Let $a$ and $b$ be two positive integers such that $ab-1|a^2+b^2$.
Show that $\frac{a^2+b^2}{ab-1}=5$.
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1I did this, including the more general $ab-t$ for positive integer $t$ and $ab>t,$ see http://math.stackexchange.com/questions/829228/is-it-true-that-fx-y-dfracx2y2xy-t-has-only-finitely-many-distinct-i Where did you get this problem? – Will Jagy May 11 '16 at 03:42
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Related : http://math.stackexchange.com/questions/28438/alternative-proof-that-a2b2-ab1-is-a-square-when-its-an-integer – lab bhattacharjee May 11 '16 at 04:57
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There is a famous and difficult problem of IMO for which if $ab+1|a^2+b^2$ then $\frac{a^2+b^2}{ab+1}$ is necessarily a square. (it is clear that it is another problem but it looks like...) – Piquito May 11 '16 at 21:40
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I made a bunch of pictures which show how the argument goes for this problem. All concepts are shown in detail at Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers?
On the other hand, this one is fairly easy and may help people. The Vieta Jumping argument says that, if there are any integer points $(x,y)$ in the hyperbola $$x^2 - q xy + y^2 = -q,$$ with integers $x,y,q > 0,$ then there are integer points that also satisfy $$ 2x \leq qy $$ $$ 2y \leq qx. $$ This viewpoint comes from A. Hurwitz (1907) and clarifies these Vieta Jumping problems nicely.
For now, I am just going to put several diagrams for increasing $q.$ There are no integer points in the central arc for $q=3,4.$ For $q=5$ there are some, namely $(2,1)$ and $(1,2).$ For $q \geq 6$ we see that the central arc has only a very short arc with both $1 \leq x,y \leq 2,$ outside of which there are two fairly long arcs (but still inside the Hurwitz lines) that have either $0 < x < 1$ or $0 < y < 1.$ As a result, there are no integer points satisfying the Hurwitz conditions (which are just the inequalities we get after Vieta Jumping) and therefore no integer points at all.
