Originally Posted by
red_dog $\displaystyle \left\{\begin{array}{ll}x+y=z=12\\x^2+y^2=z^2+12\e nd{array}\right.$
Let $\displaystyle s=x+y, \ p=xy$
Then $\displaystyle \left\{\begin{array}{ll}s=z+12\\s^2-2p=z^2+12\end{array}\right.$
Replace s in the second equation. Then $\displaystyle p=12z+66$. The quadratic equation having the roots such as their sum is s and their product is p is $\displaystyle t^2-st+p=0$
That means $\displaystyle t^2-(z+12)t+12z+66=0$. The roots of this equation are x and y and are integers. Then the discriminant must be a perfect square.
$\displaystyle \Delta=(z-12)^2-264=k^2\Rightarrow(z-k-12)(z+k-12)=264$
$\displaystyle \left\{\begin{array}{ll}z-k-12=2\\z+k-12=132\end{array}\right.\Rightarrow z=79$
Replace z in t-equation: $\displaystyle t^2-91t+1014=0$ with the roots 13 and 78. So, $\displaystyle x=13, \ y=78, \ z=79$ or $\displaystyle x=78, \ y=13, \ z=79$
You can try other cases:
$\displaystyle \left\{\begin{array}{ll}z-k-12=4\\z+k-12=66\end{array}\right.$ or $\displaystyle \left\{\begin{array}{ll}z=k-12=8\\z+k-12=33\end{array}\right.$ etc.