided by two, which returns an integer and closure is nice and happy. and that hypotenuse is just a number plus a number. Thus IF diamond is a square triangular number, THEN the formula is good, and IF some number satisfies the formula, THEN it must be both square and triangular. This completes the proof.

Now, some thoughts about AIRAs, as they relate to Euler bricks and similar forms, and the missive >>7497736 .

What is an Euler brick? We of course have our geometric definition >>7496579 , but consequently we can think of "a brick" as a system of three coupled diophantine equations, or as a point in R^3 whose coordinates are simply the edge lengths. As an aside, coordinates in R^3 generally follow a signing-permuting convention, sketched out in pic related. But by way of generality, how could AIRAs conceivably fit into an Euler brick (or similar forms) at this point?

If the "thing" is a cube, then it's no Euler brick as we already know (all edge lengths are equal). But it's worth noting the triviality that for a cube, it's also impossible that any of its (one with multiplicity three) associated pythagorean formula(e) can be an AIRA -none of the legs differ by a unit of course, so that's one degenerate case.

Another degenerate case is the situation (a,a,b) (or, for that matter, (a,b,b) ) with nonzero a and b, This has two equations (one with multiplicity two), of which one (the square one) is definitely not an AIRA, but the other two could be: imagine either of (3,3,4) or (3,4,4), say. But then of course, these are not Euler bricks, having some two edges equal (and thus entailing some √2 term along that hypotenuse).

So we meander back towards the meat of the subject, but some other degeneracy is first entertained: for an Euler brick (a,b,c) with no two of nonzero a-c equal, it certainly could /not/ be the case that all three of its equations are AIRAs, as this would entail the absurdity that c-a = c-b = b-a = 1 (you can't have three unequal naturals all having a distance of one from each other, lines don't work that way); this is simply due to the way in which the equations (and the attendant legs, and the natural number line itself) are coupled.

On the other hand, just look at some primitive known bricks! (cont.)

>>7498397

https://en.wikipedia.org/wiki/Euler_brick#Examples

None of THESE perfectly good, primitive bricks entails an AIRA! So zero happens all the time, three is RIGHT OUT, and we are left to consider the situations where there is either exactly one AIRA among an Euler brick's equations, or two. Let's take the latter first.

Let (a,b,c) be an Euler brick, which satisfies the characteristic equations given in the OP >>7496579 . Suppose that this brick has EXACTLY TWO AIRAs among its three equations, in particular, suppose that b=a+1 by assumption. Then WLOG the only thing that's left, to get exactly two AIRAs, is to have c=b+1=a+2, or else set c=a-1, but we won't do this out of our preferred convention that a<b<c. Either way, it doesn't matter. The assumption that an Euler brick has exactly two AIRAs among its equations requires that all three of (a,b,c) are /consecutive/ natural numbers, of some (-1,0,+1) form, however we like. And the "exactly two" thing is all well and good, since the remaining equation involves two legs with a difference of two, and is thus not an AIRA. Moreover, this alleged brick is primitive, since gcd(n,n+1)=1 for some natural n. But this is a bit beside the point, it's just a thought. I'm now in uncharted territory, but I'm feeling a major case of the conjecture >>7496684 getting very close...

What all of this furthermore entails is that there are /consecutive AIRA triangles/ of the form (m,m+1,2s+1) , (m+1, m+2, 2v+1) for some natural m,s,v. But such a possibility looks veeeery bogus in light of our strong result about the AIRA triangles in general, per >>7497923 ...!!!

So I think an attentive reader can tell where I'm going, but in my chosen (inelegant) approach, the grunt work to actually show such-and-such will involve some onerous calculation, for which I need more blank paper, and to think a bit more. :^)

A polite bump to keep the thread going. The OP is still thinking about how to completely characterize evenly spaced natural triples, as suggested in >>7497736 , as well as how to formally establish the so-far developed contradiction by means of the sparsity of AIRA triples as defined by >>7497923.

Hints are welcome! Any help with the two items in >>7496773 is especially wanted, esp maximizing a cuboid's product over some surface.

Found two bricks that share a face:

(1008,1100,1155)
(1008,1100,12075)

>>7499844

This is /very/ interesting. So interesting in fact, that it deserves inspection.

