TITLE: Find all pairs of integers $(a,b),~ b\ne 1$ such that $\frac{a^4-b+1}{ab}$ is an integer
QUESTION [5 upvotes]: Find all pairs of integers $(a,b)$ such that $\frac{a^4-b+1}{ab}$ is an integer.
$b=1$ trivially gives infinitely many solutions as the expression becomes $a^3$. I am not able to find any more solutions. I tried Fermat's infinite descent to prove there are no solutions and got stuck... Also I have started reviewing Vieta's root jumping. Do I get some help on how to proceed... Thanks!

REPLY [4 votes]: Just a start.
You need $a\mid b-1$ and $b\mid a^4+1$. Since this implies $a,b$ are relatively prime, this is necessary and sufficient.
We know that $b$ has to be the product if primes $\equiv 1\pmod 8$, and possibly one factor of $2$.
For any $a$ you need to find a $k$ such that $(ak+1)\mid a^4+1$. Such $k$ come in pairs.
For $a\leq 7$, $a^4+1$ is prime or twice a prime, so there are only the trivial solutions $b=1,a^4+1.$ 
For $a=8$, $8^4+1=17\cdot 241$, so $b=17$ or $b=241$ is a solution.
$9^4+1=2\cdot 17\cdot 193$, which has no non-trivial divisor $\equiv 1\pmod 9$, so no nontrivial $b$.