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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!

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    $\begingroup$ Some solutions, if that helps: $(1,-2), (1,-1), (1,2), (2,-17), (2,-1), (2,17), (3,-41), (3,-2), (3,82), (4,257), (8,17),$ $(8,241), (9,-386), (9,-17), (14,-41), (27,82), (30,241), (43,-386), (64,257)$ $\endgroup$
    – Alexey Burdin
    May 30, 2015 at 5:11
  • $\begingroup$ Oh nice! you're setting the numerator equal to 0 is it $\endgroup$
    – AgentS
    May 30, 2015 at 5:17
  • $\begingroup$ Nah, just a python script (e.g. $\frac{30^4-241+1}{30\cdot241}=112$). I'm at lack of ideas how to solve this. $\endgroup$
    – Alexey Burdin
    May 30, 2015 at 5:21
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    $\begingroup$ Thank you so much! I can finish off the rest... (8, 17) also gives a nonzero integer $\endgroup$
    – AgentS
    May 30, 2015 at 5:23
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    $\begingroup$ No, $(2,-1)$ makes the fraction $\frac {18}{-2}=-9$. Others as well do not make the numerator $0$. $\endgroup$
    – Ross Millikan
    May 30, 2015 at 5:23

1 Answer 1

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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 $b$.

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    $\begingroup$ Why select this as the correct answer? It might be the best we can do, but it seems premature to select it until people have had a chance to try a bit more. $\endgroup$
    – Thomas Andrews
    May 30, 2015 at 5:36

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