Iranian Mathematical Olympiad

Let $L(z)$ be the smallest prime divisor of $z$.

Lemma. If $a$, $b$, $c\in \mathbb{N}$ such that $a\ne b$ we have $$L\left(\frac{a^c-b^c}{a-b}\right)\ge L(abc)$$ Proof. Assume the contrary, then there should be an integer $p$ such that $\gcd (p,a)=\gcd (p,b)=1$ and $p\mid \frac{a^c-b^c}{a-b}$. Now consider two cases, if $a\equiv b(\mathrm{mod}p)$ then by lifting the exponent lemma ${\nu }_p(a^c-b^c)={\nu }_p(a-b)$ which is a contradiction. If $a\not\equiv b(\mathrm{mod}p)$ then $or{d}_p(a{b}^{-1})\ne 1$ ($b^{-1}$ is multiplicative inverse of $b$ modulo $p$). But we know that $or{d}_p(a{b}^{-1})\mid c$ and we have $p>{\text{ord}}_p(a{b}^{-1})>L(c)$ which is a contradiction.

Consider a set $$A_k=\{x\in \mathbb{N}\mid L(x)\ge k\}$$ now if $a$, $b$, $c\in A_k$ and $d\mid \frac{a^c-b^c}{a-b}$ then by lemma we have $L(d)<L(abc)<n$ and hence $d\in A_k$. Now it is enough to take $k>n$. ■