/sci/ - Science & Math

[Anonymous No.11761675]

Here's a fun little exercise I thought of. The Pythagorean triples are those triples [math](a,b,c)[/math] of positive integers such that [math]a^2 + b^2 = c^2[/math]) Some examples are (3,4,5), (6,8,10), and (5,12,13). In fact there is an infinite amount of these. For convenience we shall always list them such that [math]a < b < c[/math].

If we specifically we look at the ones where the first number is a prime we find an interesting pattern. Let's take a look at the first few!
(3,4,5)
(5,12,13)
(7,24,25)
(11,60,61)
It seems as though if the first number is a prime, then the last number is just one more than the middle one!

Prove this.

[Anonymous No.11761681]

>>11761675
Does this mean prime numbers are the geometric means of b,c in Pythagorean triples?

[Anonymous No.11761686]

>>11761675
Proof: think.

[Anonymous No.11761691]

>>11761675
Every square number has an infinite array of 3-AP that fan out from it.

[Anonymous No.11761695]

>>11761681
Not all Pythagorean triples contain any primes, so no. Although you could probably quite easily show that if the difference between the squares of two sequential numbers is itself a square, then the square root of that is a prime. That would be the converse to the question in the OP.

[Anonymous No.11761732]

>>11761695
The converse to OPs question is false as the example (9, 40, 41) shows.

[Anonymous No.11761740]

>>11761675
Take a prime number [math] a \neq 2 [/math]
[math] a [/math] is odd
therefore [math] a^2 [/math] is odd
therefore we can write [math] a^2 = 2b + 1 [/math] where [math] b [/math] is an integer
rewriting gives:
[math] a^2 = 2b + 1 [/math]
[math] a^2 + b^2 = b^2 + 2b + 1 [/math]
[math] a^2 + b^2 = (b+1)^2 [/math]
which is the same as [math] a^2 + b^2 = c^2 [/math] where [math] c [/math] is one more than [math] b [/math]

>but what if a = 2?
there is no such pythagorean triple

[Anonymous No.11761743]

>>11761732
Ah, very good.

[Anonymous No.11761760]

>>11761740
Good job!

[Anonymous No.11761772]

>>11761740
>>11761760
This has very little to do with what was asked. All you've done is constructed a single Pythagorean triple (a,b,b+1) for any odd a.

[Anonymous No.11761787]

p^2 + a^2 = b^2
p^2 = b^2 - a^2 = (b-a)(b+a)
Since b,a assumed positive, we have b+a>1 so either b-a=1 or b-a=p. If b-a=p then b+a must also be p which is impossible if the triple is assumed nontrivial.
Thus b=a+1.
Ez

[Anonymous No.11761792]

>>11761772
Yes, but since it is possible for any odd a, it is also possible for all primes (except 2).

[Anonymous No.11761797]

>>11761792
Please go back to elementary school, learn to read simple sentences, and then come back

[Anonymous No.11761807]

>>11761797
Please think. We have shown that for _any_ odd a, we can construct a triple (a,b,b+1). The primes above 2 are a subset of all odd a, and thus we can construct such a triple for them as well. The original question doesn't mention it's ONLY true for the primes.

[Anonymous No.11761814]

>>11761807
>The original question doesn't mention it's ONLY true for the primes.
No, it says it's ALWAYS true for the primes. This is not the same thing as "true once", unless you want to go ahead and also prove that there's exactly one pythagorean triple containing any prime.

[Anonymous No.11761819]

>>11761772
Yeah I guess you're right, since I didn't show that there might exist a different pythagorean triple with the same [math] a [/math] where it's not the case that c = b +1.

>>11761807
nah dude I'm the guy who wrote the answer, he's right

[Anonymous No.11761821]

>>11761814
Oh shit yeah, you're right.

[Anonymous No.11761852]

>>11761740
OP here. You're close. You just need to show now that no triples other than the ones you can construct like that exist.

[Anonymous No.11761860]

>>11761787
This is probably the best one.

