China Mathematical Olympiad

If $mn=1$, then the claim is valid. For $mn\ge 2$, since $$n^a(a{m}^a+b{n}^b)=(a+b)n^{a+b}+a((mn{)}^a-n^{a+b}),$$ it is sufficient to prove the existence of infinitely many coprime number pairs $a$, $b$, such that $$a+b\mid (mn{)}^a-n^{a+b},(a+b,n)=1.$$ Let $p=a+b$, we only need to prove that there are infinitely many prime numbers $p$ and positive integer $1\le a\le p-1$ such that $$p\mid (mn{)}^a-n^p.$$ By Fermat's theorem, i.e., when $a_1\equiv a_2(\mathrm{mod}p-1)$, $a_1\ge 1$, $a_2\ge 1$, $(mn{)}^{a_1}\equiv (mn{)}^{a_2}(\mathrm{mod}p)$. So we only need to prove that there are infinitely many prime numbers $p$ and positive integer $a$ such that $$p\mid (mn{)}^a-n.\stackrel{◯}{1}$$ If there are only finitely many such primes, say $p_1,p_2,\dots ,p_r$ (as $mn\ge 2$, the existence of such primes is obvious). Suppose that $$(mn{)}^2-n=p_1^{a_1}{p}_2^{a_2}\dots p_r^{a_r},{\alpha }_i\text{’s are non-negative integers}(1\le i\le r).\stackrel{◯}{2}$$ Let $a=p_1^{a_1}{p}_2^{a_2}\dots p_r^{a_r}(p_1-1)\dots (p_r-1)+2$, and suppose that $$(mn{)}^a-n=p_1^{{\beta }_1}{p}_2^{{\beta }_2}\dots p_r^{{\beta }_r},{\beta }_i\text{’s are non-negative integers}(1\le i\le r).\stackrel{◯}{3}$$ If $p_i\nmid n$, then, by ③ and $a\ge 2$, we know that $p_i^{{\beta }_i}\nmid n$, hence $p_i^{{\beta }_i}\mid (mn{)}^2-n$, and by ② we have ${\beta }_i\le {\alpha }_i$. If $p_i\nmid n$, then $p_i\nmid m$, so $(p_i^{a_i+1},mn)=1$. By Euler's Theorem (as $\phi (p_i^{a_i+1})=p_i^{a_i}(p_i-1)$ is a factor of $a-2$) $$(mn{)}^a-n\equiv (mn{)}^2-n(\mathrm{mod}{p}_i^{a_i+1}).$$ Because $p_i^{a_i+1}\nmid (mn{)}^2-n$, the congruence relation above implies that $p_i^{a_i+1}\nmid (mn{)}^a-n$. So ${\beta }_i\le {\alpha }_i$. Hence, $$(mn{)}^a-n=p_1^{{\beta }_1}{p}_2^{{\beta }_2}\cdots p_r^{{\beta }_r}\le p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_r^{{\alpha }_r}=(mn{)}^a-n,$$ which is in contradiction with $a>2$. So there are infinitely many primes $p$ and positive integers $a$ such that $p\mid (mn{)}^a-n$.