13. Let $f$ be a function from the set of real numbers $\mathbf{R}$ to the set of real numbers $\mathbf{R}$, satisfying
$$
f(x+y)=f(x)+f(y)+2 x y .
$$

If the graph of $f(x)$ has a line of symmetry $x=k$ and in the interval
Three, 13. Let $x=0$, we get $f(0)=0$. From the given condition, we have
$$
0=f(0)=f(x-x)=f(x)+f(-x)-2 x^{2} \text {, }
$$

which means $f(x)+f(-x)=2 x^{2}$.
$$
\begin{array}{l}
\text { Also, } f(k+x)=f(k)+f(x)+2 k x, \\
f(k-x)=f(k)+f(-x)-2 k x,
\end{array}
$$

Subtracting the above two equations, and noting that $f(k+x)=f(k-x)$, we get
$$
f(x)-f(-x)=-4 k x \text {. }
$$

From equations (1) and (2), we have $f(x)=x^{2}-2 k x$.
This is a parabola opening upwards, and the interval of monotonic decrease is $(-\infty, k]$. Therefore, $k \geqslant 3$.