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!

  • 1
    $\begingroup$ 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 '15 at 5:11
  • $\begingroup$ Oh nice! you're setting the numerator equal to 0 is it $\endgroup$ – AgentS May 30 '15 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 '15 at 5:21
  • 1
    $\begingroup$ Thank you so much! I can finish off the rest... (8, 17) also gives a nonzero integer $\endgroup$ – AgentS May 30 '15 at 5:23
  • 1
    $\begingroup$ No, $(2,-1)$ makes the fraction $\frac {18}{-2}=-9$. Others as well do not make the numerator $0$. $\endgroup$ – Ross Millikan May 30 '15 at 5:23

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 $b$.

  • 6
    $\begingroup$ 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 '15 at 5:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.