If $a,b,c\in\mathbb{Z} \text{ and } n\in\mathbb{Z}\ge2$ where $$\begin{cases} a+b-c=n \\ a^2+b^2-c^2=n \end{cases}$$ Prove that there is at least $1$ solution and finitely many solutions.

This question is probably one of the hardest I have attempted yet. I have tried to solve it but couldn't. Is there any way I could get help here?

  • $\begingroup$ What did you attempt and where did you get stuck? $\endgroup$ – Alex Jan 27 at 15:05
  • $\begingroup$ @Alex I attempted many things but felt that they were too silly and wrong to be added. $\endgroup$ – user587054 Jan 27 at 15:06
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    $\begingroup$ You mean, for fixed $n\ge 2$, there is at least one (but not infinitely many) solution in $a,b,c$? Or there are only finitely many solutions overall? $\endgroup$ – Hagen von Eitzen Jan 27 at 15:09
  • $\begingroup$ @HagenvonEitzen for fixed $n\ge2$, there is at least one (but not infinitely many) solutions in $a,b,c$ $\endgroup$ – user587054 Jan 27 at 15:16

For even $n$,

$$a=2n-1, b=\frac{3n}2, c=\frac{5n}2-1$$

is a solution, while for odd $n$,

$$a=2n, \space b=\frac{3n-1}2, \space c=\frac{5n-1}2$$

is one.

To prove that there are only finitely many solutions for a given $n\ge2$, we observe that

$$b-c=n-a, \space b^2-c^2=n-a^2$$

and hence $n-a|n-a^2$. Since $n-a|n^2-a^2$ it follows that $n-a|n^2-n$. Since $n \ge 2$, we have $n^2-n > 0$ and thus it has only finitely many integer devisors and thus only finitely many $a$ are possible. It remains to be shown that for each such $a$, only finitely many $b,c$ can exist.

$a=n$ is impossible, as that would imply $0|n^2-n$, which is impossible. So we get

$$b+c = \frac{b^2-c^2}{b-c} = \frac{n-a^2}{n-a}$$

Together with $b-c=n-a$ this gives exactly one solution $(b,c)$ in rationals, so at most one for integers, which concludes the proof. $\blacksquare$

These equations for $b-c$ and $b+c$ are also the way I found the special solutions given above. $n-a|n^2-n$ implies looking for divisors of $n^2-n$, where $n$ and $n-1$ obviously stand out. A little experimentation with signs lead to considering $a-n=n$ and $a-n=n-1$, and the equations for $b-c$ and $b+c$ then lead straightforward to the solutions.


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