The equation of the term cruzado

1. The term 2xy aparece to the expandir (x+and)^2.
2. Try subtracting x^2 from f(x) and define a new function.
3. A function aditiva and continua in R has that ser lineal.

Answer: all and only functions $$
f(x)=x^2+cx,\qquad c\in\mathbb{R}.

$$ The key is to recognize the crossover term: $$

(x+y)^2=x^2+2xy+y^2.

$$ That is why we define $$

g(x)=f(x)-x^2.

$$ So: $$

g(x+y)=f(x+y)-(x+y)^2.

$$ Using the equation from the statement: $$

g(x+y)=f(x)+f(y)+2xy-(x^2+2xy+y^2).

$$ By simplifying: $$

g(x+y)=(f(x)-x^2)+(f(y)-y^2)=g(x)+g(y).

$$ So $g$ is additive. Since $f$ is continuous and $x^2$ is also continuous, $g$ is continuous. Every continuous additive function in $\mathbb{R}$ is linear, therefore there exists a real constant $c$ such that $$

g(x)=cx.

$$ Therefore: $$

f(x)=x^2+cx.

$$ We now verify that all these functions work. If $f(x)=x^2+cx$, then $$

f(x+y)=(x+y)^2+c(x+y),

$$ and $$

f(x)+f(y)+2xy=(x^2+cx)+(y^2+cy)+2xy.

$$ Both expressions are the same: $$

x^2+2xy+y^2+cx+cy.
$$ Therefore, all and only the functions $f(x)=x^2+cx$, with $c\in\mathbb{R}$, satisfy the equation.