# Show that $a^n-b^n$ has a prime factor which does not divide $a-b$ for all $n>1$ .

I was asked to prove the following using the lifting the exponent lemma.

Show that $$a^n-b^n$$ has a prime factor which does not divide $$a-b$$ for all $$n>1$$ .

Using the first lemma, what I got was this:
if $$p$$ is any prime greater than $$2$$, then we have

$$V_p(a^n-b^n)= V_p(a-b) + V_p(n)$$

where $$V_p(x)$$ is the highest power of $$p$$ that divides $$x$$ and $$p|a-b$$ but does not divide a or b.
I don't know how to approach this and would welcome some hints.

• See this related question. – Dietrich Burde Mar 7 at 19:11
• @DietrichBurde thanks. i am trying what i can but the post helped. – user0111 Mar 7 at 19:14
• Following the pattern; $a^2-b^2=(a-b)(a+b)\text{, } a^3-b^3=(a-b)(a^2+ab+b^2)\text{, } a^4-b^4=(a - b)(a + b)(a^2 + b^2)$, etc. we need to show that some sum of any combination of powers of $a \text{ and } b$ is prime. – poetasis Mar 7 at 19:26

## 1 Answer

Hint:

$$\frac {a^n-b^n}{a-b}=\sum_{k=1}^n a^{k-1}b^{n-k}>n\ge\prod_{p|(a-b)} p^{V_p(n)}.$$

$$\implies a^n-b^n>\prod_{p|(a-b)} p^{V_p(a^n-b^n)}$$