Coprime Divisors of Square Number are Square

Theorem

Let $r$ be a square number.

Let $r = s t$ where $s$ and $t$ are coprime.

Then both $s$ and $t$ are square.

Proof

Let $p$ be a prime factor of $s$.

Then for some $n \in \Z_{>0}$:

$p^{2 n} \divides s t$

where $\divides$ denotes the divisibility relation.

But we have that $s \perp t$ and so $p \nmid t$.

Thus $p^{2 n} \divides s$.

This holds for all prime factors of $s$.

Thus $s$ is the product of squares of primes.

Thus $s$ is square.

The same argument holds for $t$.

Hence the result.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Some Properties of $\Z$: Exercise $2.13$