(given to Thomas Cruzel and Rémi Vergnaud). For which values of $n$, is the number $P_{n}=$ $36^{n}+24^{n}-7^{n}-5^{n}$ divisible by 899?
. On a $899=900-1=31 \times 29$, and 31 and 29 are primes. For all $n, a^{n}-b^{n}$ is divisible by $a-b$, so $36^{n}-5^{n}$ is divisible by $36-5=31$, and $24^{n}-(-7)^{n}$ is also divisible by $24+7=31$. For even $n$, since $(-7)^{n}=7^{n}$, the term $P_{n}$ is indeed divisible by 31, but for odd $n$, it is $36^{n}+24^{n}+7^{n}-5^{n}$ that is divisible by 31, so $P_{n}$ is not divisible by 31 (nor a fortiori by 899), because the difference $2 \times 7^{n}$ is not divisible by 29. Similarly, for even $n$, $36^{n}-7^{n}$ is divisible by $36-7=29$, and $24^{n}-5^{n}$ is divisible by $24+5=29$, so $P_{n}$ is divisible by 29, whereas if $n$ is odd, $36^{n}-7^{n}$ is still divisible by 29, but $24^{n}-5^{n}$ is not divisible by 29 (since 29 does not divide $2 \times 5^{n}$), so $P_{n}$ is not divisible by 29. $P_{n}$ is divisible by 29 and by 31 (hence by 899) if and only if $n$ is even.