First of all, none of the edges in either brick are pairwise relatively prime. (not even 1155 and 12075 are relatively prime, the only remaining curiosity worth checking). But, each /triple/ has a gcd of 1, and are consequently relatively prime, primitive bricks.

Furthermore, notice that both bricks have components which are multiples of 252 and 275, two pairwise relatively prime numbers which are components of an interesting low brick (240,252,275):

https://en.wikipedia.org/wiki/Euler_brick#Examples

This brick is interesting as an early counterexample, because its /longest edge/ c=275 is /shorter/ than its /shortest face diagonal/ d=348, which serves as a spoiler of the global inequality, referenced in the BRIGHT RED cells in >>7496745 . (That is, opposite what the above image said, MOST early Euler bricks DISOBEY this inequality, with the brick currently under discussion being an exception, meaning that you can have either c>(=?)d or vice verse).

(240,252,275) is also interesting just for its simple geometry. It certainly seems to be a lot more "cube-like" and less eccentric than the other bricks listed.

Now compare (240,252,275) with (1008,1100,1155). The latter is NOT a simply scaled multiple of the former, of the form >>7497051 . And yet, in both cases, the long edge c is to the short edge a as in the other, that is, c/a : c/a , or

<span class="math"> \displaystyle \frac{1155}{1008} = \frac{275}{240} = \frac{55}{48} [/spoiler]

Meanwhile, the other ratios transpose. the LONG TWO edges of the SMALLER brick are to each other as the SHORT TWO edges of the LARGER brick are to each other...

<span class="math"> \displaystyle \frac{1100}{1008} = \frac{275}{252} = \frac{275}{252} [/spoiler]

and the SHORT TWO edges of the SMALLER brick are to each other as the LONG TWO edges of the LARGER brick! viz.

<span class="math"> \displaystyle \frac{1155}{1100} = \frac{252}{240} = \frac{21}{20} [/spoiler]

>>7500151

And the 20 and 21 in the latter term are so interesting, in that they are precisely the legs of an small AIRA triangle (20,21,29), just as I have been discussing!

The train of thought is such that I would reach out to program-anon to set up some sort of script to give prime factorizations of a brick's terms, look for interesting ratios, etc. I can set up simple data checks in excel by myself and model things in that environment, but I know almost nothing about actual coding.

I'm going to complete some observations about the two bricks shown in >>7499844 , then go back to my thing.

What the fuck is the point in this shit? This is back-of-the-newspaper-puzzle-tier. People actually get PhDs studying this bullshit? Universities actually fund research of this shit? This is why we have gotten no where in stuff like fusion. Back in the 20th century the government forced the spergs to concentrate on matters of national importance such as atomic weapons and ballistic missiles. Now autists have returned to wasting time on fancy puzzles.

Let's do a raw, straight-up check of the two bricks given in >>7499844 :

<span class="math"> 1008^2 + 1100^2 = 1492^2 \rightarrow 1016064+1210000=2226064 [/spoiler]

<span class="math"> 1008^2 + 1155^2 = 1533^2 \rightarrow 1016064+1334025=2350089 [/spoiler]

<span class="math"> 1100^2 + 1155^2 = 1595^2 \rightarrow 1210000+1334025=2544025 [/spoiler]

<span class="math"> 1008^2 + 1100^2 + 1155^2 = \sqrt{3560089}^2 = g^2 [/spoiler]

-------------------------------------------------------------------------------------------------------------------------

<span class="math"> 1008^2 + 1100^2 = 1492^2 \rightarrow 1016064+1210000=2226064 [/spoiler]

<span class="math"> 1008^2 + 12075^2 = 12117^2 \rightarrow 1016064+145805625=146821689 [/spoiler]

<span class="math"> 1100^2 + 12075^2 = 12125^2 \rightarrow 1210000+145805625=147015625 [/spoiler]

<span class="math"> 1008^2 + 1100^2 + 12075^2 = \sqrt{148031689}^2 = g^2 [/spoiler]

-------------------------------------------------------------------------------------------------------------------------

Each brick has seven components (a,b,c), (d,e,f) and g(^2) , which are the three edge lengths from shortest to longest, the three face diagonals from shortest to longest, and the spatial diagonal. It may happen that c>d or c<d, as these two different bricks themselves demonstrate. In light of the curious irreducibility of each brick's spatial diagonal term, all the components' prime factorizations now follow:

