Determine all functions $\Bbb R _{\ge 0} \rightarrow \Bbb R_{\ge 0}$ such that $f(x)+f(y)+2xy=f(x+y) $

Question

Determine all functions $\Bbb R _{\ge 0} \rightarrow \Bbb R_{\ge 0}$ such that $f(x)+f(y)+2xy=f(x+y) $

The only nice progress I could make was $f(x)=x^2$ is a solution and $f(0)=0 $

when we take $f(x)=x^2$ , we get $f(x)+f(y)+2xy=x^2+y^2+2xy =f(x+y) $

and for $(x,y)=(0,0)$, we get $2f(0)=f(0)$ $\implies f(0)=0$

I tried making substitution , but couldn't make a nice one , also I think Cauchy might help since we have the FE similar to $f(x)+f(y)=f(x+y) $

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If possible, can someone post hints rather than solution, it helps me a lot.

Thanks in advance!

Answer 1

Hint: If $f$ and $g$ are two solutions, what can you say about $f-g$?

Answer 2

Hint. As you noted out that $f(x)=x^2$ satisfies the conditions, its natural to define the function $g(x):=f(x)-x^2$ or simply $f(x)=g(x)+x^2$.

Is this an additive Cauchy in $g$?

Can you show $g$ is bounded in a non-trivial interval?

Can you use the fact that $f$ is bounded?

Note that $g(x)> -10$ in the interval $[0,3]$ and so $g$ is linear but as $f(0)=0$ we get, $f(x)=kx+x^2$ for some $k$.