Korean Mathematical Olympiad Final Round

Let $f(x)$ be such a polynomial. Define $$g(x):=\frac{f(x)-f(-x)}{2},h(x):=\frac{f(x)+f(-x)}{2}.$$ Then $g(x)$ is a polynomial consisting of odd degree monomials of $f(x)$, $h(x)$ is a polynomial consisting of even degree monomials of $f(x)$ (including the constant term), and $f(x)=g(x)+h(x)$. For any positive integers $a$ and $b$, since $$a+b\mid a^{2k+1}+b^{2k+1}$$ for any nonnegative integer $k$, we have $$a+b\mid f(a)+f(b)⟺a+b\mid h(a)+h(b).$$ Hence we may assume that $f(x)=h(x)$. Suppose that $f(x)=c{x}^{2n}$ for a nonnegative integer $n$ and $a+b\mid f(a)+f(b)$ for some positive integers $a,b$. Note that $$a^{2n}+b^{2n}=a(a^{2n-1}+b^{2n-1})-b^{2n}(a+b)+2b^{2n}.$$ Hence $$a+b\mid f(a)+f(b)⟺a+b\mid 2f(b)=2c{b}^{2n}.$$ Furthermore if $\gcd (a,b)=1$, $\gcd (a+b,b^{2n})=1$. Therefore $a+b\mid 2c$, which implies that there exist only finitely many such positive integer pairs $(a,b)$. Now assume that $f(x)=a_{2n}{x}^{2n}+t(x)$, where $a_{2n}\ne 0$ and $t(x)$ is a nonzero polynomial consisting of monomials of even degrees greater than $2n$. Let $b$ be any sufficiently large positive integer such that $\gcd (a_{2n},b)=1$ and $a$ be any positive integer such that $$|a+b|=a_{2n}+\frac{t(b)}{b^{2n}}.$$ Note that $\gcd (a,b)=1$ and there exist infinitely many such integer pairs $(a,b)$. Since $2f(b)=2b^{2n}|a+b|$, we have $a+b\mid f(a)+f(b)$. Therefore $f(x)$ satisfies the above property if and only if $f(x)$ contains no even degree monomials (that is, it consists of odd degree monomials only) or contains at least two even degree monomials. $\square$