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Find infinitely many pairs of integers $(a,b)$ such that $1<a<b$ and $ab$ divides $a^{2}+b^{2}-1$. Also find all positive integers $k$ such that there exists $(a,b)$ such that $\frac {a^{2}+b^{2}-1}{ab}=k$.

SANZEED wrote:Find infinitely many pairs of integers $(a,b)$ such that $1<a<b$ and $ab$ divides $a^{2}+b^{2}-1$. Also find all positive integers $k$ such that there exists $(a,b)$ such that $\frac {a^{2}+b^{2}-1}{ab}=k$.

Take $a>1,\; b=a^2-1$. Then $\dfrac {a^2+b^2-1}{ab}=a=$ Any integer greater than $1$. $\dfrac {a^2+b^2-1}{ab}=1$ is not possible because it will imply $(a-b)^2=1-ab<0$ since $ab>1$. So all the $\{k\}=\mathbb N\backslash 1$

SANZEED wrote:Find infinitely many pairs of integers $(a,b)$ such that $1<a<b$ and $ab$ divides $a^{2}+b^{2}-1$. Also find all positive integers $k$ such that there exists $(a,b)$ such that $\frac {a^{2}+b^{2}-1}{ab}=k$.

We use Vieta Jumping to solve this problem.