Why is the following statement false: $ p ^{m} = q^{n}$ where $\ m,n $ are natural numbers.

Question

Question:

Assume $p$ and $q$ are distinct prime numbers. Why is the following statement false?

$ p ^{m} = q^{n}$ where $\ m,n $ are positive integers.

My attempt:

The fundamental theorem of arithmetic tells us that every natural number greater than $1$ is either prime or can be expressed as a product of prime and this product is unique. So, $ \ p ^{m} \neq q^{n}$ because both sides have a different product of primes.

Answer 1

Assuming that $p^m=q^n$:

Certainly then $p\mid q^n$ and $q\mid p^m$

$p$ prime $\implies p\mid q$

$q$ prime $\implies q\mid p$

Giving $p=q$ and this contradicts the specification that $p$ and $q$ are distinct primes.

Answer 2

If you'd like a more elementary tack: By Bezout, there exist integers $x$, and $y$ such that $px+qy=1$, so we can plut $1-px$ in for $q$ to get

$$p^m = (1-px)^n = 1 -{n \choose 1} px + {n \choose 2}(px)^2- \cdots.$$

Or

$$p^m +{n \choose 1} px -{n \choose 2}(px)^2- \cdots = 1.$$

The left side is a multiple of $p$, so $p\mid 1$, contradiction.