[Anonymous No.11761922]

>>11761852
that's just 2n-1 < n^2 for all n>2

[Anonymous No.11761978]

>>11761922
Well, considering e.g. (9,12,15) and (9,40,41) both exist, you would have to use the primality somehow. I would suggest as this:
>>11761787

[Anonymous No.11761987]

>>11761978
Actually you're wrong and >>11761922
is right. You don't need to use primality at all.

[Anonymous No.11762124]

>>11761978
No, the original guy's list already includes (a+1)^2 = a^+2a+1

If the gaps between (a+1)^2 and a^2 are the set of all odds and the primes over 2 are all odd, all he needs to add is that a^2 > 2a-1 for all a>2.

[Anonymous No.11762147]

>>11762124
That doesn't prove it. Read the OP again. You're dead wrong.

[Anonymous No.11762171]

>>11761675
Not with that gay anime picture.

[Anonymous No.11762189]

>>11762147
Sure it does. The gaps between n^2 and (n+1)^2 are the set of all odd numbers and the squares of all p>2 are odd. Therefore p^2 corresponds to a gap for every p>2.

[Anonymous No.11762387]

Another pattern that follows is that [math]c=\textstyle\frac{p^2+1}{2}[/math] for all p>2.

[Anonymous No.11762393]

Another pattern that follows is that [math]c=\frac{p^2+1}{2}[/math] for all p>2.

[Anonymous No.11762401]

suppose x is an element in Z_p^2, then for all y in (a,b,c) there exists a p such that p^2 => a^2+b^2 = c^2, then by definition x > y. but this contradicts our hypothesis statement, namely that (a,b,c) is a subset of the pythagorean triples.

Thus, x is prime by the Cantor-LeGrange Theorem

[Anonymous No.11762413]

>>11762189
>The gaps between n^2 and (n+1)^2 are the set of all odd numbers and the squares of all p>2 are odd.
Are you actually retarded? That doesn't matter. You don't care what the gaps between n^2 and (n+1)^2 are.
I'll state the problem to you again:
You are asked to prove that if p,b,c are positive integers, p is a prime and
p^2 + b^2 = c^2, then c=b+1.
How do you prove that?
I'm sorry for being rude, I'm just annoyed by people like you who pretend to be smarter than they are and spew bullshit.

[Anonymous No.11762420]

i wrote and proof and then gave up and deleted it and posted this instead

[Anonymous No.11762422]

>>11762420
How is this relevant to the thread?

[Anonymous No.11762424]

>>11762422
i'd say its more relevant than the no less than five anime girls that have been posted itt

[Anonymous No.11762425]

>>11762424
>i'd say its more relevant than the no less than five anime girls that have been posted itt
How?

[Anonymous No.11762435]

>>11761675
a^2 = (c+b)(c-b)
a is a prime number, so either
c+b = c-b = a
or
c+b = a^2,
c-b = 1.
Since b would be zero in the first case, only the second case is relevant, therefore c = b+1.

[Anonymous No.11762441]

>>11762435
>>11761787
The only correct answers ITT. The other attempts are by low IQ kids.

[Anonymous No.11762442]

>>11762413
I think you're the one who needs to read OP again. a<b<c and p=a are given.

[Anonymous No.11762446]

If you expand [math](n+p)^2=n^2+2np+p^2[/math] for some prime [math]p[/math], the only value of [math]n[/math] such that [math]2np+n^2[/math] is a square is 1. Q.E.D.

[Anonymous No.11762479]

>>11762441
>low IQ kids
Projection much?

There's only one possible right triangle for any shortest leg p, and we're GIVEN IN THE OP that a = p = shortest leg.

All you need to prove from there is that p maps to a right triangle for all p.

[Anonymous No.11762488]

counterexample: 9, 12, 13
you can send my fields medal to the office

[Anonymous No.11762489]

>>11762441
>low IQ kids
Projection much?

There's only one possible right triangle for any shortest leg p, and we're GIVEN IN THE OP that a = p = shortest leg.

