## 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 $1 \leq n \leq 10^{11}$.

### To solve this problem, you first need to understand the properties of numbers of the form $n^{15}+1$. By applying the sum of cubes factorization and Fermat’s Little Theorem, you can see that if $n > 1$, then $n^{15}+1$ must be composite since it can be factored as $(n^5-1)(n^{10}+n^5+1)$.

However, calculating all the prime factors for $n^{15}+1$ for $n$ ranging from $1$ to $10^{11}$ is infeasible due to the high computational complexity. You would need more advanced mathematical and computational techniques to solve this problem.

First, notice that $s(n,m)$ contributes to the sum only if $n^5-1$ is divisible by a prime less than m, or equivalently if $n$ is a $5$th power modulo a prime less than m.

Now, consider all the $5$th powers modulo p, for a prime p between $2$ and $10^8$. For a fixed prime p, the solutions to $x^5 = a \mod p$ form a cyclic group because $5$ does not divide $p-1$ (otherwise p would be of the form $5k+1$ and $p > 10^8$). It follows all the $5$th powers form a single cycle whose length divides $p-1$. More concretely, if we list all the $5$th powers modulo p, the list first increases strictly up to a peak and then decreases strictly back to $1$ (modulo p).

There are two cases to consider. If $p$ has the form $5k+3$ or $5k+4$, the cycle has length $p-1$, and there are exactly $2$ values of $n < p$ for which $n^5 = 1 \mod p$. It is clear that these $2$ values, call them $x$ and $p-x$, have no common factor with $p$ except possibly $2$. Moreover, exactly one of $x$ and $p-x$ is even, so if $p > 3$, they are both less than $10^{11}$ and should be counted exactly once. If $p = 3$, $x$ and $p-x = 2$ and $1$ should be counted $33$ and $66$ times respectively.

If $p$ has the form $5k+1$ or $5k+2$, the cycle has length $\frac{p-1}{2}$, and there are exactly $4$ values of $n < p$ for which $n^5 = 1 \mod p$. Among these $4$ values, there are $2$ pairs that add to $p$ modulo $k$. Similarly as before, we count each pair exactly once. Here the tricky case $p = 2$ has to be worked out separately. Here, $x$ and $p-x = 1$ and $1$ should be counted $50,000000050$ and $50,000000000$ times respectively. In both cases, we find that the sum by powers of $5$ modulo $p$ increases initially and eventually cycles. By considering how many times each term of the form $n^5$ modulo $p$ appears for $1 \le n \le 10^{11}$, we can sum all these terms effectively by transforming the problem into one of counting special residues modulo $p$. By summing this result over all such primes $p \le 10^8$, this gives the requested sum of $s(n,10^8)$ for $1 \leq n \leq 10^{11}$ as stated in the problem. However, it should be noted that this problem is beyond the scope of what can be reasonably computed by hand, and solving it in practice would typically involve using number theory algorithms implemented in a high-precision computational software.

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