<span class="math"> \displaystyle (1008,1100,1155) \rightarrow (2^4 3^2 7,2^2 5^2 11,3*5*7*11), (2^2 373,3*7*73,5*11*29), 13*17*89*181 [/spoiler]

<span class="math"> \displaystyle (1008,1100,12075) \rightarrow (2^4 3^2 7,2^2 5^2 11,3*5^2*7*23), (2^2 373,3*7*577,5^3*97), 13*29*41*61*157 [/spoiler]

A worthwhile exercise, though nothing too interesting pops out apart from the irreducibility of g^2, and that 12125 term. As a final note on this, note also that the smaller Euler brick (240,252,275) has a spatial diagonal g(^2) with an irreducible prime factorization 13*37*409, just like the other two - that is, there are no even, let alone repeated powers that could come out.

Wouldn't each number base have it's own set of bricks?

>>7500170

Calm down, your own 'tism is showing.

Anyway, this attitude is exactly why G.H. Hardy (rightfully) liked the conceit that his discoveries probably could not be used for killing. Quoth the excellent Hardy, from pic related:

"There is one comforting conclusions which is easy for a real mathematician. Real mathematics has no effects on war. No one has yet discovered any warlike purpose to be served by the theory of numbers or relativity, and it seems very unlikely that anyone will do so for many years. It is true that there are branches of applied mathematics, such as ballistics and aerodynamics, which have been developed deliberately for war and demand a quite elaborate technique... They are indeed repulsively ugly and intolerably dull... So a real mathematician has his conscience clear; there is nothing to be set against any value his work may have... The trivial mathematics, on the other hand, has many applications in war. The gunnery experts and aeroplane designers, for example, could not do their work without it... Mathematics facilitates (if not so obviously as physics or chemistry) modern, scientific, ‘total’ war."

Even if some detail of the above is now factually incorrect, the /conceit/ is still right, which is that it is generally regrettable that any science should be turned toward killing. Even coders don't like the idea of their stuff being used maliciously. Nor is it worth mentioning that you yourself find this subject "intolerably dull". Because, in a definite sense, your feelings don't matter. Because...

What really matters is, that every aspect of this thread falls within 4chan's global rules (legal, work-safe, etc), the board-specific rules, overall encouragements for quality content creation, and even the current mod sticky (no homework). In literally every sense, this thread has a perfect rule-following privilege to exist on this board, any anyone who doesn't like it can easily hide the thread, and check out other threads!

>>7500322

No, because numbers themselves (and the relationships between them) are not affected by the arbitrary number base that we use to write statements about them. That said, you can of course also make observations about numbers written out in some particular base. e.g. nice dubs. If you can't pace each length off with some unit (1), then it's not a brick. The base used to represent the brick is immaterial, apart from our comfort with working with such, or desire to condense information into a fairly compact symbol which is still tractable (which is why hexadecimal is used in computer science).

For example, suppose you have some candidate brick (1,2,7) written as (1,10,111) in binary. we have

1x1 + 10x10 = 101
1x1 + 111x111 = 110010
10x10 + 111x111 = 110101

What is still necessary, in binary, just as in decimal? That each right hand side should have some some smaller natural number (written in binary, of course) such that its square is equal to it. but 1x1=1 (a square in binary), 10x10 =100 (a square in binary, not to be confused with ten times ten equals one hundred in decimal), and 11x11=1001. But we've just skipped over our above 101 by taking consecutive squares, so 101 in binary (that is, 5) is (still) not the square of some other natural number in binary. Base don't matter...

...Except when we can make certain quick observations in such-and-such a base just by looking at a number, (owing to the number property of the base itself), or else glancing at an RRGGBB code which densely encodes 16^6 different colors in hex, just to get a feel for what the color must be like. "FF0000" is bright red, for example:

https://en.wikipedia.org/wiki/Web_colors#HTML_color_names

111,111,111^2 is still equal to 12345678987654321 no matter which base you use, it just has a cute representation in decimal, again, owing to the decimal base itself.

>>7500151
Nice. Glad you found interest. I think that a paper in arxiv might interest you. It's a table of bricks and factorizations

http://arxiv.org/abs/math/0111229

I've been dancing around a special case of the conjecture in >>7496684 , and it's time that I just got it done by logic. Typing the problem out and saying it back to myself really helps, which is why this thread exists.

