$x^2+y^2+xy=(z+t)^2\Longrightarrow xy=0$

@Hopeoooo You might add the proposer's name M. Abduvaliev

ratatuy wrote: $x^2+y^2+xy=(z+t)^2\Longrightarrow xy=0$ this because?

+1 to the above: $x=3, y=5$ gives a square so this needs way more justification

http://matol.kz/comments/4559/show Here's a link to the official solution. Should still make sense even if you don't understand Russian or Kazakh

Here's the solution in English anyway (solution by A. Golovanov): 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 $v=\ell^2$; $k$ and $\ell$ 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 $3u+v=9k^2+\ell^2$, and the triple square is $u+3v=3(k^2+\ell^2)$. We have proved that the original equations imply the existence of coprime $k$ and $\ell$ such that $9k^2+\ell^2$ and $k^2+\ell^2$ are perfect squares. It remains to prove that there are no such $k$, $\ell$. If $k^2+\ell^2$ and $9k^2+\ell^2$ are perfect squares, the pairs $(k, \ell)$ and $(3k, \ell)$ are pairs of legs in primitive Pythagorean triples. Unfortunately, we do not know which of the numbers $k$ and $\ell$ 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 $GCD(b, c) = 1$, then there exists such positive integer $f$ that \[ a^2 + d^2 = b^2 f, \quad 9a^2 + d^2 = c^2 f. \]\[ f = GCD(a^2 + d^2, 9a^2 + d^2) \implies f \ | \ 8 a^2, \quad f \ | \ 8 d^2.\]\[ GCD(a, d) = 1 \implies f \ | \ 8. \]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 \[ a^2 + d^2 \equiv 2 \pmod{8} \implies f \ | \ 2. \]Let $f = 2$. \[ 9a^2 + d^2 = c^2 f = 2 c^2 \implies d^2 \equiv 2 c^2 \pmod{3} \implies 3 \ | \ c, \quad 3 \ | \ d \]--- a contradiction, since $GCD(c, d) = 1$. 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 $k=2u_1v_1$, $3k=2u_2v_2$, $\ell=u_1^2-v_1^2=u_2^2-v_2^2$, where $(u_1, v_1)$ and $(u_2, v_2)$ are two pairs of coprime numbers with different parity. Since $u_2v_2=3u_1v_1$, 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 $\ell$ we obtain $a^2c^2+c^2d^2=9a^2b^2+b^2d^2$, or $c^2(a^2+d^2)=b^2(9a^2+d^2)$. According to the lemma, $a^2+d^2$ and $9a^2+d^2$ are squares, and, obviously, $a<k$, $d<\ell$. Case 2. $\ell$ is even. Then $k=u_1^2-v_1^2$, $3k=u_2^2-v_2^2$, $\ell=2u_1v_1=2u_2v_2$. 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 $\ell$, 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 $a<\ell$, $d<k$. In both cases, for each pair $(k, \ell)$ such that $k^2+\ell^2$ and $9k^2+\ell^2$ 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.