Saudi Arabia Mathematical Competitions 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 $f(c)\ne 0$ and suppose $f(a)=f(b)$. Then $$a=\frac{f(f(c)+cf(a))-c}{f(c)}=\frac{f(f(c)+cf(b))-c}{f(c)}=b,$$ so $f$ is also injective.

Let $z$ be the real number such that $f(z)=0$. Then $x=z$ gives $f(zf(y))=z$. 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 $f(df(1)-d)=f(d^2)=0$. 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 $n\ne 0$, put in $x=n,y=\frac{m}{n}$ to get $f(n+nf(\frac{m}{n}))=n+m=f(n+m)$. By injectivity, $nf(\frac{m}{n})=m$, 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.

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Alternative solution.

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 $f(f(x)+x)=f(x)+xf(1)$. Thus $x+f(x)f(1)=f(x)+xf(1)$, 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.