$p=2 \implies$ true $p>2 \implies p$ is odd $\implies p<1 + 2^p + 3^p +...+ p^p \equiv 1+2+...+(p-1) \equiv 0 (mod p)$

If $p = 2$ we get $1 + 2^2 = 5$, which is a prime number. If $p > 2$, $p$ is odd so $a^p + (-a)^p \equiv 0\hspace{2pt} (mod \hspace{5pt}p)$ so $1 + ... + p^p = [1 + (p - 1)^p] + [2^p + (p - 2)^p] + ... + p^p \equiv 0 \hspace{2pt}(mod \hspace{5pt} p)$ so $p\mid LHS$ and therefore $p \mid RHS$, but since $RHS$ is prime, we get that $RHS = p$ which is impossible since $LHS >> p$. Therefore, $p = 2$ is the only prime number for which $1 + ... + p^p$ is prime ($1 + 2^2 = 5$ is prime)

For all $p>2$, By Fermat's little theorem, the sum is divisible by $p$ and greater than $p$, so it cannot be a prime. So, just $p=2$ satisfies. My doubt- why is the case $p=2$ missed out in Fermat's little theorem?

Because Fermat's little theorem yields the given expression mod p as $\frac{p(p-1)}{2}$ which is divisible by p iff p is odd, which in your doubt's case, isn't.

Ohh! Thanks

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