Iranian Mathematical Olympiad

$$a=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}{p}_3^{{\alpha }_3}\dots p_n^{{\alpha }_n}\text{and}b=q_1^{{\beta }_1}{q}_2^{{\beta }_2}{q}_3^{{\beta }_3}\dots q_m^{{\beta }_m}$$ We have $\tau (a)=({\alpha }_1+1)\dots ({\alpha }_n+1)$ and $\tau (b)=({\beta }_1+1)\dots ({\beta }_m+1)$ so it suffices to prove that for each $1\le i\le n$ there is an integer $j$ such that ${\alpha }_i={\beta }_j$. Assume the contrary, suppose that there is some integer $i$ such that ${\alpha }_i\notin \{{\beta }_1,{\beta }_2,\dots ,{\beta }_m\}$. It is well known that the sequence $(a^k-1{)}_{n\in \mathbb{N}}$ has infinitely many prime divisors. Consider a large prime $s$, such that $s>\max ({\alpha }_i)$, $s>\max ({\beta }_j)$ and $$s\mid \frac{p_i^{k{\alpha }_i+1}-1}{p_i-1}.$$ We can also assume that $\gcd (k,s)=1$ by taking $k={\text{ord}}_{p_i}(s)$. Now by Chinese remainder theorem for each $t$, there is an integer $z$ such that $z\equiv 1(\mathrm{mod}s-1)$, $zk{\alpha }_i\equiv -1(\mathrm{mod}{s}^M)$ for some positive integer $M$.

By the lifting the exponent lemma, $s^{M+1}\mid p_i^{zk{\alpha }_i+1}-1$. But because ${\alpha }_i\ne {\beta }_j$ we have $zk{\beta }_j\not\equiv -1(\mathrm{mod}{s}^M)$. Hence if $s\mid q^{k{\beta }_i+1}-1$ we have

${\nu }_s(q_j^{k{\beta }_j+1}-1)={\nu }_s(q_j^{zk{\beta }_j+1}-1)$ and if $s\nmid q^{k{\beta }_i+1}-1$ then by Fermat's little theorem $s\nmid q^{zk{\beta }_i+1}-1$ and we again have $${\nu }_s(q_j^{zk{\beta }_j+1}-1)={\nu }_s(q_j^{zk{\beta }_j+1}-1).$$ In concise $${\nu }_s(\sigma (b^k))={\nu }_s(\sigma (b^k)),{\nu }_s(\sigma (a^{zk}))>M+1,$$ so if we consider $M>{\nu }_s(\sigma (b^k))$ we would obtain ${\nu }_s(\sigma (a^{zk}))>{\nu }_s(\sigma (b^{zk}))$, a contradiction. ■