All you need to prove from there is that p maps to a right triangle for all p>2.

[Anonymous No.11762498]

>>11762488
damn how did mathematics survive this long without anyone realizing 9 was a prime number?

[Anonymous No.11762499]

>>11762442
Ok and how is that relevant? Provide a full proof or shut up (because you can't).
>>11762488
9 is not a prime.

[Anonymous No.11762502]

>>11762489
>There's only one possible right triangle for any shortest leg p,
Prove it.

[Anonymous No.11762503]

>>11762499
see>>11762489

[Anonymous No.11762505]

>>11762499
Already done

[Anonymous No.11762511]

>>11762503
>>11762505
Holy shit you're low IQ. Read
>>11762502

[RllySmart person No.11762517]

science is 4 incells bro.. imagine science, mc2 my ass

[Anonymous No.11762519]

>>11762511
>suppose there were a right triangle with a shortest leg that wasn't p
>then, that would be impossible
>therefore shut the fuck up

QED

[Anonymous No.11762543]

>>11762519
It's OK, not everyone is smart.

[Anonymous No.11762585]

>>11762489
You need to prove this though. It's not true for non-primes, e.g.
>>11761978

[Anonymous No.11762657]

>>11762511
I'm sorry your lack of spatial ability annoys you so much. Why would you imagine that's even possible? The only interesting part for me was the injection.

If you have p,n,n+1 where p is the shortest leg, no other a>n fits. That's the whole point of the trivial statement I gave you in the first place: 2n-1 < n^2 for all n>2.

[Anonymous No.11762674]

>>11762657
>If you have p,n,n+1 where p is the shortest leg, no other a>n fits
What's a?

[Anonymous No.11762685]

>>11762657
Dude just accept it, you're wrong. Your argument doesn't work. If it did, you would be able to write it out in full but you haven't because you can't, because there's no actual argument. Just stop.

[Anonymous No.11762706]

anime niggers ruin every thread they touch

[Anonymous No.11762708]

>>11762685
I'm curious if you'll ever see it or not. Here's a "write it out" hint.
[math]\sqrt{p^2+(n+k)^2}<\sqrt{(n+1+k)^2}[/math]

[Anonymous No.11762710]

>>11762674
The next integer > n such that p,a,a+1 holds.

[Anonymous No.11762717]

I still can't decide if this is a pretentious 13 year old who doesn't know what a proof is or if he's just started damage-control trolling

[Anonymous No.11762724]

>>11761675
p^2 = c^2-b^2 = (c+b)(c-b)

But if p is prime then c-b = 1 and c+b = p

Trivial.

[Anonymous No.11762727]

>>11762717
Not your proof army.

[Anonymous No.11762733]

>>11762724
*p^2

[Anonymous No.11762771]

>>11762708
Idiot.
>>11762710
Showing that p, a, a+1 holds for no other a DOES NOT PROVE what OP is asking for, retard. Holy shit you're a low IQ moron.
>>11762724
>>11762733
Correct.
>>11762717
I know right? It's like we're being invaded by retards. Very similar people were posting false "proofs" and general nonsense in /mg/ today.

[Anonymous No.11762781]

>The retards actually think that proving p^2 + a^2 = (a+1)^2 has a unique solution for prime p and integer a is the same as proving that if p^2 + a^2 = b^2 then b=a+1.

[Anonymous No.11762806]

>>11762781
which one is what OP asked?

[Anonymous No.11762811]

>>11762806
>It seems as though if the first number is a prime, then the last number is just one more than the middle one!
>Prove this.
Can you even read?

[Anonymous No.11762815]

>>11762811
dont be rude, nigger, it was a simple question

[Anonymous No.11762826]

>>11762815
A simple question by a simple mind.

[Anonymous No.11762829]

>>11762771
Sure it does. You just accepted that there's only one possible right triangle for any shortest leg p. I don't think you've made the connection to
>2n-1 < n^2 for all n>2
yet, but if you accept that both of these are true:
1. p, n, n+1 hols for all p
2. p, a, a+1 holds for no other a
then OP is proven.

