Corrected copy of my solution on MathLinks, using Gaussian integers: Write $xy=z^2+1=(z+i)(z-i)$. Now factor $z+i = \pi_1\cdot \pi_2\cdot ...\cdot \pi_n$ into Gaussian primes and get ${z-i = \overline{z+i} = \overline{\pi_1\cdot \pi_2\cdot ...\cdot \pi_n} = \overline{\pi_1} \cdot \overline{ \pi_2}\cdot ...\cdot \overline{\pi_n}}$ (the primes $\pi_i$ are not necessary distinct). Now by uniqueness of factorisation (and $xy=z$), we w.l.o.g. get that $x=(\pi_1\cdot \pi_2\cdot ...\cdot \pi_k)(\overline{\pi_1}\cdot \overline{\pi_2}\cdot ...\cdot \overline{\pi_k})=(a+bi)(a-bi)=a^2+b^2$ and $y=(\pi_{k+1}\cdot \pi_{k+2}\cdot ...\cdot \pi_n)(\overline{\pi_{k+1}}\cdot \overline{\pi_{k+2}}\cdot ...\cdot \overline{\pi_n})=(c-di)(c+di)=c^2+d^2$. Now $(a+bi)(c-di)=(z+i)$ by definition, which gives $z=ac+bd$.

Won't there be a nice elementary solution?

Peter wrote: Won't there be a nice elementary solution? Actually, there is one elementary solution that uses Fermat's method of finite descent.

Can you post it for us? Additional solutions are always nice.

Quote: Suppose that $ x,y$ and $ z$ are positive integers with $ x^2 + 1 = yz$.Prove that there exist integers $ a,b,c,d$,such that $ x = ac + bd,z = a^2 + b^2,y = c^2 + d^2$. -Suppose to the contrary that there exist some triplets $ (x,y,z)$ that satisfies to equation $ xy = z^2 + 1$,but they can not be represented in a needed form. -Let's take among of them a triplet with minimum sum,i.e $ x + y + z$ reaches minimum value. WLOG assume $ y > z$,obviously $ x < y$ and $ x\geq z$.If $ x = z$ then $ y = 2,z = 1,x = 1$,for this triplet we can take $ a = 1,b = 0,c = 1,d = 1$. -Now we can assume that $ y > x > z$: Then consider triplet $ (x - z,y + z - 2x,z)$,these numbers are obviously positive and satisfies to equation $ xy = z^2 + 1$,but their sum is equal to $ y + z - x < y + z + x$,hence there exist positive integers $ a,b,c,d\in\mathbb{N}$,such that $ x - z = ac + bd$,$ z = a^2 + b^2$,$ y - 2x + y = c^2 + d^2$,but this system implies that $ x = a(a + c) + b(b + d)$,$ y = (a + c)^2 + (b + d)^2$,$ z = a^2 + b^2$.Now it is enough to take $ a_0 = a,b_0 = b,c_0 = a + c,d_0 = d + b$.Thus contradiction obtained.

An alternative solution using Minkowski's Theorem

Let's take $a,b,c,d$ pairwisely co-prime such that $ac-bd=1$.Bezout's identity confirms that such $a,b,c,d$ exists.Now using $\text {Brahmagupta-Fibbonacci Identity}$ we get that $x,y$ both are bi-squares i.e. sum of two squares.So let $x=a^2+b^2,y=c^2+d^2$.Then $xy=(ad+bc)^2+(ac-bd)^2$ by the identity.So now let's take $z=ad+bc$.

I don't see where you have proven anything related to the problem: You start with special $a,b,c,d$ and define $x,y,z$ from them. That's completely the wrong order.

ZetaX wrote: I don't see where you have proven anything related to the problem: You start with special $a,b,c,d$ and define $x,y,z$ from them. That's completely the wrong order. I wrote about $a,b,c,d$ first but it will come after we deduce $x,y$ are sum of two-squares.We need to show the existence of such $a,b,c,d$,and I have found out a triple of $(x,y,z)$ to satisfy this.Hope it's enough.

