Is the famous problem number #6 solvable in first order Peano arithmetic?

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I just came accross the famous "very difficult" problem 6 of the 1988 International Mathematical Olympiad:

Let $a$ and $b$ be positive integers and $k=\frac{a^2+b^2}{1+ab}$. Show that if $k$ is an integer then $k$ is a perfect square.

The solution involves Vieta jumping or a descending sequence, for those interested the exact solution can be read here.

Question: Is this kind of descending sequence proof possible under first order Peano axioms?

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Yes, this is easily provable in PA. Indeed, at a glance the tiny fragment I$\Sigma_1$ is already enough.

PA, remember, gives us induction along $\mathbb{N}$ for all formulas expressible in the language of arithmetic. Principles like "every Goodstein sequence terminates" are proved via "long induction" (e.g. induction along $\epsilon_0$), but something basic like Vieta jumping - where the statements involved are all arithmetic, and the induction is just along $\mathbb{N}$ with the usual ordering - goes through in PA without any difficulty.