Prove that no integer $x$ exists where $y=n^2-m^2-x^2$ has solutions:
- For all even integer values of $y$ in the range $2\le y \le 2x+1$ where $x$ is odd.
- For all odd integer values of $y$ in the range $1\le y \le 2x+1$ where $x$ is even.
Assume $n,m\in\Bbb{Z}$.
I've noticed that:
- $x<n<\frac{x^2+2x+1}{4}$ and $\sqrt{n^2-(x+1)^2}<k<\sqrt{n^2-x^2}$
- $x \rightarrow odd \Rightarrow (k \rightarrow even,n \rightarrow odd) ||(k \rightarrow odd,n \rightarrow even)$
- $x \rightarrow even \Rightarrow (k \rightarrow even,n \rightarrow even) ||(k \rightarrow odd,n \rightarrow odd)$
Per @individ 's answer here, there is a set of solutions at:
$k=1\pm{b}$
$x=\frac{(b^2+y\pm{2b})}{2}$
$n=\frac{(b^2+2+y\pm{2b})}{2}$
But this doesn't seem to represent all solutions for any given $y$.
The statement can also be expressed
as related right triangles.