What am I trying to do, RIGHT NOW? I am trying to show that that an Euler brick with consecutive edges (e.g. (2,3,4) ) does not exist.

How do I propose to do this? If an Euler brick has three consecutive edges (a,b,c), then EXACTLY two of its three equations have to be distinct AIRA Pythagorean triples. The converse is also obviously true, and so the two propositions rise and fall together. But since I know the explicit formula for the nth AIRA Pythagorean triple, and since they are spread far apart, this seems to contradict the above. I just have to do it formally.

The above scenario also described in >>7498498 is better characterized as a biconditional: (a,b,c) is an Euler brick with consecutive edges IFF exactly two of its equations are AIRAs. The thing goes both ways, there, which we can use. Recall per >>7497923 that if you have consecutive AIRA triangles as our assumption requires, then for some m, m+1 components of consecutive AIRA triangles n, n+1, we have

<span class="math"> \displaystyle (m+1)_{n+1} > m_{n+1} = (m+1)_{n} > m_{n} [/spoiler]

So, the long leg of the previous triangle is the short leg of the next-up triangle, by assumption. In terms of >>7497923 , (and after a bit of cancelling), this equation is rendered as

<span class="math"> \displaystyle 4 \sqrt{\diamond_{n+1}} + \sqrt{8 \diamond _{n+1}+1} = 4 \sqrt{\diamond_{n}} + \sqrt{8 \diamond _{n}+1} + 2 [/spoiler]

At which point we must investigate a comparison between the square triangular numbers themselves. We are once again in the VERY tall weeds, and I do not know for a fact, but I sense a happy outcome...

>>7496705
>>7496720

I'm looking into the derivation of Euler's parameterization of bricks. See relevant pic. You can see the starting point is Euclid's parameterization of Pythagorean triples. Where does (2.1) come from? Why should the product of the ratios be 1? Author doesn't say... is it obvious?

>>7501854
Sorry for posting, it's obvious.
For any non-zero x,y,z,
y/x* z/y * x/z = 1

>>7501854
>>7501887

Looks like my hunch was correct! I'll be needing a dollar now. :^)

Please continue your investigation of Euler's parametrization of bricks as far as you can take it. Also I didn't properly cite-thank this person for their script >>7497600 , also thanks to whoever posted >>7501317 for this interesting paper.

I'm getting an idea of the literature which is related to Euler bricks, although I don't know all the pertinent derivations: Euclid, Euler (some tiny subset of his writings, of course), also "Pell", and on my triangle thing, Sierpinski, Nyblom, the above paper, and some wiki links to "Durango Bill" and from there to one other very interesting link.

>>7502154
Here's the whole argument. I haven't been able to follow the whole thing.

>>7498498
>>7501496

The train of thought developed in these posts can now be concluded.

By use of Nyblom's explicit formula for square triangular numbers given toward the end of the link in >>7497830 , we can re-arrange a comparison (an assumed equality, which is shown to be absurd) of consecutive square triangular numbers, and be forced to conclude that for natural n (inclusive of zero in this case, but there are other situations coming back up the pipe where we have to be careful about degenerate zero-cases)

<span class="math"> \displaystyle \diamond_{n+1} > \diamond _{n} [/spoiler]

This /is/ admittedly "obvious" by inspection, but it needed review so I could internalize it. I can now show it to be true on demand, and therefore leave it as an exercise (hint: x^n(x-1) > y^n(1-y) ).

From this, we can finally, finally start going back up the pipe, in a manner not unlike Nyblom's proof of the theorem! With the above, we can re-visit the form at the end of >>7501496 , re-arrange so that a big mess on the LHS is equal to two, observe that like terms are necessarily un-equal in a certain way (recall the properties of a square triangular number as satisfying both radical forms, you can set up side inequalities as checks beginning with the result in this post), in particular, that differences of like terms are necessarily /strictly positive natural numbers/, such that the LHS is necessarily greater than two in any event (as a minimal case, the absurdity 6=2 is reached). THEN, in light of the necessary inequality, you can un-do the re-arrangement in >>7501496 (this time with an inequality), put Humpty-Dumpty together again, and observe that the assumption's required equality is false. From this, the comparison (inequality) of legs in successive AIRA triangles instead works out to

<span class="math"> \displaystyle (m+1)_{n+1} > m_{n+1} > (m+1)_{n} > m_{n} [/spoiler] .

