See that $p=2,3,5$ works. Let $p\ge 7$. See that $3^p+4^p\equiv 5^p+9^p\equiv -98\equiv 0\pmod{7}$ as $p$ is odd. Also, $11|3^p+4^p+5^p+9^p-98$ as $p$ is odd and not divisible by $5$. Since this number has at most $6$ positive divisors, we find that $3^p+4^p+5^p+9^p-98\leq 11^2\cdot 7=847$. But, this contradicts the fact that $p\ge 7$, done.

A bit easy for P8

BarisKoyuncu wrote: See that $p=2,3,5$ works. Let $p\ge 7$. See that $3^p+4^p\equiv 5^p+9^p\equiv -98\equiv 0\pmod{7}$ as $p$ is odd. Also, $11|3^p+4^p+5^p+9^p-98$ as $p$ is odd and not divisible by $5$. Since this number has at most $6$ positive divisors, we find that $3^p+4^p+5^p+9^p-98\leq 11^2\cdot 7=847$. But, this contradicts the fact that $p\ge 7$, done. how did you find $11|3^p+4^p+5^p+9^p-98??$

lian_the_noob12 wrote: BarisKoyuncu wrote: See that $p=2,3,5$ works. Let $p\ge 7$. See that $3^p+4^p\equiv 5^p+9^p\equiv -98\equiv 0\pmod{7}$ as $p$ is odd. Also, $11|3^p+4^p+5^p+9^p-98$ as $p$ is odd and not divisible by $5$. Since this number has at most $6$ positive divisors, we find that $3^p+4^p+5^p+9^p-98\leq 11^2\cdot 7=847$. But, this contradicts the fact that $p\ge 7$, done. how did you find $11|3^p+4^p+5^p+9^p-98??$ We can prove it by Fermat's little theorem.