Prime factors of $5^n+6^n+7^n+8^n+9^n+10^n$

Question

I currently run an integer factoring project of the numbers of the form $$5^n+6^n+7^n+8^n+9^n+10^n$$ where $n$ is a non-negative integer.

Do the prime factors have a particular form as it is the case for the Mersenne-numbers with prime exponent or the Fermat-numbers ? Is there a reason why the largest prime upto $n=70\ 000$ occurs for $n=176$ ?

For the second question , we only have to consider the case $4\mid n$ because otherwise $3$ or $5$ is a prime factor. Nevertheless , it is somewhat surprising that the prime after $n=176$ occurs so late , if there is a further such prime.

I thought about using the Faulhaber-formula , but this becomes too complicated for large exponents.

I am pretty sure that the factorizations of those numbers can be accelerated because of the large occuring exponents. An overview of the first not fully factor cases is here. How can I modify the available siqs-programs to make use of this ?

Answer 1

Comment: Some modifications:

$176=11\times 16$

$A=5^n+6^n+7^n+8^n+9^n+10^n$

For $p=17$ we have:

$a^{16}\equiv 1\bmod 17\Rightarrow (a^{16})^{11}\equiv 1 \bmod 17; a= 5, 6, 7, 8, 9, 10$

$\Rightarrow$ for $n=176$:

$ A\equiv 6\bmod 17$

$b^{11}\equiv b\bmod 11; b=5^{16}, 6^{16}, 7^{16}, 8^{16}, 9^{16}, 10^{16}$

$\Rightarrow$ for $n=176$:

$ A\equiv (6\bmod 17)\bmod 11 $

In this way the linear form of $A$ for $n=176 $ can be:

$A=17 k + 11m +6$

this is the form of a prime generator by $A$. To make it simpler, let $k=11s$ and $m=17t$ , so $A$ can be written as:

$A=187 u + 6$

This progression mostly gives composite numbers,for example for u up to $13$ only four numbers $193, 941, 2063$ and $2437$ are primes. probably the density of primes reduces for large values of u. Can this be the reason for the occurring large prime when $n=176$?

Answer 2

HINT.- Never $2$ is a factor because the expression is always odd.

$3\text { is a factor }\iff \text { n is odd}$.

$5\text { is a factor }\iff n\not\equiv0\pmod4$.

$7\text { never is a factor }$ because if $f(n)=5^n+6^n+1+2^n+3^n$ then $f(\mathbb Z/7\mathbb Z)=\{3,5,6\}$

For greater prime factors the problem becomes harder but we can keep in mind that if$f(n)\equiv0\pmod p$ then also $f(n+k(p-1))\equiv0\pmod p$ because $a^{k(p-1)}\equiv1\pmod p$ when $(a,p)=1$. For example in order to see when $11$ is a factor of $f(n)$ we look first at the first eleven values of $n$ and we see that $n=7$ is the only for which $f(7)\equiv0\pmod{11}$ so, consequently we have

$11\text { is a factor }\iff n=7+10k$.

With $13$ the only $n$ analogue to the $7$ for the factor $11$ above is $3$ so we have

$13\text { is a factor }\iff n=3+12k$.

And so on if we follow this procedure in which obviously it becomes harder when the considered prime factor increase.