Iranian Mathematical Olympiad

Let $P$ be the set of all prime divisors of members of $S^{\prime }$. Suppose, to the contrary, that $P$ is finite and $P=\{p_1,p_2,\dots ,p_n\}$. Define $$N=4\prod _{i=1}^n{p}_i(p_i-1).$$ $S$ is infinite, hence there exists an infinite subset $S_1$ of $S$ such that every two members of $S_1$ are congruent to each other modulo $N$. Let $x,y>1$ be members of $S_1$ such that $y$ is large enough to satisfy $\frac{y}{{\log }_{2}y}>x$. It's easy to see that $$x<\frac{y}{{\log }_{2}y}\le \frac{y}{{\log }_{x}y}⟹y>x{\log }_{x}y⟹x^y>y^x$$ Since $x^y+y^x\in S^{\prime }$, it can be written in the form $x^y+y^x=q_1^{{\alpha }_1}{q}_2^{{\alpha }_2}\dots q_k^{{\alpha }_k}$ where $q_1,q_2,\dots ,q_k\in \mathbb{P}$ and ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_k\in {\mathbb{Z}}^{+}$. For every $1\le i\le k$ we have $$\begin{array}{rl}{q}_i\mid N,x\equiv y(\mathrm{mod}N)& ⟹x\equiv y(\mathrm{mod}{q}_i)\\ & ⟹x^y\equiv y^y(\mathrm{mod}{q}_i)\end{array}(1)$$ $$\begin{array}{rl}{q}_i-1\mid N,x\equiv y(\mathrm{mod}N)& ⟹x\equiv y(\mathrm{mod}{q}_i-1)\\ & ⟹y^x\equiv y^x(\mathrm{mod}{q}_i)\end{array}(2)$$ According to (1), (2) $$\begin{array}{l}{x}^y\equiv y^x(\mathrm{mod}{q}_i)\\ x^y+y^x\equiv 0(\mathrm{mod}{q}_i)\end{array}\}⟹2x^y\equiv 0(\mathrm{mod}{q}_i)$$ So for every odd $q_i$ we have $q_i\mid x$. For $q_i=2$ if at least one of $x$ and $y$ is even, then the other one is even too; And if they both are odd we have $$\begin{gathered} \begin{array}{l}2\mid x-y⟹y^y\equiv y^x(\mathrm{mod}4)\\ y\equiv x(\mathrm{mod}N)\\ 4\mid N\end{array}\}⟹x^y\equiv y^y\equiv y^x(\mathrm{mod}4) \\ ⟹x^y+y^x\equiv 2x^y\equiv 2(\mathrm{mod}4)⟹{\alpha }_i=1(3) \end{gathered}$$ Now let $$x=q_1^{{\beta }_1}{q}_2^{{\beta }_2}\dots q_k^{{\beta }_k}{x}^{\prime },y=q_1^{{\gamma }_1}{q}_2^{{\gamma }_2}\dots q_k^{{\gamma }_k}{y}^{\prime }$$ so that $${\beta }_i,{\gamma }_i\in {\mathbb{Z}}^{+}\cup \{0\},\text{gcd}(x^{\prime },q_1,q_2,\dots ,q_k)=\text{gcd}(y^{\prime },q_1,q_2,\dots ,q_k)=1$$ For every $1\le i\le k$ if $q_i\ne 2$, it was proven that ${\beta }_i,{\gamma }_i>0$. Furthermore, if $q_i=2$ and one of $x,y$ is even, then ${\beta }_i,{\gamma }_i>0$. So if these conditions hold, we have ${\beta }_{i}y\ge y$. From the definition of $y$ we have $${\beta }_{i}y\ge y>x{\log }_{2}y\ge x{\log }_{q_i}y\ge x{\gamma }_i.$$ Also if $x,y$ are odd and $q_i=2$, the inequality $0={\beta }_{i}y\ge {\gamma }_{i}x=0$ still holds. Now we have $$\begin{array}{rl}{q}_1^{{\alpha }_1}{q}_2^{{\alpha }_2}\dots q_k^{{\alpha }_k}& =x^y+y^x\\ & =x^{\prime y}\prod _{1\le i\le k}{q}_i^{y{\beta }_i}+y^{\prime x}\prod _{1\le i\le k}{q}_i^{x{\gamma }_i}\\ & =\prod _{1\le i\le k}{q}_i^{x{\gamma }_i}\left(y^{\prime x}+x^{\prime y}\prod _{1\le i\le k}{q}_i^{y{\beta }_i-x{\gamma }_i}\right)\end{array}$$ If $q_i=2$ and $x,y$ are odd, from (3) we have ${\alpha }_i=1$. Otherwise we have $$y{\beta }_i-x{\gamma }_i>0⟹\left(y^{\prime x}+x^{\prime y}\prod _{1\le i\le k}{q}_i^{y{\beta }_i-x{\gamma }_i},q_i\right)=(y^{\prime x},q_i)=1$$ But we also have y'^x + x'^y \prod_{1\le i\le k} q_i^{y\beta_i - x\gamma_i} \quad \left| \quad x^y + y^x = q_1^{\alpha_1} q_2^{\alpha_2} \dots q_k^{\alpha_k} So $$\begin{array}{rl}& y^{\prime x}+x^{\prime y}\prod _{1\le i\le k}{q}_i^{y{\beta }_i-x{\gamma }_i}|2\\ ⟹& y^{\prime x}+x^{\prime y}\prod _{1\le i\le k}{q}_i^{y{\beta }_i-x{\gamma }_i}\le 2\\ ⟹& y^{\prime x}\le 1⟹y^{\prime x}=1⟹y^x=\prod _{1\le i\le k}{q}_i^{x{\gamma }_i}\\ ⟹& x^y+y^x=q_1^{{\alpha }_1}{q}_2^{{\alpha }_2}\dots q_k^{{\alpha }_k}=y^x\left(y^{\prime x}+x^{\prime y}\prod _{1\le i\le k}{q}_i^{y{\beta }_i-x{\gamma }_i}\right)\le 2y^x\\ ⟹& x^y\le y^x\end{array}$$ Which contradicts the way $y$ was chosen, and completes our proof. ■