# 4. CONDITION

The square of a natural number $a$ when divided by a natural number $n$ gives a remainder of 8. The cube of the number $a$ when divided by $n$ gives a remainder of 25. Find $n$.
Solution. Note that the number $\mathrm{x}=\mathrm{a}^{6}-8^{3}=\left(\mathrm{a}^{2}\right)^{3}-8^{3}=\left(\mathrm{a}^{2}-8\right)\left(\mathrm{a}^{4}+8 \mathrm{a}^{2}+64\right)$ is divisible by $n$. Also note that the number $\mathrm{y}=\mathrm{a}^{6}-25^{2}=\left(\mathrm{a}^{3}\right)^{2}-25^{2}=\left(\mathrm{a}^{3}-25\right)\left(\mathrm{a}^{3}+25\right)$ is divisible by $n$. Then the difference $\mathrm{x}-\mathrm{y}=\left(\mathrm{a}^{6}-8^{3}\right)-\left(\mathrm{a}^{6}-25^{2}\right)=625-512=113$ is divisible by $n$. The number 113 is a prime number, it has only two divisors: 1 and 113. Since by the condition $\mathrm{n}>25$, then $\mathrm{n}=113$.

Answer: $\mathrm{n}=113$.