Proof that $a^n \ne b^{n+1}, (a, b) \in \mathbb{Z}$

Question

I was wondering if there exists a proof that:

$$a^n \ne b^{n+1},$$ $$a, b \in \mathbb{Z}$$

It would really help with a project that I am working on.

Thank you very much!

$8^2=4^3$, $16^3=8^4$, $32^4=16^5$, .... – Gerry Myerson May 06 '15 at 09:48 — Gerry Myerson, May 06 '15 at 09:48

score 3 · Accepted Answer · answered May 06 '15 at 09:48

3

$$8^4 = 4096 = 16^3$$ or in general, for any $k \in \mathbb N$, $$ (k^n)^{n+1} = (k^{n+1})^n $$ gives a counterexample with $a = k^{n+1}$, $b = k^{n}$.

answered May 06 '15 at 09:48

martini

84,101

Thank you very much! I am sorry that it was such an easily answered question! – Taylor May 06 '15 at 09:53

Proof that $a^n \ne b^{n+1}, (a, b) \in \mathbb{Z}$

1 Answers1