11-th Czech-Slovak-Polish Match, 2011

Let $k$ be an arbitrary integer. Note that $$(9a^2{k}^3{)}^2+3=3(27a^4{k}^6+1)$$ and $$(9a^3{k}^4{)}^3-a=a(3^6{a}^8{k}^{12}-1)=a(27a^4{k}^6-1)(27a^4{k}^6+1)$$ It follows that for every $k\in \mathbb{Z}$ the number $27a^4{k}^6+1$ is a common divisor of the numbers $n^2+3$ and $m^3-a$ with $n=9a^2{k}^3$ and $m=9a^3{k}^4$. So it is enough to prove that there are infinitely many primes $p$ such that $p\mid 27a^4{k}^6+1$ for some integer $k$. Suppose that there are only finitely many such primes and these are $p_1,p_2,\dots ,p_r$. If we take $k=p_1{p}_2\dots p_r+1$, then it is clear that the number $27a^4{k}^6+1$ is not divisible by any $p_i$ for $1\le i\le r$ and that it is also greater than 1. It follows that it has a prime divisor $p$, which is different from every $p_i$ for $1\le i\le r$. We have obtained a contradiction, which finishes the proof. $\square$