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.

  • 1
    $\begingroup$ See this related question. $\endgroup$ – Dietrich Burde Mar 7 '19 at 19:11
  • $\begingroup$ @DietrichBurde thanks. i am trying what i can but the post helped. $\endgroup$ – user0111 Mar 7 '19 at 19:14
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    $\begingroup$ 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. $\endgroup$ – poetasis Mar 7 '19 at 19:26


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

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


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