# Prove that there is at least $1$ solution and finitely many solutions.

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?

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