5. Calculate $\sqrt{n}+\sqrt{n+524}$, given that this number is rational and that $n$ is a natural number.
Answer: 262.

Solution. Let the desired number be $a$. We have $\sqrt{n+524}=a-\sqrt{n}, n+524=a^{2}-2 a \sqrt{n}+n$. By the condition, $a$ is rational, so $\sqrt{n}$ is also rational. Therefore, $n=k^{2}, k \in \mathbb{N}$. Then the number $\sqrt{n+524}$ is also rational, so $n+524=m^{2}, m \in \mathbb{N}$. Thus, $m^{2}-k^{2}=524,(m-k)(m+k)=4 \cdot 131$. Note that the numbers $m-k$ and $m+k$ have the same parity, and the number 131 is prime. Therefore, $m-k=2$ and $m+k=2 \cdot 131$. Both equations are satisfied when $m=132, k=130$. Therefore, $a=m+k=262$.