Here is a nice problem:
Let $f:\mathbb R\to\mathbb R$ be a function, $\mathbb R$ is the set of real numbers, satisfying the following properties: $f(1)$ is an integer and $$xf(y)+yf(x)=f(x+y)^2-f\left(x^2\right)-f\left(y^2\right)\text,$$ for all real numbers $ x , y $.
$f(x)=0$ is a solution, another is $ f(x)=x $. These are all solutions? Better asking: determine all functions that satisfy the above conditions.
I would like to see a complete solution! Thank you!
$x=y$ $$2xf(x)=f(2x)^2-2f(x^2)\\(x=y=0)0=f(0)^2-2f(0)\implies f(0)=0 \lor 2\\xf(0)=f(x)^2-f(x^2)-f(0)\\(x+1)f(0)=f(x)^2-f(x^2)\\2f(0)=f(1)^2-f(1)$$ From this we can conclude $f(0)=0$ since if $f(0)=2$ then by replacing $x=f(1)$,we get quadratic equation which has no integer solutions($x^2-x-4$),so $$0=f(x)^2-f(x^2)\\f(x)^2=f(x^2)\\f(x)=x^c\\2x^{c+1}=(2x)^{2c}-2x^{2c}\\2x^{c+1}=2x^{2c}(2^{2c-1}-1)\\1=x^{c-1}(2^{2c-1}-1)$$ since $2^{2c-1}-1$ is a constant means $x^{c-1}$ has to be a constant,and since the equation is valid for all x in reals,means c=1,so the solutions are $f(x)=x$ and $f(x)=0$
Note the $f(x)=x^c$ is what wolfram gave me,though I guess i could prove it(just would take me more time)