Another nice problem by Mr.Safaei! Note that $x^2+y^2=z^2+xyz-2$, or $(x+y)^2-2=xy(z+2)+z^2-4$ and $(x-y)^2-2=xy(z-2)+z^2-4$, so there exists integers $a,b$ such that $z+2\mid a^2-2$ and $z-2\mid b^2-2$ ($a=x+y$ and $b=x-y$). Thus for every odd prime divisor of $z-2$ and $z+2$ like $p$ we have that $(\frac{2}{p})=1$, by the following lemma we have that $p\equiv \pm 1$ (mod $8$). If $2\mid z$, since $4\nmid a^2-2$ we have that $v_2(z+2)=1$, so $z=4k$. By the following lemma we have $8\mid 2k$ so $k=4t$. Now $z=16t$ so $x^2+y^2\equiv 14$ (mod $16$) which is impossible since $x^2\equiv 1,9$ (mod $16$) because $x$ is odd. So $z$ is odd and by the lemma we have $(z+2,z-2)=(1,7)$ (mod $8$) and its impossible, thus we are done! Lemma: For an odd prime $p$, $(\frac{2}{p})=1\iff p\equiv \pm 1 (mod 8)$ Please correct me if I'm wrong

\[z^2+xyz-x^2-y^2-2=0\]$\Delta=(xy)^2+4x^2+4y^2+8$ \[t^2=x^2y^2+4x^2+4y^2+8\iff t^2+8=(x^2+4)(y^2+4)\]If both $x,y$ are even, then $16|t^2+8$ which is impossible. WLOG let $x$ be odd. Take a prime $p|x^2+4$. We have $(\frac{-1}{p})=(\frac{-4}{p})=1$ hence $p\equiv 1(mod \ 4)$. Also $p|t^2+8$ gives $(\frac{2}{p})=(\frac{-2}{p})=(\frac{-8}{p})=1$ so $p\equiv 1,7(mod \ 8)$. These two give that $p\equiv 1(mod \ 8)$ but since $x^2+4 \not \equiv 1(mod \ 8),$ there exists a prime $\not \equiv 1(mod \ 8)$. Thus, we get contradiction as desired.$\blacksquare$

Suppose there exists such a triple. It's easy to check that none of them can be zero. Note we can flip signs of any two and still keep equation valid, so wlog say $x, y > 0$. Note that if $z$ works, then $\frac{-2-x^2-y^2}{z}$ also works (and is an integer, by vieta), so we can also assume that $z > 0$. Now consider a triple with all $x,y,z$ positive with $x+y$ minimal. But now, note that (wlog $x \geqslant y$) if $x$ works, by vieta so does $\frac{y^2 - z^2 + 2}{x}$. Since $x$ was minimal, we have that $\frac{y^2 - z^2 + 2}{x} \geqslant x$ or that $x^2 \geqslant y^2 \geqslant x^2 + z^2 - 2$ so we must have $z = 1$, but again in this case it's easy to see there are no solutions.

L567 wrote: if $x$ works, by vieta so does $\frac{y^2 - z^2 + 2}{x}$. Couldn't it happen that this second solution is negative? (And then you can't continue to argue by minimality...)

We can view this as a quadratic in $z$. The discriminant of this equation must be a perfect square, say $s^2$. \begin{align*} \Delta=s^2 & =x^2y^2+4x^2+4y^2+8 \\ s^2 & =(x^2+4)(y^2+4)-8 \\ s^2+8 & =(x^2+4)(y^2+4) \end{align*}It is standard to show that if some odd prime $p$ divides $s^2+8$ then $p\equiv 1,3 \pmod{8}$ and if some odd prime $p$ divides $x^2+4$ then $p\equiv 1,5 \pmod {8}$. Thus for all odd prime factors $p$ of both sides $p\equiv 1 \pmod{8}$. Now notice that if both $x$ and $y$ are even then $16|(x^2+4)(y^2+4)$ but it is not hard to show that $16 \nmid (x^2+4)(y^2+4)$. So assume that $x$ is odd. Then $x^2+4\equiv 3,5 \pmod{8}$, a contradiction.

Let x•y = even no. (Like 2,4,6,8,...) Now after you solve the equation considering xy=even then it's never possible to get such x,y,z integers. Let now consider x•y=odd no. Which again proves it's impossible