Japan 2013 Final Round

Let $f(1)=a$. By letting $n=1$, we get from the given equation that $f(m)+f(1)=f(m)+f(2m+1)$ is valid for any integer $m$. Thus we see that $f(d)=a$ must hold for any odd integer $d$.

An arbitrary non-zero integer can be written in the form $2^{k}d$ where $d$ is an odd integer and $k$ is a non-negative integer. By taking $m=d$ and $n=2^k$, we obtain from the given equation that $f(d)+f(2^k)=f(2^{k}d)+f(2^k(d+1)+d)$. Since both $d$ and $2^k(d+1)+d$ are odd integers, we get from the preceding identity that $a+f(2^k)=f(2^{k}d)+a$, from which it follows further that $f(2^{k}d)=f(2^k)$. Consequently, we see that if we can determine $f(2^k)$ for all $k=0,1,2,\cdots$ and $f(0)$, then $f(n)$ is determined completely for all integers $n$.

For $k\ge 2$, substitute $m=2^k,n=2$ into the given equation we obtain $f(2^k)+f(2)=f(2^{k+1})+f(2^k\cdot 3+2)$. Since $2^k\cdot 3+2$ is an odd multiple of 2, we have $f(2^k\cdot 3+2)=f(2)$ by what we observed in the preceding paragraph. Hence, we have $f(2^k)=f(2^{k+1})$ for any $k\ge 2$. If we put $f(2^2)=b$, we then get $f(2^2)=f(2^3)=\cdots =b$. Letting $m=n=2$ in the given equation, we get $2f(2)=f(4)+f(8)=2b$, which says that $f(2)=b$ holds as well. What we have seen so far tells us that $f(n)=b$ holds for all even integer different from 0. Finally, by letting $m=n=-2$ in the given equation, we get $2f(-2)=f(4)+f(0)$, which implies that $f(0)=b$ holds as well. Thus, we can conclude that any function $f$ satisfying the given equation must have the form $$(\ast )f(n)=\{\begin{array}{ll}a& (n\text{ odd})\\ b& (n\text{ even or }0).\end{array}$$ for some pair $a,b$ of real numbers.

Conversely, let us show that for an arbitrary pair $a,b$ of real numbers the function defined by $(\ast )$ satisfies the given equation.

When both $m$ and $n$ are even, $mn$ and $m+n+mn$ are both even, so both sides of the given equation are equal to $2b$.

When both $m$ and $n$ are odd, $mn$ and $m+n+mn$ are both odd, so both sides of the given equation are equal to $2a$.

* If one of $m,n$ is even and the other is odd, $mn$ is even and $m+n+mn$ is odd, so both sides of the given equation are equal to $a+b$.

Thus, the function defined by $(\ast )$ satisfies the given equation for any choice of a pair $a,b$ of real numbers and thus the desired functions $f$ are all functions of this form.