BMO 2019 Shortlist

Obviously, the identical function $f(p)=p$ for all $p\in \mathbb{P}$ is a solution. We will show that this is the only one.

First we will show that $f(2)=2$. Taking $q=2$ and $p$ any odd prime number, we have $$f(p{)}^{f(2)}+2^p=f(2{)}^{f(p)}+p^2.$$ Assume that $f(2)\ne 2$. It follows that $f(2)$ is odd and so $f(p)=2$ for any odd prime number $p$.

Taking any two different odd prime numbers $p,q$ we have $$2^2+q^p=2^2+p^q\Rightarrow p^q=q^p\Rightarrow p=q,$$ contradiction. Hence, $f(2)=2$.

So for any odd prime number $p$ we have $$f(p{)}^2+2^p=2^{f(p)}+p^2.$$ Copy this relation as $$2^p-p^2=2^{f(p)}-f(p{)}^2.(1)$$ Let $T$ be the set of all positive integers greater than 2, i.e. $T=\{3,4,5,\dots \}$. The function $g:T\to \mathbb{Z}$, $g(n)=2^n-n^2$, is strictly increasing, i.e. $$g(n+1)-g(n)=2^n-2n-1>0(2)$$ for all $n\in T$. We show this by induction. Indeed, for $n=3$ it is true, $2^3-2\cdot 3-1>0$. Assume that $2^k-2k-1>0$. It follows that for $n=k+1$ we have $$2^{k+1}-2(k+1)-1=(2^k-2k-1)+(2^k-2)>0$$ for any $k\ge 3$. Therefore, (2) is true for all $n\in T$.

As consequence, (1) holds if and only if $f(p)=p$ for all odd prime numbers $p$, as well as for $p=2$.

Therefore, the only function that satisfies the given relation is $f(p)=p$, for all $p\in \mathbb{P}$.

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