IMO Shortlist 2009 A7
Posted: Sat Mar 29, 2014 9:09 pm
Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all $x$, $y$ the identity \[ f\left(xf(x+y)\right) = f\left(yf(x)\right) +x^2\]
Let \(P(x,y)\) denote the given equation.
1. \(f(0)=0\). If \(f(0)\neq 0\) then \(P\left(0,\frac{y}{f(0)}\right)\Longrightarrow f(y)=f(0)\) so \(f\) is constant. But no constant function works. So \(f(0)=0\). If \(f(n)=0\) for some \(n\) then \(P(n,0)\Longrightarrow 0=f(nf(n))=n^2\) so \(n=0\).
2. \(f\) is Bijective.
(I) Surjectivity: \(P(x,0)\Longrightarrow f\left(xf(x)\right)=x^2\) and \(P(x,-x)\Longrightarrow f\left(-xf(x)\right)=-x^2\). Therefore \(\forall ~ y\in\mathbb{R}~\exists~ x\in\mathbb{R}\mid f(x)=y\), hence \(f\) is surjective.
(II) Injectivity: Suppose \(f(x)=f(y)\) then \(f(xf(x))=f(xf(x+(y-x)))+x^2=x^2\). This leads to \(f((y-x)f(x))=0\). Hence \((y-x)f(x)=0\Longrightarrow x=y\). So \(f\) is injective.
This completes proving bijectivity of \(f\).
3. \(f\) is odd. Notice that \(f(-nf(n))=-n^2=-f(nf(n))\) hence \(f(-x)=-f(x)\) by bijectivity.
4. \(f(xf(y))=xy\). Firstly, \(P(y,x)\Longrightarrow f(xf(y))=f(yf(x+y))-y^2\). Now notice that
\[f(yf(x+y))-y^2=(x+y)^2-y^2-\left((x+y)^2+f(-x(x+y))\right)\] and by the given equation this is \(x^2+2xy-f\left((x+y)f(x)\right)\). But now \[x^2+2xy-f\left((x+y)f(x)\right)=\left((-x)^2+f((x+y)f(-x))\right)+2xy=2xy-f(xf(y)).\] So we finally have \(f(xf(y))=2xy-f(xf(y))\) from where the result follows.
5. \(f(x)\in\{x,-x\}\). Since bijective, take \(n\) such that \(f(n)=1\). Then last result implies \(f(x)=nx\). Setting in the equation \(f(xf(x))=x^2\) gives \(n^2x^2=x^2\Longrightarrow n\in\{1,-1\}\). Hence we have two solutions \(f(x)=x\) and \(f(x)=-x\). Verify that each solution works. \(\square\)
1. \(f(0)=0\). If \(f(0)\neq 0\) then \(P\left(0,\frac{y}{f(0)}\right)\Longrightarrow f(y)=f(0)\) so \(f\) is constant. But no constant function works. So \(f(0)=0\). If \(f(n)=0\) for some \(n\) then \(P(n,0)\Longrightarrow 0=f(nf(n))=n^2\) so \(n=0\).
2. \(f\) is Bijective.
(I) Surjectivity: \(P(x,0)\Longrightarrow f\left(xf(x)\right)=x^2\) and \(P(x,-x)\Longrightarrow f\left(-xf(x)\right)=-x^2\). Therefore \(\forall ~ y\in\mathbb{R}~\exists~ x\in\mathbb{R}\mid f(x)=y\), hence \(f\) is surjective.
(II) Injectivity: Suppose \(f(x)=f(y)\) then \(f(xf(x))=f(xf(x+(y-x)))+x^2=x^2\). This leads to \(f((y-x)f(x))=0\). Hence \((y-x)f(x)=0\Longrightarrow x=y\). So \(f\) is injective.
This completes proving bijectivity of \(f\).
3. \(f\) is odd. Notice that \(f(-nf(n))=-n^2=-f(nf(n))\) hence \(f(-x)=-f(x)\) by bijectivity.
4. \(f(xf(y))=xy\). Firstly, \(P(y,x)\Longrightarrow f(xf(y))=f(yf(x+y))-y^2\). Now notice that
\[f(yf(x+y))-y^2=(x+y)^2-y^2-\left((x+y)^2+f(-x(x+y))\right)\] and by the given equation this is \(x^2+2xy-f\left((x+y)f(x)\right)\). But now \[x^2+2xy-f\left((x+y)f(x)\right)=\left((-x)^2+f((x+y)f(-x))\right)+2xy=2xy-f(xf(y)).\] So we finally have \(f(xf(y))=2xy-f(xf(y))\) from where the result follows.
5. \(f(x)\in\{x,-x\}\). Since bijective, take \(n\) such that \(f(n)=1\). Then last result implies \(f(x)=nx\). Setting in the equation \(f(xf(x))=x^2\) gives \(n^2x^2=x^2\Longrightarrow n\in\{1,-1\}\). Hence we have two solutions \(f(x)=x\) and \(f(x)=-x\). Verify that each solution works. \(\square\)