SILK ROAD MATHEMATICS COMPETITION XX

Assume that $xyzt\ne 0$. If three of the numbers $x$, $y$, $z$, $t$ have a common divisor $d$, the remaining number is also a multiple of $d$. The division of all the four numbers by $d$ preserves the validity of our equations. Since $x$, $y$, $z$, $t$ are not zero, we can divide them by their greatest common divisor and henceforth assume that they are coprime.

The numbers $x$ and $y$ can not be both odd (since $xy$ is even) or both even (since in this case $zt$ is also even and at least three of the numbers $x$, $y$, $z$, $t$ are even). Thus $x$ and $y$ and therefore $z$ and $t$ have different parity.

Denote $s=x^2+y^2$, $n=zt$. Without loss of generality, $n>0$. The condition on $x$, $y$, $z$, $t$ means that $s-4n$, $s-2n$, $s+2n$, $s+4n$ are perfect squares. We will prove that this is not possible for any odd $s$ and positive $n$. First, we find a statement equivalent to these four numbers being squares. Their product $(s^2-4n^2)(s^2-16n^2)=(s^2-10n^2{)}^2-36n^4$ is a square of some odd $T$. The numbers $T$, $6n^2$, $s^2-10n^2$ form a primitive Pythagorean triple, hence there exist coprime $u$, $v$ of different parity such that $2uv=6n^2$ and $u^2+v^2=s^2-10n^2$. Since $u$ and $v$ are coprime, it follows from $uv=3n^2$ that one of these numbers is a square and another is a triple square. Without loss of generality, assume that $u=3k^2$ and $v=l^2$; $k$ and $l$ are coprime. Multiplying $u^2+v^2=s^2-10n^2$ by $3$ we get $3s^2-30n^2=3s^2-10uv=3u^2+3v^2$, that is, $3s^2=3u^2+10uv+3v^2=(3u+v)(3v+u)$, where $3u+v$ and $3v+u$ are coprime (they have different parity, and their common divisor must also divide $8u=3(3u+v)-(3v+u)$ and, similarly, $8v$). It follows that one of the numbers $3u+v$ and $3v+u$ is a square and another is a triple square. Since $3$ does not divide $3u+v$, the square is $3u+v=9k^2+l^2$, and the triple square is $u+3v=3(k^2+l^2)$. We have proved that the original equations imply the existence of coprime $k$ and $l$ such that $9k^2+l^2$ and $k^2+l^2$ are perfect squares.

It remains to prove that there are no such $k$, $l$. If $k^2+l^2$ and $9k^2+l^2$ are perfect squares, the pairs $(k,l)$ and $(3k,l)$ are pairs of legs in primitive Pythagorean triples. Unfortunately, we do not know which of the numbers $k$ and $l$ is even, so two cases are possible. But let's prove the following lemma first.

Lemma. Let $a$, $b$, $c$, $d$ be pairwise coprime positive integers such that $c^2(a^2+d^2)=b^2(9a^2+d^2)$. Then, $a^2+d^2$ and $9a^2+d^2$ are perfect squares.

Proof. Since $\text{GCD}(b,c)=1$, then there exists such positive integer $f$ that $$a^2+d^2=b^{2}f,9a^2+d^2=c^{2}f.$$ $$f=\text{GCD}(a^2+d^2,9a^2+d^2)⟹f\mid 8a^2,f\mid 8d^2.$$ $$\text{GCD}(a,d)=1⟹f\mid 8.$$ If $a$, $d$ are of different parity, then $a^2+d^2$ is odd, and therefore $f=1$. Let both $a$ and $d$ be odd (they can't be simultaneously even). Then $$a^2+d^2\equiv 2(\mathrm{mod}8)⟹f\mid 2.$$ Let $f=2$. $$9a^2+d^2=c^{2}f=2c^2⟹d^2\equiv 2c^2(\mathrm{mod}3)⟹3\mid c,3\mid d$$ — a contradiction, since $\text{GCD}(c,d)=1$. Thus, $f=1$, $a^2+d^2=b^2$, $9a^2+d^2=c^2$. The lemma is proven.

Case 1. $k$ is even. Then $k=2u_1{v}_1$, $3k=2u_2{v}_2$, $l=u_1^2-v_1^2=u_2^2-v_2^2$, where $(u_1,v_1)$ and $(u_2,v_2)$ are two pairs of coprime numbers with different parity. Since $u_2{v}_2=3u_1{v}_1$, then there exist pairwise coprime positive integers $a$, $b$, $c$, $d$ such that $u_2=3ab$, $v_2=cd$, $u_1=ac$, $v_1=bd$ (the case when $u_2$ is not divisible by $3$ can be considered similarly). Substituting this in the expression for $l$ we obtain $a^2{c}^2+c^2{d}^2=9a^2{d}^2+d^2{c}^2$, or $c^2(a^2+d^2)=d^2(9a^2+c^2)$. According to the lemma, $a^2+d^2$ and $9a^2+c^2$ are squares, and, obviously, $a<k$, $d<l$.

Case 2. $l$ is even. Then $k=u_1^2-v_1^2$, $3k=u_2^2-v_2^2$, $l=2u_1{v}_1=2u_2{v}_2$. Since $(u_2-v_2)(u_2+v_2)=3(u_1-v_1)(u_1+v_1)$, then we have $u_1-v_1=ab$, $u_1+v_1=cd$, $u_2-v_2=ac$, $u_2+v_2=3bd$ for some pairwise coprime positive integers $a$, $b$, $c$, $d$ (the case when $u_2+v_2$ is not divisible by $3$ can be considered similarly). Expressing $u_1$, $v_1$, $u_2$, $v_2$ in terms of $a$, $b$, $c$, $d$ and putting the result in the formula for $l$, we get $(ab+cd)(cd-ab)=(3bd+ac)(3bd-ac)$, that is, $c^2(a^2+d^2)=b^2(a^2+9d^2)$. Hence, by the lemma, we obtain that $a^2+d^2$ and $a^2+9d^2$ are perfect squares, and $a<l$, $d<k$.

In both cases, for each pair $(k,l)$ such that $k^2+l^2$ and $9k^2+l^2$ are perfect squares we constructed a pair of smaller numbers with the same property. It follows that such numbers do not exist. Thus, the original system does not admit a solution in non-zero integers.