The diophantine equation $a^n - 1 = (a-1)^m$. 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.
 A: 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 $a>2?$
A: You can prove the following more general result without using Catalan's Conjecture (i.e. Mihăilescu's theorem), but still using Zsigmondy's theorem, which is difficult too, but is provable using cyclotomic polynomials.
More general result: The only positive integer solutions to $a^n-1=(a-1)^m$ are $(a,n,m)=(1,n,m),(2,1,m),(t,1,1),(3,2,3), n,m\ge 1, t\ge 3$.
Proof: If $a\in\{1,2\}$, then $(a,n,m)=(1,n,m),(2,1,m)$. Let $a\ge 3$. 
If $n=1$, then $1=(a-1)^{m-1}$, so $(a,n,m)=(t,1,1), t\ge 3$. 
If $n=2$, then $a+1=(a-1)^{m-1}$. If $m\in\{1,2\}$, then no solutions. If $m\ge 3$, then $a+1\ge (a-1)^2=a^2-2a+1$, i.e. $3a\ge a^2$, so $a=3$, so $(a,n,m)=(3,2,3)$.
If $a,n\ge 3$, then by Zsigmondy's theorem $a^n-1$ has a prime divisor that does not divide $a-1$, contradiction.
