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Here is a nice problem: Let $f:R\rightarrow R$ be a function, R is the set of real numbers, satisfying the following properties: $ f(1)$ is an integer and $xf(y)+yf(x)=(f(x+y))^2-f(x^2)-f(y^2)$, for all x, y real numbers. $f(x)=0$ is a solution, another is $ f(x)=x $. These are all solutions? better asking: determine all functions that satisfy the 2 conditions. I would like to see a complete solution! Thank you!

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$x=y$ $$2xf(x)=f(2x)^2-2f(x^2)\\(x=y=0)0=f(0)^2-2f(0)\implies f(0)=0 \lor 2\\xf(0)=f(x)^2-f(x^2)-f(0)\\(x+1)f(0)=f(x)^2-f(x^2)\\2f(0)=f(1)^2-f(1)$$ 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)

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  • $\begingroup$ I suspect you need some additional assumptions on continuity to be able to conclude that there are no other "strange" solutions to $f(x^2) = f(x)^2$. EDIT: woops I just realized this was 3 years ago. $\endgroup$ – Tob Ernack Jan 2 '18 at 20:45

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