13.230. There are three positive two-digit numbers with the following property: each number is equal to the incomplete square of the sum of its digits. It is required to find two of them, knowing that the second number is 50 units greater than the first.
## Solution.

Since one number is 50 greater than the other, the number of units in both numbers is the same. Let $10 x_{1}+y, 10 x_{2}+y$ be the desired numbers.

By the condition $10 x_{1}+y-\left(10 x_{2}+y\right)=50$, from which $x_{1}-x_{2}=5$ (1), we also have the system

$$
\left\{\begin{array}{l}
10 x_{1}+y=x_{1}^{2}+x_{1} y+y^{2} \\
10 x_{2}+y=x_{2}^{2}+x_{2} y+y^{2}, \text { from which }
\end{array}\right.
$$

$10\left(x_{1}-x_{2}\right)=x_{1}^{2}-x_{2}^{2}+\left(x_{1}-x_{2}\right) y$ or $50=\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)+5 y$, from which $10=x_{1}+x_{2}+y(2)$.

From (1) we have $x_{1}=5+x_{2}$. Substituting the value of $x_{1}$ into (2):

$$
10=5+2 x_{2}+y, \text { from which } x_{2}=\frac{5-y}{2} .
$$

Since $1 \leq x_{2} \leq 9$ and $x_{2}$ is a natural number, then $y=1$ or $y=3$.

If $y=1$, then $x_{2}=2$ and from (1) $x_{1}=7$, i.e., we get the numbers 71 and 21. But for them, our system is not satisfied.

If $y=3$, then $x_{2}=1$ and $x_{1}=6$, i.e., we have the numbers 63 and 13, for which our system is true.

Answer: 13 and 63.