I have to show that the fraction
$$\frac{u^{pq}-1}{(u^p-1)(u^q-1)}$$ is of the from $$\frac{f_{p,q}(u)}{u-1}.$$
I'm searching for a short proof. If we factorize, the p'th and q'th roots of unity cancel, except for 1, which canceles only once. Is there a nicer way to explain this?
Consider the expressions as polynomials in $u$.
Then $u^{pq}-1$ is a multiple of both $u^p-1$ and $u^q-1$ and so is a multiple of their lcm: $$ u^{pq}-1 =f(u)\operatorname{lcm}(u^p-1,u^q-1) $$ Since $\gcd(u^p-1,u^q-1)=u-1$, we get $$ \operatorname{lcm}(u^p-1,u^q-1)=\frac{(u^p-1)(u^q-1)}{u-1} $$ and so $$ u^{pq}-1 =f(u)\operatorname{lcm}(u^p-1,u^q-1) =\frac{f(u)(u^p-1)(u^q-1)}{u-1} $$
This argument only needs that $\gcd(p,q)=1$, not that $p,q$ are primes.
More generally, if $m=\operatorname{lcm}(p,q)$ and $d=\gcd(p,q)$, then $$ \frac{u^m-1}{(u^p-1)(u^q-1)} = \frac{f(u)}{u^d-1} $$ since $\gcd(u^p-1,u^q-1)=u^d-1$.
