Coprime Divisors of Square Number are Square
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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 $2.13$