$x^2+y^2 =n +z^2$ where $x,y,z$ are different natural numbers. What values can $n$ assume?
What if x,y,z>0 considering the confusion of 0 as natural number?
Any odd number : $0^2+(n+1)^2=(2n+1)+n^2$
Any positive even number : $1^2+(n+1)^2=(2n+2)+n^2$
Any negative odd number : $0^2+n^2=-(2n+1)+(n+1)^2$
Any negative even number : $1^2+n^2=-2n+(n+1)^2$
If you really need to have different numbers x y z just remark that
and $3^2+5^2= -2 + 6^2$
If you need x, y, z to be non zero for odd numbers :
$2^2+(n+1)^2=(2n-3)+n^2$ for example... $2^2+n^2=-(2n-3)+(n+1)^2$
and for the case 1 (all different) $4^2+7^2=1+8^2$ and $4^2+8^2=-1+9^2$...