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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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Spencer Connaughton
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  • https://math.stackexchange.com/questions/2803972/which-integers-can-be-written-as-x22y2-3z2/2804058#2804058 – individ Jul 30 '18 at 04:26
  • https://math.stackexchange.com/questions/351491/integral-solutions-of-hyperboloid-x2y2-z2-1/709219#709219 – individ Jul 30 '18 at 04:26
  • 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 00:43
  • 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
  • For odd y, we also have solutions: $5+4^2=21=5^2-2^2$ with $y=5, x=4$ – user25406 Jul 31 '18 at 19:28
  • @user25406 fixed upper limit. You can actually use the inequalities from point 1 above to generate valid values for $y$ for a given $x$ by sweeping $n$ and subsequent valid values of $k$ to get them. – Spencer Connaughton Aug 01 '18 at 00:03
  • I don't understand why you start with "Prove that no integer $x$ exists where $y=n^2−m^2−x^2$ has solutions" when we can easily find solutions for both odd and even y. – user25406 Aug 01 '18 at 17:44
  • Looking for a proof that no integer $x$ exists with solutions for all even or odd values of $y$ over that range, not just a few. Rewrote to make that slightly clearer. – Spencer Connaughton Aug 01 '18 at 20:18

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