NUMBER THEORY(unsolvable)

Here are many solutions .

My approach

We notice that for $n=1,2$ $p= 2,5$ which are obviously the prime numbers . Further we have that $p<10^{19}\implies n^n+1 <10^{19}$ . Taking $\log$ on both sides we get that $\frac{\log(n^n+1)}{\log(10^{19})} > 1$ for $n=16$ and $\frac{\log(n^n+1)}{\log(10^{19})} < 1$ for $n=15$ . Thus, the range of $n$ is $1\leq n <16$ . .

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If $n = kb$ where $k$ is an odd number greater than 1. Then $n^n+1 = (n^b)^k+1$ is divisible by $n^b+1$ ( by factorization ) of $a^n+b^n$ . Hence we arrived that $n$ must be an even number.

Let $n =2l$ and primes number are expressed as $2k_1+1 = n^n+1\implies {\color{#3D99F6}l = \frac{\sqrt[2l]{2k_1}}{2}}$ . This must be in form power of 2 if not then $n^n+1$ is composite. Working out with $2,4,8$ . Only $n= 2,4$ are possible values. Hence , the required primes we have $1^1+1,2^2+1, 4^4+1$ . Hence , the sum of primes we have is $\boxed{2+5+257 =264}$ .

Note $8$ doesn't work since $8^8$ has odd divisors in it power tower.