Professor Antônio discovered an interesting property related to the integer $x$ that represents his age. He told his students that $x^{2}=\overline{a b a c}$ and that $x=\overline{a b}+\overline{a c}$. What is the professor's age?

Note: We are using a bar to distinguish the decimal representation of the four-digit number $\overline{a b c d}$ from the product $a \cdot b \cdot c \cdot d$. For example, if $\overline{a b c d}=1267$, then $a=1, b=2, c=6$ and $d=7$. The notation is the same for numbers with other quantities of digits.
Solution

Given that $x^{2}$ has 4 digits, we have $31<x<100$. Note that $\overline{a b c d}=\overline{a b} \cdot 100+\overline{c d}$, and thus it follows that

$$
\begin{aligned}
x^{2} & =\overline{a b} \cdot 100+\overline{c d} \\
& =99 \overline{a b}+\overline{a b}+\overline{c d} \\
& =99 \cdot \overline{a b}+x
\end{aligned}
$$

Therefore,

$$
\begin{aligned}
x^{2}-x & =11 \cdot 9 \cdot \overline{a b} \\
x(x-1) & =11 \cdot 9 \cdot \overline{a b}
\end{aligned}
$$

Since 11 is a prime number, it follows that 11 divides $x$ or $x-1$. The possibilities for $x$ are:

i) When 11 divides $x: 33,44,55,66,77,88,99$.

ii) When 11 divides $x-1: 34,45,56,67,78,89$.

We also know that 9 divides $x(x-1)$, so we are left with the following possibilities for $x$: $45,55,99$.

By squaring these three values, we can conclude that $x=45$. Indeed, $45^{2}=$ 2025 and $20+25=45$.