Alternative proof that $(a^2+b^2)/(ab+1)$ is a square when it's an integer

Solution 1:

This is, IMHO, one of the most popular (and actually the most beautiful) problems in number theory. It is IMO 1988 Problem 6.

You can find it in these links (most of the solutions are the same as the two you remarked, but you may find some more facts about these problems if you see below):

1st (master link)

2nd

3rd

4th

5th

This is a cool generalization of this problem (you may see this, too):

Problem. Let $a,b$ be positive integers satisfying $$(ab)^{n-1}+1 \mid a^n +b^n.$$ Then the number $\frac{a^n +b^n}{(ab)^{n-1}+1}$ is a perfect $n^{th}$ power.

Here are some nice problems related to this one:

1st

2nd

3rd

4th

And, for the sake of completeness about this problem, see Vieta Jumping Method. (IMO 1988 is the best example for Vieta Jumping!)

Solution 2:

Let us prove the general case: if $a, b, n\in \mathbb{N}$ and $$c:=\frac{a^n+b^n}{(ab)^{n-1}+1}$$ then $c\in\mathbb{N}$ is an $n$th power.

Proof: In order to show, it is clear that we need to express $c$ as $c=t^n$ for some $t\in\mathbb{N}$, isn't it?

So let us consider $a=t^{\alpha}, b=t^{\beta}$. We are given that $a, b$ are integers (WOLOG we have chosen as positive integers) and hence if $c$ is supposed to be as $t^n$ then $a, b$ have to be some power of $t$; Otherwise, contradiction will arrive.

So we have $$c=\frac{a^n+b^n}{(ab)^{n-1}+1}$$ $$i.e. c =\frac{(t^\alpha)^n+(t^\beta)^n}{(t^{\alpha+\beta})^{n-1}-1}$$ $$i.e. c =\frac{t^{n\beta}(t^{n\alpha-n\beta})+1}{t^{(\alpha+\beta)(n-1)}+1} $$

So let $\beta=\beta_0$. Now in order to make $c$ as integer, we must have $$t^{(\alpha+\beta_0)(n-1)}+1 \mid t^{n\alpha-n\beta_0}+1$$ which is true iff $$(\alpha+\beta_0)(n-1)\mid (n\alpha-n\beta_0)$$ $$\text{iff}\; (\alpha+\beta_0)(n-1)s=n\alpha-n\beta_0 \; \text{for some}\; s\in\mathbb{N}$$ $$\text{so}\; \alpha=\frac{-\beta_0(2n-1)}{(n-1)s-n}$$

We choose $s$ in such a way that $\alpha$ is in $\mathbb{N}$. For the sake of simplicity we take $s=1$ then $\alpha=\beta_0(2n-1)$.

Thus we have $a=t^{\alpha}=t^{\beta_0 (2n-1)}, b=t^{\beta_0}, \beta_0 \in\mathbb{N}.$ And consequently $c=t^{n\beta_0}=(t^{\beta_0})^n$. We are done.