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$x,n,m >1$ and $x,n,m \in \mathbb{Z}$

I've tried to solve it myself, but I'm getting nowhere, so apologies if it's an irritatingly basic question.

Whilst I'm on it, is it true that $y^x \ne m^n+1$ ($y>1, y \in \mathbb{Z}$)? There's probably a glaringly obvious counterexample I'm overlooking.

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up vote 9 down vote accepted

The Diophantine equation is $$2^x = 1 + m^n.$$

  • Since the left hand side is even $m$ must be odd, let $m = 2r+1$.
  • Since $m^n + 1$ factors into $(m+1)(\cdots)$ when $n$ is odd, $n$ must be even, let $n = 2u$.

Putting these two facts together and applying the binomial theorem gives $$m^n = (2r+1)^{2u} = (4r(r+1)+1)^u = \sum_{k=0}^u \binom{u}{k} [4r(r+1)]^k$$

Plugging this into our original equation, and pulling out the first term of the sum we have $$\begin{array}{rcl} 2^x &=& 1 + \binom{u}{0} [4r(r+1)]^0 + 4 \sum_{k=1}^u \binom{u}{k} 4^{k-1}[r(r+1)]^k \\ &=& 2 + 4 M \\ \end{array}$$ for some $M$, this is impossible unless $M=0$, then $x=1$ and $u=0$ and $m$ can be anything.

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Pretty proof. [spacefiller] – Alyosha Nov 23 '12 at 22:09
+1 from me , although it could be longer and clearer. – mick Nov 23 '12 at 22:56
Actually, could you elucidate as to why it can't factor like that? – Alyosha Nov 24 '12 at 11:52
@Alyosha, it forces $m$ to be $2^r-1$ which makes the other factor $m^{n-1} - m^{n-2} + \ldots + 1$ odd (it's just $1 + 1 + 1 + \ldots + 1 = 1$ when viewed mod 2) which can only happen if it's $1$ (in that case we have the solution $2^1 = 1 + 1^1$). – user50336 Nov 24 '12 at 13:35
Is that argument equivalent to saying: $m^n+1=(m+1)(m^{n-1}-m^{n-2}+m^{n-3}-m^{n-4}....-m^1+m^0)$ only if $n-1$ is odd (so both the $m^{n-1}$ part and the $m^0$ part do not have a $-1$ coefficient? – Alyosha Nov 24 '12 at 23:03

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