Problem Archive

Numbers of the form $n^{15}+1$ are composite for every integer $n \gt 1$.
For positive integers $n$ and $m$ let $s(n,m)$ be defined as the sum of the distinct prime factors of $n^{15}+1$ not exceeding $m$.

E.g. $2^{15}+1 = 3 \times 3 \times 11 \times 331$.
So $s(2,10) = 3$ and $s(2,1000) = 3+11+331 = 345$.

Also $10^{15}+1 = 7 \times 11 \times 13 \times 211 \times 241 \times 2161 \times 9091$.
So $s(10,100) = 31$ and $s(10,1000) = 483$.

Find $\sum s(n,10^8)$ for $1 \leq n \leq 10^{11}$.

Solution
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