Find all continuous functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x+y) = f(x)+f(y)+2xy$.
We have that $f(0) = 0$ and $f(x+1) = f(x)+f(1)+2x$ and thus $f(x+1) - f(x) = f(1)+2x$. Then we see that $f(0) = f(x)+f(-x)-2x^2 \implies f(x)+f(-x) = 2x^2$. Then I get stuck since I don't see what I can do next.
Define $g(x)=f(x)-x^2$. Then by the above equation, $$g(x+y)+(x+y)^2=g(x)+x^2+g(y)+y^2+2xy \implies g(x+y)=g(x)+g(y)$$
To solve the latter, note that $g(0)=0$, and $g(cx)=cg(x)$ for all integers $c$. Let $\frac{p}{q}$ be any rational number, then replace $x$ by $\frac{x}{c}$ in the equation $g(cx)=cg(x)$ to get: $$g(x)=cg\biggl(\frac{x}{c}\biggr).$$
Now, $$ g\biggl(\frac{px}{q}\biggr) = pg\biggl(\frac{x}{q}\biggr) = \frac{p}{q}g(x) $$ Given that $g$ is continuous, it follows that $g(cx)=cg(x)$ for all real numbers (use the sequential definition of continuity). Thus, suppose $g(1)=k$ for some constant $k$, then for every $x$, $g(x)=kx$. It follows that $g(x)$ is always of this form, and hence $f(x)$ is of the form $kx+x^2$ for some constant $k$. Let's confirm $kx+x^2$ for fun: $$ f(x+y) = k(x+y)+(x+y)^2 = kx + x^2 + ky + y^2 + 2xy = f(x)+f(y)+2xy $$ Hence the answer.