# Positive even numbered integer solutions of $y=n^2-m^2-x^2$

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:

1. $$x and $$\sqrt{n^2-(x+1)^2}
2. $$x \rightarrow odd \Rightarrow (k \rightarrow even,n \rightarrow odd) ||(k \rightarrow odd,n \rightarrow even)$$
3. $$x \rightarrow even \Rightarrow (k \rightarrow even,n \rightarrow even) ||(k \rightarrow odd,n \rightarrow odd)$$

Per @individ 's answer here, there is a set of solutions at:

### $$n=\frac{(b^2+2+y\pm{2b})}{2}$$

But this doesn't seem to represent all solutions for any given $$y$$.
The statement can also be expressed as related right triangles.

• math.stackexchange.com/questions/2803972/… – individ Jul 30 '18 at 4:26
• math.stackexchange.com/questions/351491/… – individ Jul 30 '18 at 4:26
• That's helpful @individ. How did you derive the particular solutions in the integral solutions problem? And will they produce all possible solutions for $X$, $Y$, and $Z$? – Spencer Connaughton Jul 31 '18 at 0:43
• you can rewrite the equation as: $y + x^2 = n^2-m^2 = (n-m)(n+m)$. So you are looking for numbers that can be represented as $N=y+x^2$. For example, $6+3^2=15=4^2-1^2$ – user25406 Jul 31 '18 at 12:19
• you change the upper limit on x from $2x+1$ to $2x$ with even y. The solution I gave in my previous comment is still valid for $y=6$ and $x=3$. – user25406 Jul 31 '18 at 19:19