Let $a,n,m$ be odd integers larger than one. The diophantine equation $a^n - 1 = (a-1)^m$ fascinates me.
I know that Catalan's conjecture has been proven and that Pillai's conjecture has not been proven yet.
See : http://en.wikipedia.org/wiki/Catalan%27s_conjecture
However the proof of Catalan's conjecture is not so easy ( for beginners ).
SO I was wondering about easy proofs for the equation $a^n - 1 = (a-1)^m$.
I assume they exist because I think this special case of Catalan's conjecture is less difficult than the whole Conjecture.
Also - though probably a poor argument - because this equation looks familiar to me. Maybe I have seen it , or something that resembles it - before.
I know that we can expand $(a-1)^m$ with the binomium theorem and I also know that $\dfrac{a^n-1}{a-1}$ can be reduced. But Im not sure if that knowledge helps here.
It is not immediately clear how infinite descent or mod arithmetic helps.
because $m$ is odd we get $a^n -1 = (a-1)^m$ mod $a$.
This reduces to $-1 = (-1)^m$ mod $a$ which is Always true.
My guess is that we use mod $p^2$ for some prime $p$ but Im not sure how.
Since you want all $a,n,m$ be odd and greater than one, it makes it easier to prove that there's no solution.
Hint: Show that in general, $\gcd(\frac{a^n-1}{a-1},a-1)=\gcd(n,a-1).$
Now can you see the contradiction for $a>2?$