[Anonymous No.11762834]

>>11762826
i never claimed otherwise

[Anonymous No.11762839]

>>11762829
>It seems as though if the first number is a prime, then the last number is just one more than the middle one!
>Prove this.

[Anonymous No.11762845]

>>11762829
That's not what OP is asking for, moron.

[Anonymous No.11762863]

>>11762839
>>11762845
Lmao. Okay last hint, although you basically have the full solution at this point.

[math]0<\frac{(n+k)^{2}}{(n+1+k)^2}<1[/math]

[Anonymous No.11762899]

>>11762863
You actually, genuinely don't get it, do you?

[Anonymous No.11762911]

>>11762899
I understand that
>p, n, n+1 holds for all p
>p, a, a+1 holds for no other a
proves OP
You actually, genuinely don't?

[Anonymous No.11762916]

>>11762899
I understand that
>p, n, n+1 holds for all p
>p, a, a+1 holds for no other a>n
proves OP
You actually, genuinely don't?

[Anonymous No.11762923]

>>11762911
Holy fucking shit anon, have you ever taken a course in basic logic? Here, let me enlighten your retarded monkey brain.
OP is asking:
>It seems as though if the first number is a prime, then the last number is just one more than the middle one!
>Prove this.
That means we have to prove that if p is prime, a,b positive integers, then p^2 + a^2 = b^2 => b=a+1.
What you've proved: if b=a+1, then p^2 + a^2 = b^2 has a unique solution for a.
Now the issue is that b could be another number, not just a+1. For example, you could potentially have |b-a|=3, or 5, or 7.
You've only proven that p^2 + a^2 = (a+1)^2 has only one solution. But that doesn't prove that p^2 + a^2 = b^2 has only one solution because the latter is way more general.

[Anonymous No.11762925]

>>11762923
you're being trolled, retarded animefaggot

[Anonymous No.11762936]

>>11762923
Holy fucking shit anon, do you not understand that 2n-1 < n^2 for all n>2 means that there WILL NEVER BE A SOLUTION TO a>n because
[math]0<\frac{(n+k)^{2}}{(n+1+k)^2}<1[/math]

[Anonymous No.11762997]

>>11762936
Dude, you have to use the fact that p is prime. For non-primes you have many triples with the same shortest leg.

[Anonymous No.11763025]

>>11762997
a=p and a<b<c are given in the OP.
But I'd love to hear why you think non-prime triples should apply to a question that rules them out per se.

[Anonymous No.11763097]

>>11761675
We must have [math]p^2 = (c-b)(c+b)[/math].
Hence, either [math]c-b = c+b = p[/math] or [math]c-b = 1, c+b = p^2[/math].
Because [math]0 < b < c[/math], we have [math]c - b < c + b[/math], hence [math]c-b = 1[/math].
The triplets containing a prime must therefore be the [math]\left(p, \frac{p^2-1}{2}, \frac{p^2+1}{2}\right)[/math].

[Anonymous No.11763223]

Imagine the nonsense you make other people imagine every day.

[Anonymous No.11763450]

>>11761852
Because the difference between two integer square a, b | a =/= b and a, b > 2 are always over 4

[Anonymous No.11763487]

>>11761675
Literally, all you have to do is prove that all Pythagorean triples can be expressed with a simple, constructive formula, and then you just have to use the difference of two squares formula, and apply the definition of a prime number.

[Anonymous No.11763553]

a^2 = c^2-b^2=(c+b)(c-b)

if c-b>1 a^2 has two different factors and a cannot be prime. Therefore if a is prime c-b has to be equal to 1.
t. engineer

[Anonymous No.11763633]

>>11762899
>>11762923
>>11762997
Bueller? Bueller?

[Anonymous No.11764163]

>>11762925
Fuck off, retard.

[Anonymous No.11764171]

>>11762420
>non-moving .gif
into the trash it goes anon