, 2012

Solution 1. We show $f(x) = x$ is the only solution. It is easy to check that it works.

Put in $y = 0$ to obtain that $f$ is surjective. Let $c$ be a real with $f(c) \neq 0$ and suppose $f(a) = f(b)$. Then
$$a = \frac{f(f(c) + c f(a)) - c}{f(c)} = \frac{f(f(c) + c f(b)) - c}{f(c)} = b,$$
so $f$ is also injective.

Let $z$ be the real number such that $f(z) = 0$. Then $x = z$ gives $f(z f(y)) = z$. Choosing $y$ such that $f(y) = 1$ or $f(y) = 0$, possible by surjectivity, shows that $f(z) = f(0) = z$. By injectivity $z = 0$, so $f(0) = 0$.

Put in $x = y = -1$ to obtain $f(-1) = -1$. Put in $y = -1$ to get $f(f(x) - x) = x - f(x)$. Let $d = f(1) - 1$; with $x = 1$ our previous equation this gives us $f(d) = -d$. Put in $x = d, y = 1$ to get $f(d f(1) - d) = f(d^2) = 0$. By injectivity, $d = 0$ and so $f(1) = 1$.

Put in $y = 0$ to obtain $f(f(x)) = x$. Now put in $x = 1$ and change $y$ to $f(y)$ to get $f(1 + y) = 1 + f(y)$. This implies $f(n) = n$ for all integers $n$. Finally, for arbitrary integers $m, n$ with $n \neq 0$, put in $x = n, y = \frac{m}{n}$ to get $f(n + n f(\frac{m}{n})) = n + m = f(n + m)$. By injectivity, $n f(\frac{m}{n}) = m$, so $f(\frac{m}{n}) = \frac{m}{n}$. Since $\frac{m}{n}$ is an arbitrary rational number, $f(x) = x$ for all $x$ in the domain.

Solution 2. Put in $x = 0$ to obtain $f(f(0)) = f(0)y$ for all rational $y$. Thus $f(0) = 0$ since $f(0)y$ is constant. Now put in $y = 0$ to get $f(f(x)) = x$.

Let $y = f(1)$. We get $f(f(x) + x) = x + f(x)f(1)$. If we replace $x$ with $f(x)$, then by $f(f(x)) = x$ we get $f(f(x) + x) = f(x) + x f(1)$. Thus $x + f(x)f(1) = f(x) + x f(1)$, or $(f(1) - 1)(x - f(x)) = 0$. Putting in $x = 1$ gives us $f(1) = 1$.

Put $x = 1$ into the given equation to get $f(y) + 1 = f(y + 1)$. This implies $f(y + n) = f(y) + n$ for all integers $n$. Since $f(0) = 0$, $f(n) = n$ for all integers $n$. Finally, let $x = \frac{2}{q}$ and $y = q$ where $\frac{2}{q}$ is an arbitrary rational. This gives us $f(f(\frac{2}{q}) + p) = \frac{2}{q} + f(\frac{2}{q})q$. But $f(f(\frac{2}{q}) + p) = f(f(\frac{2}{q})) + p = \frac{2}{q} + p$, so $\frac{2}{q} + p = \frac{2}{q} + f(\frac{2}{q})q$ and $f(\frac{2}{q}) = \frac{2}{q}$. Since $\frac{2}{q}$ was arbitrary, $f(x) = x$ for all $x$, and this clearly satisfies the equation.