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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. $x<n<\frac{x^2+2x+1}{4}$ and $\sqrt{n^2-(x+1)^2}<k<\sqrt{n^2-x^2}$
  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:

$k=1\pm{b}$

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

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

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

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  • $\begingroup$ math.stackexchange.com/questions/2803972/… $\endgroup$ – individ Jul 30 '18 at 4:26
  • $\begingroup$ math.stackexchange.com/questions/351491/… $\endgroup$ – individ Jul 30 '18 at 4:26
  • $\begingroup$ 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 $Z$? $\endgroup$ – Spencer Connaughton Jul 31 '18 at 0:43
  • $\begingroup$ 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, $6+3^2=15=4^2-1^2$ $\endgroup$ – user25406 Jul 31 '18 at 12:19
  • $\begingroup$ 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 $x=3$. $\endgroup$ – user25406 Jul 31 '18 at 19:19

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