You are given that triple!

Erken wrote: Quote: ...Then consider triplet $ (x - z,y + z - 2x,z)$,these numbers are obviously positive and satisfies to equation $ xy = z^2 + 1$,... The triplet should be $(x, x+y-2z, y-z)$.

@vexation : can you explain your way please ?

About Fermat's method, one could make some progress by noticing the identity: ((x+y+2z)/2).((x+y-2z)/2)-((x-y)/2)^2=xy-z^2. With this one, we may prove that all possible counterexamples with z as small as possible have this number being odd. Could one prove this last case?

Hey Erken! Could you post the reference from which you`ve found this elementary proof?

So I have the proof of the case where "z"is even. With this, we easily note that every prime p such that (p-1)/4 is integer, is a sum of two integer Square numbers

@escadainfinita If I am not mistaken I came up with this solution based on the Fermat's Descent Procedure which served as a key idea to solving one of the famous IMO 30 years back .There is a huge variety of similar problems out there utilising the very same idea. P.S: It has been almost 8 years since I posted that solution

I am posting the statement I have proved. Suppose x, y, z are positive integers such that: xy=z^2+1, z being even with z the smallest value which is able to lead to a factorization, which shows a counterexample to the assertion of the problem. One can find positive integers: a, b, c such that: (1) x= (a+b+2c)/2; (2) y= (a+b-2c)/2; (3) z= (a-b)/2, since: x+y>2z, because this turns into: (x-z)^2+1>0. Plugging these relations into the first one we get: ab= c^2+1. Now: c< (a-b)/2, because this yields to: a^2-6ba+b^2+4>0, which is true, since the polynomial: f(x) = x^2-6bx+b^2+4 has the discriminant being: 32b^2-16, which is always positive for positive integer b. Now, c is even, since: x-y is divisible by 4 (because both x, y are congruent to 1 (mod 4)). Since z is the smallest value which leads to some factorization which would shows a counterexample, we have: a= s^2+t^2; b= v^2+u^2; c= sv+tu, some integers: s, t, u, v. Plugging these into the formulas (1); (2); (3), we get: a= ((s+v+t+u)/2)^2+((s+v-t-u)/2)^2; b= ((s-v+t-u)/2)^2+((s-v-t+u)/2)^2; c= (s^2+t^2-u^2-v^2)/2= MP+NQ, where: M= (s+v+t+u)/2; N= (s+v-t-u)/2; P= (s-v+t-u)/2; Q= (s-v-t+u)/2. Since a, b have the same parity, we easily get that {s,t,u,v} can be particioned in two subsets, forming two pairs, each of them having elements of same parity. This imples: M, N, P, Q, are integers. We have reached a contradiction, which shows there is no counterexample. This solves a particular case of the problem, namely: if x, y, z, are positive integers e xy=z^2+1, z being even, there exists integers a,b,c,d with: x= a^2+b^2; y= c^2+d^2; z=ac+bd Note: notice this yields a very simple proof of the fact that every prime p such that (p-1)/4 is an integer, is the sum of the squares of two integer numbers. Just observe that for such primes p, p is a divisor of: [((p-1)/2)!]^2+1 (in view of the fact that: [(p-1)!+1]/p is an integer, for every prime p) and (p-1)/2 is even, since p>4. Apllying the statement I proved, we have: p=a^2+b^2, for some integers: a,b. Does anyone has the proof for the case of z being odd?

Now, the uniqueness of: a,b, can be explained by other problem from the divisibility section

Observe that if we can show that $x,y$ are sum of two squares then $z$ takes the required form.Choose any prime divisor $p$ of $z^2+1$ then $p \equiv 1 \mod 4$. By Fermat's Christmas theorem, $\exists m,n \in \mathbb{N}$ such that $p=m^2+n^2$. Repeat this for all primes dividing $z^2+1$. Use the identity $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$ repeatedly to express $x,y$ as sum of two squares. Some manipulation therefore leads to the conclusion.