Prove that if $m$ and $n$ are relatively prime and $(m+n)(m-n)$ is odd, then $(m+n)$ and $(m-n)$ must also be relatively prime.

We can use an indirect argument here.

Assume that $(m+n)$ and $(m-n)$ are not relatively prime. Then they share some common divisor other than one. Let's call this divisor $d$. But if $d \mid m+n$ and $d \mid m-n$, then $d$ must divide both their sum and their difference: $$d \mid (m+n) + (m-n)$$ $$d \mid (m+n) - (m-n)$$ After collecting like terms, we see that $$d \mid 2m\quad \textrm{and} \quad d \mid 2n$$ The first requires that either $d \mid 2$ or $d \mid m$, while the second requires $d \mid 2$ or $d \mid n$.

Notice, however, considering the first case in both possibilities: $d \mid 2$ (with $d \neq 1$), implies $d=2$. Consequently, since $d \mid m+n$ and $d \mid m-n$, both $m+n$, $m-n$, and their product $(m+n)(m-n)$ must be even, which would contradict the given statement that $(m+n)(m-n)$ is odd.

So instead, the second case in both possibilities must be true: $d \mid m$ and $d \mid n$. This, however, means that $m$ and $n$ are not relatively prime. Again, we reach a contradiction with the given information.

So we reject our assumption that $(m+n)$ and $(m-n)$ are not relatively prime. Instead, the opposite must be true: $(m+n)$ and $(m-n)$ are relatively prime.