(cont.)

Will the Euler Brick solve the Riemann hypothesis?

>>7502473

Although I sincerely doubt it, the mere idea that the two could somehow be related, is perhaps not quite so farcical as you may have thought. That is, another relatively simple object in number theory, Farey sequences, are employed in formulations which have been shown to be /equivalent/ to the Riemann Hypothesis! viz.

https://en.wikipedia.org/wiki/Farey_sequence#Riemann_hypothesis

So if you really want to settle RH one way or the other, then perhaps you should git gud at Farey sequences. And I wouldn't completely rule out some relationship between RH and Euler bricks (although I kindly doubt it).

(cont. from) >>7502465

NOW. Since all of the "m's" just listed are strictly ordered natural numbers, at this point, the closest conceivable arrangement (leaving data aside) of m_n+1 and m_n is precisely

<span class="math"> \displaystyle m_{n+1} - m_{n} \ge 2 [/spoiler]

But even the MINIMAL hypothetical situation of equality to two, let alone the rest, shows the absurdity of our other, larger assumption >>7501496 , which is that there exist consecutive AIRA triangles such that

<span class="math"> \displaystyle m_{n+1} - m_{n} =1 [/spoiler]

Which is impossible, due to the above. Therefore, we are forced to conclude 1) that it is impossible for an Euler brick to have exactly two distinct AIRA triangles among its three equations (for an Euler bricks' terms are coupled in such a way as to force these hypothetical, impossible triangles to share a fixed number which must pull double duty as short leg in one equation, long leg in another, impossible under the 2-AIRA constraint as shown) AND SO WE MAKE A VERY IMPORTANT

Lemma: It is impossible for an Euler brick to be composed of some consecutive edges of the form (p,p+1,p+2), where p is a natural number.

Corollary i: Due to the lemma >>7497051 , no natural multiples of a triple of consecutive natural numbers can possibly be an Euler brick. Therefore, a certain large class of triples (NOT ALL SUCH TRIPLES) satisfying c-b = b-a cannot possibly be Euler bricks, and consequently cannot possibly have their "opposite corner" representations in R^3, developed in >>7496684 , lying on the "minor diagonals" of the associated octahedral face.

Corollary ii: It is impossible for exactly two of an Euler bricks' characteristic equations >>7496579 to be AIRA triangles.

I'm not even done yet, but I'm officially pleased.

>>7502559

As can be thought through, the lemma just shown (and its corollaries) do not cover the situation where some evenly-spaced triple is NOT a multiple of some primitive consecutive triple. For example, (1,5,9) is not a multiple of such a triple, nor is (2,6,10). So in order to prove the remainder of the conjecture, we must account for evenly spaced natural triples in general, as was sketched out in >>7497736 . We'll have to build up some apparatus of a set of (a,b,c)s such that c-b = b-a, define a,b,c scalings which respectively leave each of a,b,c unchanged (a notion of a "b" scaling is at the bottom of the image) in addition to conventional scaling, and show that some subset of these operations cover the whole set. This is the next significant step to be taken.

And, as an aside, there remains the unaddressed idea >>7498498 that an Euler brick has exactly one AIRA among its characteristic equations. I think I'll explore this first, but I don't know/can't fill in the details as I write, so this will be a (possibly wrong) proof sketch:

-An Euler brick always has exactly one odd edge (check this premise per wiki, examples bear the notion out).

Given the premise, the AIRA necessarily has an even and odd edge. In the other two equations, the two split off to pair with the third edge. If the third edge is odd, then it constitutes a second odd edge and immediately contradicts the premise, so it's impossible for any one eqn to be AIRA.
Or, if the third edge is even, then the possibility of exactly one AIRA may still be alive...

(work to do around known conditions on Euler bricks, described by Sierpinski etc. Happily his little babby-book came today)

>>7502601

Of course, I misspoke in this post when I claimed that (2,6,10) is not a multiple of some other evenly-spaced triple, since it is obviously 2*(1,3,5).

