Question

Find all functions $f:R\to R$ such that $f(x+y)f(x-y)=\left(f\right(x)+f(y){)}^2-4x^{2}f(y)..........(1)$

for all $x,y\in R$, where $R$ denotes the set of all real numbers.

Answer 1

Put $x=y=0$ ; we get $f(0{)}^2=4f(0{)}^2$ and hence $f\left(0\right)=0$.
Put $x=y$; we get $4f(x{)}^2-4x^{2}f(x)=0$ for all $x$.
Hence for each $x$, either $f\left(x\right)=0$ or $f\left(x\right)=x^2$.
Suppose $f\left(x\right)\ne 0$.
Then we can find $x_0\ne 0$ such that $f\left(x_0\right)\ne 0$.
Then $f\left(x_0\right)=x_0^2\ne 0$.
Assume that there exists some $y_0\ne 0$ such that $f\left(y_0\right)=0$.
Then $f(x_0+y_0)f(x_0-y_0)=f(x_0{)}^2$.
Now $f(x_0+y_0)f(x_0-y_0)=0$ or $f(x_0+y_0)f(x_0-y_0)=(x_0+y_0{)}^2(x_0-y_0{)}^2$.
If $f(x_0+y_0)f(x_0-y_0)=0$, then $f\left(x_0\right)=0$, a contradiction.
Hence it must be the later so that $(x_0^2-y_0^2{)}^2=x_0^4$.
This reduces to $y_0^2(y_0^2-2x_0^2)=0$. Since $y_0\ne 0$, we get $y_0=\pm \sqrt{2}{x}_0$.
Suppose $y_0=\sqrt{2}{x}_0$.
Put $x=\sqrt{2}{x}_{0}andy=x_0$ in $\left(1\right)$;
we get $f\left(\begin{array}{c}(\sqrt{2}+1)x_0\end{array}\right)f\left(\begin{array}{c}(\sqrt{2}-1)x_0\end{array}\right)={\left(\begin{array}{c}f\left(\sqrt{2}{x}_0\right)+f\left(x_0\right)\end{array}\right)}^2-4\left(2x_0^2\right)f\left(x_0\right)$.
But $f\left(\sqrt{2}{x}_0\right)=f\left(y_0\right)=0$.
Thus we get $f\left(\begin{array}{c}(\sqrt{2}+1)x_0\end{array}\right)f\left(\begin{array}{c}(\sqrt{2}-1)x_0\end{array}\right)=f(x_0{)}^2-8x_0^{2}f(x_0)$
$=x_0^4-8x_0^4=-7x_0^4$.
Now if LHS is equal to $0$, we get $x_0=0$, a contradiction.
Otherwise LHS is equal to $(\sqrt{2}+1{)}^2(\sqrt{2}-1{)}^2{x}_0^4$ which reduces to $x_0^4$.
$\therefore x_0^4=-7x_0^4$ and this forces again $x_0=0$.
Hence there is no $y\ne 0$ such that $f\left(y\right)=0$.
$\therefore f\left(x\right)=x^2$ for all $x$.
Thus there are two solutions : $f\left(x\right)=0$ for all $x$ or $f\left(x\right)=x^2$, for all $x$.