<s>[INST] Number Theory And Vieta Jumping $\textbf{Question:}$Find all positive integers $a, b$ such that the expression $$\frac{a^2+b^2+1}{ab-1}$$ is an integer.
$$$$As the expression is symmetric in $a, b$, so let $a \geq b$. It is easy to check the cases when $a=b+k$ where $0 \leq k \leq 2$, so let $a \geq b+3$. Suppose $a, b$ satisfies the given condition with $a+b$ minimized. Let us write $$\frac{a^2+b^2+1}{ab-1}=k$$ then our equation becomes $$a^2-kab+b^2+k+1=0$$. Now this equation has a root $a$ and from the Viet's formula we get the second root $$a'=kb-a$$ and $$a'=\frac{b^2+k+1}{a}$$. Now the first equation shows that $a'$ is an integer and the second equation shows that $a'$ is a positive integer. Now to show that $a' < a$ we have to prove that $$a'=\frac{b^2+k+1}{a} < a$$. Now consider $$a^2-(b^2+k+1)=a^2-b^2-1-k=a^2-b^2-1-(\frac{a^2+b^2+1}{ab-1})=\frac{a^3b-ab^3-ab-a^2+b^2+1-(a^2+b^2+1)}{ab-1}=\frac{a(a^2b-b^3-b-2a)}{ab-1}$$. Now as $a \geq b+3$ so we have $$a^2b-2a=a(ab-2) \geq a((b+3)(b)-2)=a(b^2+3b-2) \geq a(b^2+1) > b(b^2+1) > b^3+b$$ and hence we get $\frac{a(a^2b-b^3-b-2a)}{ab-1}>0$ and hence we have $a' < a$ and hence $a'+b < a+b$ contradicting the minimality of $a+b$. So there are no solutions.
$$$$Is My Proof Correct??
 [/INST] There are infinitely many solutions with $k=3$   and $k=6.$ For 3 the generating point is $(2,2)$ For $k=6$  the points are $(2,1)$ and $(1,2)$
More solutions can be found in either case by Vieta jumping.
Ruling out larger $k$ is by inequalities.

See how, as soon as $k \geq 7,$ the arc of the hyperbola passes through the interior of the square $1 < x < 2, $ $ 1 < y < 2.$ The intersections with the bounding lines $y = \frac{k}{2} x$  and $x = \frac{k}{2} y$ have one coordinate strictly below $1.$ As a result, there are no integer points on the arc between the bounding lines, therefore no integer points in the first quadrant at all.   The method is described in HURWITZ 1907


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