Proof that $a^n-b^n$ is non prime if n is non prime

Question

Hello guys I want to know if my proof is valid, and whether it is coherent and where I can improve it.

Prove that for all natural numbers, $a,b, n$, where $a>b$, and $n>1$ is a not a prime, $a^n-b^n$ is not a prime number.

My attempt:

Since $n$ is a non-prime greater than $1$, we have $n=xy$ where $1<x,y<n. $We know that $$a^n-b^n=a^{xy}-b^{xy}=(a^x-b^x)(a^{x(y-1)}+b^xa^{x(y-2)}+...+b^{x(y-2)}a^x+b^{x(y-1)})$$ Hence $(a^x-b^x)$ will always divide $a^n-b^n$. We merely need to prove that $a^n-b^n>a^x-b^x>1$. That $a^n-b^n>a^x-b^x$ is obvious since $a^{x(y-1)}+b^xa^{x(y-2)}+...+b^{x(y-2)}a^x+b^{x(y-1)}$ is a positive integer greater than $1$ ($y>1$). Furthermore, we can write $a^x-b^x$ as $(a-b)(a^{x-1}+ba^{x-2}+...+ab^{x-2}+b^{x-1})$. Since $x>1$, $a^{x-1}+ba^{x-2}+...+ab^{x-2}+b^{x-1}$ is a positive integer greater than $1$. Therefore, we have $a^x-b^x=(a-b)(a^{x-1}+ba^{x-2}+...+ab^{x-2}+b^{x-1})>a-b \geq1$. Hence $a^x-b^x>1$. This concludes the proof.

Answer 1

I don't see anything wrong with your proof. Perhaps I could mention some shortcuts or alternatives that you might find useful.

Let $d$ be a non trivial divisor of $n$, then we may write $n = dk$ for some integer $k > 1$. From $a^d \equiv b^d \pmod{a^d - b^d}$, we obtain $a^n \equiv b^n \pmod{a^d - b^d}$ by raising each side to the $k$-th power. The latter congruence relation is of course equivalent to saying that $a^d - b^d$ is a divisor of $a^n - b^n$.

$x \mapsto x^n - x^d = x^d(x^{n-d} - 1): [1, \infty) \to \mathbb{R}$ is a strictly increasing function whenever $n > d > 0$, since it is the product of strictly increasing functions mapping only to non-negative values. Here $n$ and $d$ need not be integers. Hence both $a^n - b^n > a^d - b^d$ and $a^d - b^d > a - b$ follow from rewriting them as $a^n - a^d > b^n - b^d$ and $a^d - a > b^d - b$.