0

I am trying to solve $$x^n-1=a^m$$ equation in integers for odd n.

For the case $a=2$:

From the relation: $$(x-1)(x^{n-1}+x^{n-2}+...+1)=2^m$$

we can see that $x$ should odd, and the $n$ is even. So for odd $n$ it has no solution.

Now assume that the $a>2$ the question is if there is a solution for given odd $n$. Wondering if this is known topic and there is any paper about this.

Gevorg Hmayakyan
  • 1,698
  • 1
    It is known [and not completely simple to prove] that the only difference of perfect powers which is 1 is $3^2-1=2^3.$ However you may be looking for more elementary proofs. – coffeemath Oct 01 '23 at 19:00
  • 4
    More generally, consider the Diophantine equation $x^n-y^m=c$, see this post. The case $c=1$ was known as Catalan's conjecture. – Dietrich Burde Oct 01 '23 at 19:01
  • 1
    From the duplicate: Mihăilescu's theorem, before 2002 known as Catalan's conjecture, states that: The only natural numbers $m,n>1$ and $x,a>0$ satisfying $$x^n-a^m=1,$$are $(m,n,x,a)=(2,3,3,2)$. – Dietrich Burde Oct 01 '23 at 19:10
  • 1
    There is no easy proof of this deep result. It required massive computational power combinaed with clever handling the remaining cases. You can imagine how difficult this proof was if you consider that for no integer $d\ge 2$ , it is known whether there are only finite many pairs of perfect powers with this difference , although exactly this is the statement of Pillai's conjecture for every positive integer $d$ – Peter Oct 02 '23 at 15:14

0 Answers0