1. Find all real numbers $a$ such that $\sqrt{7-a}+\sqrt{7+a}$ is an integer.
III/1. Let $\sqrt{7-a}+\sqrt{7+a}=n, n \in \mathbb{N}$. We square the equation: $14+2 \sqrt{49-a^{2}}=n^{2}$ and rearrange it to

$$
\sqrt{49-a^{2}}=\frac{n^{2}}{2}-7
$$

Since $\sqrt{49-a^{2}} \geq 0$, from equation (3) it follows that $\frac{n^{2}}{2}-7 \geq 0$ and $n>3$. Since $\sqrt{49-a^{2}} \leq 7$, from equation (3) it also follows that $\frac{n^{2}}{2}-7 \leq 7$, which gives $n<6$. Therefore, $n$ can only be $n=4$ and $a= \pm 4 \sqrt{3}$ or $n=5$ and $a= \pm \frac{5}{2} \sqrt{3}$.

The equation without square roots $a^{2}=7 n^{2}-\frac{n^{4}}{4}$ (or an equivalent form) . . . . . . . . . . . . . . . 2 points

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-11.jpg?height=48&width=1631&top_left_y=667&top_left_x=218)

Solutions $n=4$ and $a= \pm 4 \sqrt{3}$ or $n=5$ and $a= \pm \frac{5}{2} \sqrt{3} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ 1 point each

(If the contestant provides only the positive solutions, award 1 point.)

Conclusion or verification that other numbers $n<6$ do not satisfy

1 point