European Mathematical Cup

Let $P(x,y)$ be the assertion $f(x^2)+f(2y^2)=(f(x+y)+f(y))(f(x-y)+f(y))$.

$P(0,x)$ gives us $$f(0)+f(2x^2)=2f(x)(f(x)+f(-x))(1)$$ and $P(0,-x)$ gives us $$f(0)+f(2x^2)=2f(-x)(f(x)+f(-x)).(2)$$ By combining (1) and (2) we get $$f(x{)}^2=f(-x{)}^2.(3)$$ $P(0,0)$ gives us $2f(0)=4f(0{)}^2$, thus we have two cases:

Case 1. $f(0)=\frac{1}{2}$.

$P(x,0)$ gives us $$f(x^2)={\left(f(x)+\frac{1}{2}\right)}^2-\frac{1}{2},(4)$$ while $P(-x,0)$, gives us $$f(x^2)={\left(f(-x)+\frac{1}{2}\right)}^2-\frac{1}{2}(5)$$ Combining (4) and (5) and using (3) we get $$f(x)=f(-x)(6)$$ The assertion $P(x^2,x^2)$ can be written as $$f(x^4)+f(2x^4)=f(2x^2)+f(x^2)\left(\frac{1}{2}+f(x^2)\right)(7)$$ For an arbitrary $x\in \mathbb{R}$, let us denote $a=f(x)$. Using (4) we get: $$\begin{array}{rl}f(x^2)& ={\left(a+\frac{1}{2}\right)}^2-\frac{1}{2},\\ f(x^4)& ={\left(f(x^2)+\frac{1}{2}\right)}^2-\frac{1}{2}={\left(a+\frac{1}{2}\right)}^4-\frac{1}{2}.\end{array}$$ Using (1) and (6) we get $$\begin{array}{rl}f(2x^2)& =4f(x^2)-\frac{1}{2}=4a^2-\frac{1}{2},\\ f(2x^4)& =4f(x^2{)}^2-\frac{1}{2}=4{\left({\left(a+\frac{1}{2}\right)}^2-\frac{1}{2}\right)}^2-\frac{1}{2}.\end{array}$$ Plugging the last 4 equations in (7) we get: $$(a+\frac{1}{2}{)}^2+4{\left(a+\frac{1}{2}\right)}^2-\frac{1}{2}-1=\left(4a^2-1+a+\frac{1}{2}\right){\left(a+\frac{1}{2}\right)}^2$$ which is equivalent to $$(a+\frac{1}{2}{)}^2(4a-2)=0.$$ Therefore $a=\pm \frac{1}{2}$ and $f(x)=\pm \frac{1}{2}$. Now if we use (6) in (1) we get $$f(0)+f(2x^2)=4(f(x){)}^2=1$$ so $f(2x^2)=\frac{1}{2}$ for every $x$, now using (6) we conclude $f(x)=\frac{1}{2}$ for all $x$ which is easily checked to be a solution.

Case 2. $f(0)=0$.

