Understanding vieta jumping.

Question

Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Show that $$\frac{a^2 + b^2}{ab+1}$$ is the square of an integer.

I have a few questions about the proof. First here it is ,

Assume $$\frac{a^2 + b^2}{ab+1}=k$$ is not a perfect square. Rearranging, we get a quadratic in $a$ $$a^2 −kb·a+(b^2 −k)=0. \quad (*)$$ Clearly $(*)$ has two solution the first one is $a$ so let the second be $s$. Then by Vieta we have $$s=kb-a=\frac{b^2-k}{a}$$ The first equation shows $s$ is an integer, and from $*$ $s$ is positive.

Now since $*$ symmetric we can assume $a>b$ Ignore $a=b$ because it gives a perfect square. Hence $s<a$ so we have a descent. Right?

In my book the author said this So, do we have a descent? If we repeat the process of (s, b), we would get a quadratic in s, and we pick the second root. Do you see an issue? But I don’t really see any issue.

Answer 1

Basically, we assumed that there is a smallest positive integer $a$ such that $k$ is not a perfect square. Using the quadratic you gave, if we have a root such that $a$ is not equal to $b$, then we will always be able to find a root smaller than $a$ that fits the requirements. However, this contradicts our assumption that $a$ was minimal. Thus, $k$ must be a perfect square.

This is similar to the method of descent, where we will always find a smaller root that fits the requirements. However, since there is a smallest positive integer, it is not possible to do this over and over, which shows that there are no solutions with $k$ not a perfect square.

Answer 2

Problem: Let $a$ and $b$ be positive integers such that $ab+1$ divides $a^2+b^2$. Show that $$\frac{a^2+b^2}{ab+1}$$ is the square of an integer (perfect square).

Expression $$\frac{a^2+b^2}{ab+1}$$ is symmetric with respect to $a$ and $b$, then without loss of generality we can suppose $a\geq b$.

Let consider first case $a=b$. Then $$\frac{a^2+b^2}{ab+1}=\frac{2a^2}{a^2+1}=2-\frac{2}{a^2+1}$$ This expression can be integer at integer positive $a$ only if $a=1$, then $$\frac{a^2+b^2}{ab+1}=1$$ is perfect square.

Now consider case $a>b$. Let suppose that $a$ is minimum positive integer such that exists positive integer $b<a$ such that $\frac{a^2+b^2}{ab+1}$ is integer but is not perfect square. Let consider expression $$s=\frac{b^3-a}{ab+1}$$ $$s=\frac{b^3-a}{ab+1}=b\frac{a^2+b^2}{ab+1}-a\Rightarrow s\in\mathbb{Z}$$ $$s=\frac{b^3-a}{ab+1}>-\frac{a}{ab+1}\geq-\frac{a}{a+1}>-1$$ $$s>-1 \land s\in\mathbb{Z}\Rightarrow s\geq 0$$ $$s=0\Rightarrow a=b^3\Rightarrow \frac{a^2+b^2}{ab+1}=b^2$$ which contradicts to the fact that $\frac{a^2+b^2}{ab+1}$ is not perfect square. Then $s>0$. $$s=\frac{b^3-a}{ab+1}<\frac{b^3}{b^2}=b$$ Let consider expression $$\frac{b^2+s^2}{bs+1}=\frac{a^2+b^2}{ab+1}$$ Then $b$ is positive integer such that exists positive integer $s<b$ such that $\frac{b^2+s^2}{bs+1}$ is integer but is not perfect square. Then $b<a$ contradicts with the fact that $a$ is minimum positive integer with such property. Then there is no minimum positive integer with such property, then there is no positive integers $a,b$ with such property.