lemma to the primitive divisor theorem.

Question

I want to prove a lemma to the primitive divisor theorem. Where the theorem is given as: For each pair of coprime integers $a$ and $b$, $u_{n} = a^{n} - b^{n}$ has a primitive prime factor for $n > 6$.

and the Lemma to be proved which gives arithmatic properties of $u_{n}$ is : $\gcd(u_{n},u_{m}) = u_{\gcd(n,m)}$

I have attemted it in the following way: let $d=\gcd(n,m) \implies d\lvert n, d\lvert m\\ a^d -b^d \lvert a^n -b^n \space\text{ by definition of } u_{n}\\ = (a^d)^{\frac{n}{d}} -(b^d)^{\frac{n}{d}} = (a^d -b^d)((a^d)^{\frac{n}{d}-1} + \cdots+ (b^d)^{\frac{n}{d}-1})\\ \text{and similarly for m}\\ a^d -b^d \lvert a^m -b^m \space\text{ by definition of } u_{m}\\ = (a^d)^{\frac{m}{d}} -(b^d)^{\frac{m}{d}} = (a^d -b^d)((a^d)^{\frac{m}{d}-1} + \cdots+ (b^d)^{\frac{m}{d}-1})\\ \text{so }\\ u_d\lvert\gcd(u_m,u_n)$.

Could I please get some assistance with proving the reverse implication so that I can show the lemma to be true.

Answer 1

Let gcd$(u_m,u_n)=G$ and let $d=mx+ny$ for integers $x$ and $y$ (Euclid).

Then $a^m\equiv b^m \pmod G$ and so $a^{mx}\equiv b^{mx} \pmod G$. Similarly $a^{ny}\equiv b^{ny} \pmod G$.

Then $a^{mx+ny}\equiv b^{mx+ny} \pmod G$ and $a^d-b^d\equiv 0 \pmod G$ as required.