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Find all positive integer values of $\dfrac{a^2+b^2}{ab+1}$.

Vieta-Jumping Solution:

\begin{align} &\text{let } \dfrac {a^2+b^2}{ab+1}=k. \\ \ \\ &a^2+b^2=kab+k. \\ &a^2-(kb)a+(b^2-k)=0. \\ \ \\ &\text{let $(a, b)=(a_0, b_0)$ be the root which $a_0+b_0$ is the smallest, $a_0\geq b_0$.} \tag{A}\label{A} \\ \ \\ &(a_0, b_0) \Rightarrow \bigg(\dfrac{b_0^2-k}{a_0}, b_0 \bigg) \Rightarrow \bigg(b_0, \dfrac{b_0^2-k}{a_0}\bigg). \\ \ \\ &a_0 \geq b_0, a_0b_0 \geq b_0^2, a_0b_0 >b_0^2-k, b_0>\dfrac{b_0^2-k}{a_0}. \\ \ \\ & \therefore a_0+b_0 > b_0+\dfrac{b_0^2-k}{a_0}, \text{Contradiction of } \ref{A}. \\ \ \\ &\therefore b_0^2-k\leq0, b_o^2\leq k. \\ & \text{When } b_0=k=1, a_0=1, \text{ and these are the smallest ones which satisfies $b_0^2\leq k.$} \\ \ \\ &\therefore (a, b, k)=(1, 1, 1). \end{align}

Is there another solution that doesn't use Vieta-Jumping?

RDK
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    Why don't you like the existing solution? There is not always an alternative solution. – WhatsUp Jun 26 '22 at 00:18
  • https://math.stackexchange.com/questions/1906908/math-olympiad-1988-problem-6-canonical-solution-2-without-vieta-jumping – Will Jagy Jun 26 '22 at 00:19
  • @WhatsUp Umm, as you can see in my profile, I like to find various solutions. I always am curious about 'various' things, not just finding one solution. – RDK Jun 26 '22 at 00:20
  • @RDK "Find all positive integer values of" $;-;$ How does the solution you posted answer that? – dxiv Jun 26 '22 at 00:21
  • @WillJagy Wow, thanks! Maybe I have to close this. – RDK Jun 26 '22 at 00:21
  • Does this answer your question? Math olympiad 1988 problem 6: canonical solution (2) without Vieta jumping - found from @Will Jagy's comment. – John Omielan Jun 26 '22 at 02:08
  • @JohnOmielan It's the same expression, indeed, but OP's question is not the same as the '88 one. Actually, at this point it's not clear to me what OP's question really is. – dxiv Jun 26 '22 at 05:27
  • @dxiv Thanks for the feedback. However, the OP's question is in the title, plus in the last line (i.e., "Is there another solution that doesn't use Vieta-Jumping?"), seems fairly clear to me. The other question is asking basically the same thing, although in the body text it specifies, as in it's linked question, "Is there a more direct and intuitive way to arrive at the solution". Nonetheless, the OP's comment to Will says "Wow, thanks! Maybe I have to close this.". – John Omielan Jun 26 '22 at 05:43
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    https://mathoverflow.net/questions/250172/when-is-fa-b-fraca2b21ab-a-perfect-square-rational-number/250300#250300 – individ Jun 26 '22 at 10:11
  • @JohnOmielan The original problem asks to prove that integer values are perfect squares. The OP here is asking what the possible integer values are. The questions are related, but IMHO they are not the same. Anyway, that's moot now since the wise Bot decided that my vote to close for "needs details" also counts as a vote to close for "duplicate" ;-) – dxiv Jun 27 '22 at 16:22

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