Prove that for infinitely many integers $n>2$ equation
has no integer solutions for $a,b$. Edit: I would appreciate any hints. They may concern other nonlinear diophantine equations for n>... as I think I wouldn't be able to do even another example and I guess I may not be the only one having these difficulties

  • $\begingroup$ Can someone explain why it got downvoted? $\endgroup$
    – user509482
    Dec 3, 2017 at 10:38
  • $\begingroup$ Why do you downvote? $\endgroup$
    – user509482
    Dec 3, 2017 at 10:47
  • 1
    $\begingroup$ A downvote means (official definition on this site) "This question does not show any research effort; it is unclear or not useful." Where's your research? What's your question? Imperatives like "prove", "show", "explain" aren't popular, here. $\endgroup$
    – user436658
    Dec 3, 2017 at 11:01
  • $\begingroup$ @Professor Vector I got a Diophantine exercise list with equations of exponents 2,3,4 and 5. Then there is this one, that I have no idea how to solve $\endgroup$
    – user509482
    Dec 3, 2017 at 11:09
  • $\begingroup$ @user509482 Just a curiosity. What kind of school do you attend? I ask because I am a math graduate here in Italy and I never had such assignments... $\endgroup$
    – Raffaele
    Dec 3, 2017 at 11:12

1 Answer 1


The equation doesn't have integer solutions, if $n$ is an odd prime: because of the theorem of Fermat, we have $x^{n-1}=1\pmod n$, for $x\neq0\pmod n$, so that always $x^n=x\pmod n$. That means $a^n-(a-2)^n=a-(a-2)=2\pmod n$. But $b^{n-1}\pmod n$ is $0$ if $n$ divides $b$, and $1$ otherwise, so the equation can't be satisfied.

  • $\begingroup$ b^(n-1)(mod n) is 0 when b (mod n)=1 and 1 when b (mod n) doesn't equal 1? $\endgroup$
    – user509482
    Dec 3, 2017 at 12:51
  • $\begingroup$ Sorry, that was a typo, I've edited it. $\endgroup$
    – user436658
    Dec 3, 2017 at 12:57
  • $\begingroup$ b^(n-1)(mod n) is 0 when b (mod n)=0 and 1 when b (mod n) doesn't equal, am I right? $\endgroup$
    – user509482
    Dec 3, 2017 at 13:00

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.