Suppose m and n are relatively prime integers. Prove that $m^2$ and $n^2$ are also relatively prime.

Question

I know that $am+bn=1$ for some integers $a$ and $b$. I know that I will have to use this to show that $cm^2 + dn^2 = 1$. I tried squaring both sides but I am left with the term $2ambn$ which doesn't factor with the $n^2$ and $m^2$. So I'm not sure how to do this.

Answer 1

Alternative approach.

Use the fundamental theorem of arithmetic.

If $m$ and $n$ are relatively prime, then, when you examine each of their prime factorizations, the two factorizations must have no primes in common.

Further, by definition, the only difference (for example) between the prime factorizations of $m$ and $m^2$ is that with $m^2$, each of the exponents in the prime factorization of $m$ is multiplied by $2$. That is, both prime factorizations (must) consist of the exact same list of primes.

Similar considerations apply to the prime factorizations of $n$ and $n^2$. Therefore, since the prime factorizations of $m$ and $n$ have no primes in common, the same must be true for the prime factorizations of $m^2$ and $n^2$.

Therefore, since $m,n$ are relatively prime, $m^2$, and $n^2$ must also be relatively prime.