37th Iranian Mathematical Olympiad

Assume the contrary, that there exist only finitely many prime numbers that divide an element of $S$. Let's denote them $p_1,p_2,\dots ,p_t$. Take $$x_1,x_2,\dots ,x_{2mt+1}\in S.$$ For every $x\in S$ we can write it as $$x=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\dots p_t^{{\alpha }_t}.$$ The pigeonhole principle implies that there exist two indices $i,j$ such that if $$x_i=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\dots p_t^{{\alpha }_t}\text{and}{x}_j=p_1^{{\beta }_1}{p}_2^{{\beta }_2}\dots p_t^{{\beta }_t},$$ then ${\alpha }_k\equiv {\beta }_k(\mathrm{mod}{2}^m)$ for all $k$. So we have $$x_i+x_j=p_1^{{\theta }_1}{p}_2^{{\theta }_2}\dots p_t^{{\theta }_t}(a^{2^m}+b^{2^m}),$$ where $a\ne b$, $ab\ne 0$ and $\gcd (a,b)=1$. So if $q$ is an odd prime such that $$q\mid a^{2^m}+b^{2^m},$$ then we'll have $q\equiv 1(\mathrm{mod}{2}^m)$ (it is obvious since ${\mathrm{ord}}_q\left(\frac{a}{b}\right)=2^{m+1}$). And if we take $m$ large enough such that $2^m>p_1,p_2,\dots ,p_t$, we'll have $q=1$. Contradiction.

Now it's enough to show that there exists an odd prime divisor of $a^{2^m}+b^{2^m}$ where $a\ne b$, $ab\ne 0$ and $\gcd (a,b)=1$ for every large enough $m$. For this, note that $a^{2^m}+b^{2^m}>2^m$ and $a^{2^m}+b^{2^m}\equiv 1,2(\mathrm{mod}8)$. $■$