Pinko wrote: Solve the following system of equations in integers: $\begin{cases} x^2+2xy+8z=4z^2+4y+8\\ x^2+y+2z=156 \\ \end{cases}$ Here is a partial ugly work (I miss proofs of 2.3, 2.5, 2.6) giving three solutions : $\boxed{(x,y,z)\in\{(2,150,1),(10,30,13),(10,94,-19)\}}$ And likely proving (with missing parts) that these are all the solution. Hope somebody will be able to finish it and propose a prettier solution. First equation show that $x$ is even and second equation shows then that $y$ is even. Setting $x=2x_1$ and $y=2y_1$, system becomes : $\begin{cases} x_1^2+2x_1y_1+2z=z^2+2y_1+2\\ 2x_1^2+y_1+z=78 \\ \end{cases}$ Then first equation implies $x_1\equiv z\pmod 2$ and second equation implies $y_1\equiv z\pmod 2$ If all $x_1,y_1,z$ are even, first equation becomes wrong $\pmod 4$ So all are odd and we can write $x_1=2u+1$, $y_1=2v-1$ and $z=2w+1$ and system becomes : $\begin{cases} u^2+2uv=w^2\\ 4u^2+4u+v+w=38 \\ \end{cases}$ 1) If $u=0$ This implies $w=0$ and $v=38$ and so the solution $\boxed{(x,y,z)=(2,150,1)}$ 2) If $u\ne 0$ First equation implies $u|w^2$ and so we can write $u=ab^2$ and $w=abc$ for some $a$ squarefree and $b,c$ all nonzero And system becomes $\begin{cases} v=\frac{a(c^2-b^2)}2\\ v=38-4a^2b^4-4ab^2-abc \\ \end{cases}$ And so $a(c^2-b^2)$ even and $a(c^2-b^2)=76-8a^2b^4-8ab^2-2abc$ Which is $a(b+c)^2=76-8a^2b^4-6ab^2$ So squarefree $a$ idivides $76$ and so $a\in\{\pm 1,\pm 2, \pm 19, \pm 38\}$ Setting $b+c=N$, equation is $(8ab^2+3)^2+2aN^2=617$ If $a>0$, we have $8ab^2+3\le \sqrt{617}$ and so $1\le b^2\le \frac{\sqrt{617}-3}{8a}$ and so $a\le 2$ And so $a\in\{1,2,-1,-2,-19,-38\}$ 2.1) $a=1$ implies $-8b^4-6b^2+76$ perfect square In order quadratic be positive, we need $b^2\in[1,2]$ And so $b^2=1$ and expression is $62$, not a perfect square 2.2) $a=2$ implies $-16b^4-6b^2+38$ perfect square In order quadratic be positive, we need $b^2=1$ And expression is $16$ indeed a perfect square. And so $a=2,b^2=1,(b+c)^2=16$ And so $(a,b,c)\in\{(2,1,3),(2,1,-5),(2,-1,-3),(2,-1,5)\}$ And so $(u,v,w)\in\{(2,8,6),(2,24,-10)\}$ And so $\boxed{(x,y,z)\in\{(10,30,13),(10,94,-19)\}}$ 2.3) $a=-1$ implies $(8b^2-3)^2-2N^2=617$ Seems impossible : to be proved 2.4) $a=-2$ implies $(16b^2-3)^2-4N^2=617$ And so $(16b^2-3-2N)(16b^2-3+2N)=617$ Which has no solution (this implies $2b^2=39$) 2.5) $a=-19$ implies $(152b^2-3)^2-38N^2=617$ Seems impossible : to be proved 2.6) $a=-38$ implies $(304b^2-3)^2-76N^2=617$ Seems impossible : to be proved

Hey pco, The original solution uses a slightly different method to yours with different addition of variables $u$ and $v$ (Noting something about Pythagorean triples). Here it is: We rewrite the first equation in the following way: $(x+y)^2=4(z-1)^2+(y+2)^2$. Now it is obvious that these three numbers form a Pythagorean triple. Denote $y+2=u^2-v^2$, $2z-2=2uv$, and $x+y=u^2+v^2$ where $u,v \in \mathbb{Z}$. Then expressing $x, y, z$ with $u, v$ gives: $x=2(v^2+1)$, $y=u^2-v^2-2$, $z=uv+1$. Substituting in the second equation leads to $4v^4+6v^2+(u+v)^2=152$. Hence $v^4+2v^2\leq 38$, which means that $|v|\leq 2$. Now checking for $v=0$ and $v=\pm 1$ we see that there are no solutions and for $v=\pm 2$ we note that $|u+v|=8$. Now direct calculations lead to the answers $(2,150,1), (10, 30, 13), (10, 94, -19)$ which you have already found.

Pinko wrote: Denote $y+2=u^2-v^2$, $2z-2=2uv$, and $x+y=u^2+v^2$ where $u,v \in \mathbb{Z}$.. This is not the general solution for pythagorean triplets (they miss a common multiplicator)