Subject 4. Determine the numbers $\overline{a b}$ for which $\overline{b a}+\overline{a b}$ and $\overline{b a}-\overline{a b}$ are perfect squares.
## Solution:

We have $\overline{b a}+\overline{a b}=11(a+b)$ and $\overline{b a}-\overline{a b}=9(b-a)$. For $\overline{b a}+\overline{a b}$ and $\overline{b a}-\overline{a b}$ to be perfect squares, it must be that $b+a=11$ and $b-a \in\{1,4,9\}$. We have the following cases:

1) $\left.\begin{array}{l}b+a=11 \\ b-a=1\end{array}\right\} \Rightarrow 2 b=12 \Rightarrow b=6$ and $a=11-6=5$.
2) $\left.\begin{array}{l}b+a=11 \\ b-a=4\end{array}\right\} \Rightarrow 2 b=15 \Rightarrow b \notin \mathbb{N}$, this case does not work.
3) $\left.\begin{array}{l}b+a=11 \\ b-a=9\end{array}\right\} \Rightarrow 2 b=20 \Rightarrow b=10$, since $b$ is a digit, this case does not work.

The only solution is $a=5$ and $b=6$, from which we get $\overline{a b}=56$.

| Details of the solution | Associated score |
| :--- | :---: |
| Deduce that $\overline{b a}+\overline{a b}=11(a+b)$ and $\overline{b a}-\overline{a b}=9(b-a)$ | $2 \mathrm{p}$ |
| Specify that $b+a=11$ and $b-a \in\{1,4,9\}$. | $2 \mathrm{p}$ |
| Demonstrate that the only solution is $a=5$ and $b=6$ from which $\overline{a b}=56$ | $3 \mathrm{p}$ |