We immediately see using $P(x,0)$ that $$f(x^2)=f(x{)}^2.(8)$$ By comparing $P(x,y)$ and $P(x,-y)$ and using (3) we get: $$(f(y)-f(-y))(f(x+y)+f(x-y))=0$$ If there exists $c\in \mathbb{R}$ such that $f(c)\ne f(-c)$ we have for all $x$ $$f(x+c)=-f(x-c)$$ Plugging in $x+c$ in $x$ here gives us: $$f(x+2c)=-f(x).(9)$$ Specially, $f(2c)=0$. Now, $P(2c-y,y)$: $$\begin{array}{rl}f((2c-y{)}^2)+f(2y^2)& =(f(2c)+f(y))(f(2c-y)+f(y)),\\ (-f(-y){)}^2+f(2y^2)& =f(y)f(2c-2y)+f(y{)}^2\\ f(2y^2)& =f(y)f(2c-2y)=-f(y)f(-2y)\end{array}(10)$$ Let $S(x)$ denote the statement $(x\ne 0)\wedge (f(x)=f(-x)\ne 0)$. If there is no $d\in \mathbb{R}$ such that $S(d)$ then $f(x)=-f(-x)$ for all $x\in \mathbb{R}$. $P(0,x)$ gives us $$f(2x^2)2f(x)(f(x)+f(-x))=0,$$ which gives us another solution $f(x)=0$. Now, let us assume that there exists $d\in \mathbb{R}$ such that $S(d)$ holds. Obviously, $S(-d)$ holds, as well. $P(0,d)$ gives us $$f(2d)=4f(d{)}^2$$ and (10) gives us $$\begin{array}{rl}f(2d^2)& =-f(d)f(-2d)\\ f(-2d)& =-4f(d)\\ f(2d)& =-4f(-d)=-4f(d)=f(-2d).\end{array}$$ Therefore, $S(2d)$ also holds. Inductively, we deduce that $S(2^{n}d)$ holds for every $n\in \mathbb{N}$. Also, $f(2^{n}d)=(-4{)}^{n}f(d)$, which means that $f$ is unbounded.

$P(x,c)$, using the fact $f(x^2)=f(x{)}^2$: $$f(x{)}^2+f(2c^2)=f(x+c)f(x-c)+f(c)(f(x+c)+f(x-c))+f(c{)}^2,$$ and since $f(x+c)=-f(x-c)$ and $f(2c^2)=0$ (this follows from $P(0,c)$) we have $$f(x{)}^2+f(x+c{)}^2=f(c{)}^2$$ which implies that $f$ is bounded and that is contradiction. Therefore, there is no $c\in \mathbb{R}$ such that $f(c)=-f(c)$ and therefore $$f(x)=f(-x),\text{ for all }x\in \mathbb{R}.\text{\hspace{1em}(11)}$$ $P(0,x):$ $$f(2x^2)=4f(x{)}^2=4f(x^2).$$ Therefore, using (11): $$f(2x)=4f(x),\text{ for all }x\in \mathbb{R}.\text{\hspace{1em}(12)}$$ $P(x,y)$ can now be written as follows: $$f(x{)}^2+3f(y{)}^2=f(y)(f(x+y)+f(x-y))+f(x+y)f(x-y)$$ and similarly, $P(y,x)$ can be written as: $$f(y{)}^2+3f(x{)}^2=f(x)(f(x+y)+f(x-y))+f(x+y)f(x-y).$$ Subtracting the previous two equalities: $$(f(x)-f(y))(2f(x)+2f(y)-f(x+y)-f(x-y))=0.\text{\hspace{1em}(13)}$$ Assume that for some $x,y\in \mathbb{R}$, $f(x)=f(y)=a$. Let $f(x+y)=b$ and $f(x-y)=c$. Now we have: $$4a^2=bc+ab+ac\text{\hspace{1em}(14)}$$ $P(x+y,x-y):$ $$f(x+y{)}^2+4f(x-y{)}^2=(f(2x)+f(x-y))(f(2y)+f(x-y))$$ i.e.: $$b^2+4c^2=(4a+c{)}^2.\text{\hspace{1em}(15)}$$ If we plug in $x\to x+y$, $y\to x-y$ in (13) we get: $$(f(x+y)-f(x-y))(2f(x+y)+2f(x-y)-f(2x)-f(2y))=0$$ i.e.: $$(b-c)(2b+2c-8a)=0.$$ If $b=c$ (15) gives us: $$5b^2=(4a+b{)}^2$$ $$b^2=4a^2+2ab$$ while (14) gives us: $$4a^2=b^2+2ab.$$ Thus, $ab=0$ and $a=b=c=0$ which implies $2a+2a-b-c=0$. On the other hand, if $b\ne c$ we also have $2a+2a-b-c=0$.