The underlying reason for the hasty and wrong example, is that I have some vague notion of a large, generic, evenly spaced triple, say (3,5,7). Is THIS a multiple of something else? No. I notice that all of the components (particularly the middle component) are prime, and this informs my brainstorming.

(3,6,9) is generated by (1,2,3). But (3,11,19) is not (generated by anything), nor is (3,23,43). Very briefly, a recollection of "sexy primes" and "cousin primes" (get your heads out the gutter, look them up) occurs to me, but I do quickly find (3,29,55) such that the first two terms are prime, but the latter is not.

Here's a proof:

We want to show that if (a,b,c) is a brick, with a<b<c, then a+2=c is impossible. Suppose otherwise, that (a,b,c) is a brick and a+1=b and
a+2=c. Then a and c are either both even or both odd. Euclid's formula tells us that at least one leg of a Pythagorean triple must be even.
Since (a,c) are legs of a triple, they must both be even. Thus b is odd. By Euler's formula we have a=2*k*m*n, b=k*(m^2-n^2) where, since b
is odd, k must be odd and m and n must have opposite parity (i.e. one even and one odd). Similarly, c=2*j*p*q, b=j*(p^2-q^2) with j odd and p and q having opposite parity. Then a+2=c gives 2*k*m*n+2= 2*j*p*q, or k*m*n+1=j*p*q. Since m and n have opposite parity, m*n is even. Similarly p*q is even. Thus k*m*n+1 is odd while j*p*q is even, a contradiction.

I think the same argument shows that if (a,b,c) is a primitive triple, and a<b<c, then a and c can't both be even with (c-a)=2 mod 4.

>>7502959
Sorry, I meant Euclid's formula, not Euler's.

>>7502959

This is very good, and far more elegant than what I spent a long essay on. I started to have the idea that my proof was unnecessarily long once I caught wind of the idea that an Euler brick must have an odd edge length (which I don't know yet).

I'm with you up until the point where your a and c must both be even. What am I missing? I understand that your language is in terms of the exposition at

https://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple

I've already confused (Eu)ler and (Eu)clid myself at least once on this thought process. Apart from Eugene and Europe, "Eu" doesn't crop up in proper nouns too much in America. :^)

I have to internalize pic related, because along with babby-stuff (and "Ramiel" on the cover, referencing my octahedron thing), Sierpinski also leaves the work about showing the wiki Euler Brick properties to the reader (once context is built up). Having known this ahead of time might have rendered the above onerous process unnecessary, but I'm still pleased about the relationships in my result.

>>7503058
Well, a+2 =c, so either they are both odd or both even. Euclid's formula has legs of a Pyth. triple being 2kmn and k(m^2-n^2). Since 2kmn is even, at least one leg of a Pyth triple must be even. (a,c) are the legs of a triple.. a^2+c^2=f^2. Thus at least one of a or c must be even. But then they both must be even, since they are either both even or both odd.

>>7502959

Stronger: If (a,b,c) is a primitive brick, then one of a,b,c must be odd, and the other two must be divisible by 4.

Proof: Suppose (a,b,c) is a primitive brick. Then not all of a,b and c can be even, otherwise (a/2,b/2,c/2) would be a brick and (a,b,c) would not be primitive. Thus at least one of a,b, c must be odd. Now (a,b), (b,c) and (c,a) are legs of Pythagorean triples, and by Euclid's formula at least one leg of each of these must be even. So we can't have two or more of a,b,c odd. Thus exactly one of a,b,c is odd and two are even. By Euclid's formula, the odd side equals k*(m^2-n^2) and j*(p^2-q^2) where k is odd and m and n have opposite parity and j is odd and p and q have opposite parity, while the even sides are 2kmn and 2jpq. By opposite parity, mn and pq are even, thus 2kmn and 2jpq are divisible by 4.

There are many articles about Euler Bricks and Perfect Cuboids at the Unsolved problems web site at unsolved problems.org. The articles by Matson and Raines are particularly worthy of note.

a polite benediction of this thread. I got a major case done, on my own terms.

>>7505236
I enjoyed the thread too. Wrote some code, a proof... good stuff.

found a huge list of bricks...

http://www.christianboyer.com/eulerbricks/3D30414.txt

http://www.christianboyer.com/eulerbricks/