elementary number theory - Prove that for every pair of coprime positive integers $p,q$, $(x^{pq}-1)(x-1)$ is divisible by $(x^p-1)(x^q-1)$.

Prove that for every pair of coprime positive integers $p,q$ the expression $(x^{pq}-1)(x-1)$ is divisible by $(x^p-1)(x^q-1)$.

My attempts:

$x^{pq}-1=(x^p)^q-1$ which is divisible by $x^p-1$

again, $x^{pq}-1=(x^q)^p-1$ which is divisible by $x^q-1$.

But how to prove that it is divisible by their products? Just now an idea struck me. Can I consider gcd of $x^p-1$ and $x^q-1$ as $x-1$ ?. If yes, then we are done!

Answer

To answer your question, if the exponents are not coprime, you can try to prove the general formula: $$\gcd(x^m-1,x^n-1)=x^{\gcd(m,n)}-1,$$ from which you can deduce that $\;(x^{mn}-1)(x^{\gcd(m,n)}-1)$ is divisible by $\;(x^m-1)(x^n-1)$.