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/sci/ - Science & Math

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5673184 No.5673184 [Reply] [Original]

hi
having problems with some number theory:
find all positive integers x,y,z that <span class="math">x^2+1[/spoiler] and <span class="math">y^2+1[/spoiler] are prime and
<span class="math">(x^2+1)(y^2+1)=z^2+1[/spoiler]

pic not related

>> No.5673197

Bumping for interest. Not OP.

>> No.5673206

>>5673197
it has to deal with some modular arithmetics, but have no idea how

>> No.5674480

>>5673184
bumping for interest again

>> No.5674497

let x^2+1=p,x^2+1=q be primes then z^2+1=pq implies that -1 is a square mod p and q so p=4k+1, q=4h+1.

>> No.5674499

Damn you OP, now I'm wondering whether this graph is planar or not.

>> No.5674515

>>5674499
V+F=E+2 it

>> No.5674520

>>5674499
Yeah, it's planar. If you look at the vertices on the top, the last 3 form a "square" with the vertex under the third one. Pull the third one up to make some space and everything else fits neatly in that "square".

>> No.5674528

>>5674497
How did you get 4k and 4h?

>> No.5674532

>>5674528
Theorem -1 is a square mod p iff p = 1 (4).

>> No.5674533

>>5674528
http://en.wikipedia.org/wiki/Law_of_quadratic_reciprocity

>> No.5674534

>>5674533
irrelevant you ass

>> No.5674536

>>5674534
What the hell are you talking about?

> Let p > 2 be a prime number. Then
> (Supplement 1)
> x2 ≡ −1 (mod p) is solvable if and only if p ≡ 1 (mod 4).

This is exactly what>>5674497 says.

>> No.5674539

>>5674533
Thanks. This guy >>5674534 is just a retard.

>> No.5674541

>>5674536
see how it says Supplement, you retard? QR depends on this more fundamental result.

>> No.5674540

>tfw I did a math bachelor's without taking a single course in number theory

>> No.5674553

>>5674541
I gave a link. You didn't click it. Therefore you didn't know that it actually answered the question, and started flaming.

You were wrong. The link gave the answer to the question. The fact that the title of the link made you think it didn't is irrelevant.

You are an ass, you were wrong, and for some stupid e-pride reason you will refuse to admit you were wrong, which is fine because everybody knows you either know you're wrong, or are too stubbornly stupid to be worth talking to.

>> No.5674619

>>5674553
ok, so do you suggest there is an infinite number of solutions to this?
knowing that p= 4k+1 and q=4h+1 does not solve this problem.
I mean, we have to find all of the triplets (x,y,z) that meet the requirements listed.
and the number of that triplets is finite (that's what the problem suggests)

>> No.5674634

>>5674619
what does that tell you about k and h?

>> No.5674639

>>5674634
dunno, that 4k is a perfect square (and so is k) or k =(p-1)/4
does it matter?

>> No.5674649

>>5674619
Yeah, we only have a necessary condition. I gave it a thought earlier and I didn't manage to get to anything relevant.

>> No.5674666

>>5674639
of course it matters, you also did it wrong - it changes the problem into finding (4a^2+1)(4b^2+1)=4c^2+1

>> No.5675369

>>5673184
does anyone have a x,y and z that work? wondering if the answer is there aren't any
.

>> No.5675374

>>5675369
x=1,y=2,z=3

>> No.5676137

>>5675374
nice. do you think there are many (whatever that means) or just a few... or one curious.

>> No.5676965

>>5676137
i think that's the only one

>> No.5676971

>>5676965
Obviously modulo switching x and y.

>> No.5679339

bimp solve it

>> No.5679578

>>5679339
getting nowhere... getting tired of trying

>> No.5679635

Not sure if you asked this out of curiosity or as a subtle troll, but it's basically indeterminable with modern techniques.

Let f(x) = x^2 + 1. You're asking to find all integer solutions for which f(x)f(y) = f(z), with f(x) and f(y) prime. However, it is not even known how many primes are of the form f(x) for some integer x. This question is more difficult than that, unless you have a reason why only 1 or 2 trivial solutions exist.

