i think i have a proof of this one using conditions gcd(x,y,z,w)=1 it's pretty long and i have to check it.it uses the fact that every prime of the form 4k+1 can be expressed as the sum of 2 squares . and then also uses factorization in the gauss number ring z(i) i'll post it later.(in case it turns out to be correct)

This one? Hard!!!!! I obtained the formula for w and z, but I have no idea how to obtain those for x and y. Iranian olympiad represents for me together with the russian one the ultimate challenge. These problems are simply impossible.

so we have x^2+y^2+z^2=w^2, and z,w have the same parity. firstly, if w and z are both even, we know that x^2+y^2 = 0 mod 4, or x and y are both even, in which case we can divide through by 4. So WLOG, w and z are odd. lemma 1: there exists integers u,v s.t. u+v = w and u-v = z with (u,v) = (w,z) proof: since w,z have the same parity we know that u=(w+z)/2 and v = (w-z)/2 are both integers. Now, suppose (u,v) = e and (w,z) = d first, then since u+v = w and u-v = z, e|w and e|z, so e \leq d. But also, 2u = w+z, so d|2u, but since w and z are odd, d is odd so d|u, and by similar means d|v then d \leq e => d = e. Now, using the result of lemma 1, let w = u+v, z = u-v, and subbing in we see that x^2+y^2 = 4uv. we know from lemma 1 that (u,v) = d, so let u=d*u', v=d*v' where (u',v')=1, so then x^2+y^2 = (2d)^2*u'*v'. Now, the LHS is the sum of two squares, so the RHS must have an even number of prime factors = 3 mod 4. Now, (2d)^2 is a square, so it already has an even number of prime factors = 3 mod 4, and u' is relatively prime to v' so they must each have an even number of prime factors = 3 mod 4 (otherwise the product couldn't be written as the sum of two squares). Therefore, u' and v' can be written as the sum of two squares. Then, let u' = X^2+Y^2, v' = Z^2 + W^2, subbing in and using the identity (a^2+b^2)*(c^2+d^2) = (ac - bd)^2 + (ad + bc)^2 we get that: x^2 = (2d)^2(XZ-YW)^2, y^2 = (2d)^2(XW+YZ)^2 => x = 2d(XZ-YW) and y = 2d(XW+YZ) and we know that z = d*(u'-v') = d*(X^2+Y^2-W^2-Z^2) w = d*(u'+v') = d*(X^2+Y^2+W^2+Z^2).

hey zscool, I was just wondering... is your solution similar to solving a problem from a Waring's variant? Just a thought though

what is Waring's variant?

Dear zscool I think that in your solution exist very little mistake because I solved this as well in examination three month ago but Igot 0 inthis question

I think the mistake is due to the fact that a^2+b^2=c^2+d^2 does not neccesarily imply a= \pm c and b= \pm d.

my bad, maybe somebody can finish it up along those lines?

I also tried with that idea and led nowhere.

Harazi, how did you obtain the formulae for w and z? Here's what I did: Let's assume (x,y,z,w)=1. Then we have [(w-z)/2]*[(w+z)/2]=x'²+y'</sup>2</sup> (x=2x' and y=2y', since it's easy to show that both x and y must be even). [(w-z)/2]*[(w+z)/2] can be written as sum of 2 squares, so every prime of the form 4k+3 has an even exponent in [(w-z)/2]*[(w+z)/2]. Let's assume that (w-z)/2 can't be written as sum of 2 squares. Then there is a prime p=4k+3 which has an odd exponent in (w-z)/2, so it has an odd exponent in (w+z)/2, so p|w and p|z. At the same time, p|x²+y², and from the fact that p=4k+3 we get p|x and p|y (this is well-known), so p|(x,y,z,w)=1, which is false, so the assumption that (w-z)/2 can't be written as sum of 2 squares is wrong. We also get that (w+z)/2 can be written as sum of 2 squares, so (w-z)/2=X²+Y² and (w+z)/2=W²+Z². I'm sure we can continue with some arithmetics in Z, but I'm not used to that and I think I'm missing something.

I'm not used to apply Z too ,so who can finish this problem.SOS

My friend tell me he doesn't remember very well that there exist the following lemma or not : Lemma : If $ m, n,a, b, c, d$ are positive integers such that : $ m^2+n^2=(a^2+b^2)(c^2+d^2)$ then there exists positive integers $ a_1,b_1,c_1,d_1$ such that : $ a^2+b^2=a_1^2+b_1^2, c^2+d^2=c_1^2+d_1^2, m=a_1b_1+c_1d_1, n=a_1d_1-b_1c_1$ I have checked that if that lemma is right , I will have a complete solution for that nice one But I failed to prove the lemma Can anybody help me to prove or disapprove it ?

Waclaw Sierpinski wrote: All the solutions of the equation x²+y²+z²=t² (37) in natural numbers x, y, z, t, with even y, z, are obtained from the formulae $ x=\frac{l^2+m^2-n^2}{n},y=2l,z=2m,t=\frac{l^2+m^2+n^2}{n}$ l, m being arbitrary natural numbers, and n being the divisors of l²+m² less than $ \sqrt{l^2+m^2}$.Every solution is obtained exactly once in this way. ... It is worth-while to notice that, as has been proved by R. D. Carmichael [4], pp. 39-43, all the solutions of equation (37) in natural numbers can be obtained from the identity (2d(XZ-YW))²+(2d(XW+YZ))²+(d(X² +Y² -Z² -W² ))²=(d(X² +Y² +Z² +W² ))² CARMICHAEL, R.D., [4] Diophantine Analysis (New York 1915, reprint New York 1959).

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