If we square both equations and then add them we get: x^2 + y^2 +(xz)^2 + (yz)^2 =290 Then (z^2 + 1)(x^2 + y^2) = 290 We can take the prime factors of 290 and then solve the cases which I won't do here.

vicentev wrote: Find all triples of integers \( (x, y, z) \) such that \[ x - yz = 11 \]\[ xz + y = 13 \] Plugging $x=yz+11$ (from first equation) in second, we get $11z+yz^2+y=13$ and so $y=\frac{13-11z}{z^2+1}$ This implies either $13-11z\ge z^2+1$ (and so $z\in\{-12,-11,...,0,1\}$, either $11z-13\ge z^2+1$ and so $z\in\{2,3,...,8,9\}$ Checking these $22$ cases, we easily find six solutions : $\boxed{(x,y,z)\in\{(-1,1,-12),(-1,12,-1),(11,13,0),(12,1,1),(-3,7,-2),(5,-2,3)\}}$