Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$x,n,m >1$ 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.

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
1  
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
1  
@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
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.