>> No.5680406

>>5679635
It's true that it's unknown how many 1+n^2 primes there are, but it's not clear why the current problem is harder. Not OP, so unsure of motivations. I was curious about the number of small n solutions because if there is only one (or a few) it might indicate the problem is tractable.

Anybody have any insights, or progress?

>> No.5680413

OP is a troll. This is out of book which contains unsolved problems in number theory

>> No.5680584

Sketchy proof that probably has some obvious error if >>5680413 is right:

Rewrite the equation with Gaussian integers: (x + i)(x - i)(y + i)(y - i) = (z + i)(z - i), where i is the imaginary number sqrt(-1).

As z^2 + 1 is co-prime, this means that, WLOG, (z + i) must equal either (x + i)(x - i), (x + i)(y - i) or (x + i)(y + i). It is trivial to conclude that z + i =/= (x + i)(x - i) as x^2 + 1 has no imaginary part.

Casework:

Case #1: (z + i) = (x + i)(y + i)
It follows that (z - i) = (x - i)(y - i)
Expand both out, we get:

z + i = xy + ix + iy - 1
z - i = xy - ix - iy - 1

Add the two equations together:

2z = 2xy - 2
z = xy - 1
xy = z + 1

Test: (x^2 + 1)(y^2 + 1) = z^2 + 1
(xy)^2 + x^2 + y^2 + 1 = z^2 + 1
(z + 1)^2 + x^2 + y^2 + 1 = z^2 + 1
z^2 + 2z + 1 + x^2 + y^2 + 1 = z^2 + 1
2z + x^2 + y^2 + 1 = 0

But this is impossible, as for this equation to work z must be negative when the question states that x, y and z are all positive integers. Thus, we turn to:


Case #2: (z + i) = (x + i)(y - i)
It follows then that (z - i) = (x - i)(y + i)
Expand both out, we get:

z + i = xy - ix + iy + 1
z - i = xy + ix - iy + 1

Add the two equations together:

2z = 2xy + 2
z = xy + 1
xy = z - 1

Test: (xy)^2 + x^2 + y^2 + 1 = z^2 + 1
Sub: (z-1)^2 + x^2 + y^2 + 1 = z^2 + 1
z^2 - 2z + 1 + x^2 + y^2 + 1 = z^2 + 1

This simplifies out to: x^2 + y^2 + 1 = 2z. Now, if both x and y are even integers then there is a contradiction in the given formula. Thus, WLOG, we conclude that x must be equal to 1 and x^2 + 1 equal to 2, as there are no other primes of the form x^2 + 1 such that x is odd.

>> No.5680588

>>5680584
Continued:

Then: 1(y) = z - 1
y = z - 1

Next: (1^2 + 1)(y^2 + 1) = z^2 + 1
2((z-1)^2 + 1) = z^2 + 1
2(z^2 - 2z + 1 + 1) = z^2 + 1
2z^2 - 4z + 4 = z^2 + 1
z^2 - 4z + 3 = 0
(z - 3)(z - 1) = 0
z = 1, z = 3

Obviously, z cannot be equal to 1, as then y would have to be equal to 0, and unless you consider 1 as a prime number, this is not a valid solution. Therefore, the only possible solution set {x, y, z} is {1, 2, 3}.

>> No.5680855

>>5680584
Wow!
Let me see if I follow. Each of (x+i), (x-i), (y+i) and (y-i) is a Gaussian prime because their real and imaginary parts are all non-zero and their norms are all prime by assumption. Then we can partition this set of four factors into the set of factors of (z+i) and the set of factors of (z-i). Consider the factors of (z+i), and go through each case. Is that right?

>> No.5681008

>>5680584

This looks great to me. Statements like "z + i = x +/- i" and "z + i = y +/- i" are impossible since clearly x,y<z. And z+i = (x-i)(y-i) works out the same as your Case #1.

Something good in 4chan?!? I salute you! Kudos!

>> No.5681011

>>5680584
You do realize solving this problem would most likely make you hunted by the governments of the world?

>> No.5682237

>>5680413
Is this true or just trolling? If so, what book? In any case, discussing unsolved problems is a good thing.

>> No.5682941

>>5682237

It's not true. The proof given here

>>5680584

is correct.