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Is there a method to decide whether a given function of the form $f(n)=a^n+b^n+c^n$ ($a,b,c$ fixed positive integers , $n$ running over the positive integers) satisfies the following property : If $p$ is a prime number with $p\mid f(s)$ for some positive integer $s$ and $k$ is any positive integer , then there is a positive integer $t$ with $p^k\mid f(t)$. In other words , if a prime can divide this expression, every power of this prime can it as well.
I am particular interested in the case $(a,b,c)=(5,6,10)$ , so $f(n)=5^n+6^n+10^n$ because in this case my prime search revealed a probable prime quite late (for $n=3168$) , but no other probable prime upto $n=60\ 000$. Furthermore I try to factor numbers of this form the last months (see the linked question)
I tried to use that $p\mid f(n)$ is equivalent to $p\mid u^n+2^n+1$ where $u$ is one of $\frac{p+6}{5},\frac{2p+6}{5},\frac{3p+6}{5},\frac{4p+6}{5}$ depending on which of those is an integer. But this led to nowhere.
For this special case $(a,b,c)=(5,6,10)$ I found out the following :
- For $p\le 2\cdot 10^5$ and $k=2$ the conjecture is true.
- For $p\le 14\ 000$ and $3\le k\le 5$ the conjecture is true.
This is a partial answer that's too long for a comment, also I will assume $p\ne 2$ and $a,b,c$ are not divisible by $p$ for convenience and try to simplify some of what I described in my comment so it's less to digest.
Hensel's lemma can be extended to p-adic power series. I should have been more clear in my comment earlier, breaking up the exponentials how I described allows us to get a convergent power series expansion.
So instead of looking at the discontinuous (as a p-adic function) $f(n)=a^n+b^n+c^n$ we instead break it up into $p-1$ separate functions $F_t(s)=f(t+(p-1)s)$ for $0\le t <p-1$. The reason for this is to turn the exponentials into a nice, differentiable p-adic power series.
Specifically if we focus on just one of the exponentials, we can expand it as
$$g(s)=a^ta^{(p-1) s} = \exp(s \log(a^{p-1})) = \sum_{n\ge 0}\frac{a^t\log(a^{p-1})^n}{n!}s^n$$
A quick diversion for a moment, this next part can be skipped. At this point it should be somewhat plausible that this is looking similar to a polynomial in $s$ and how Hensel's lemma will start to fit in. One way to evaluate the logarithm is with the regular power series you know of from real analysis,
$$\log(a^{p-1}) = \log(1-(1-a^{p-1})) = -\sum_{n\ge 1} \frac{(1-a^{p-1})^n}{n}$$
You can approximate this to arbitrary accuracy by just computing terms $\mod p^N$ for some fixed $N$ and eventually the powers $(1-a^{p-1})^n$ will all end up all being $0\mod P^N$ for large enough $n$.
Now back from the diversion and on track, for your case here all that you probably need to use is the fact that the derivative is $g'(s) = g(s)\log(a^{p-1})$ as you might have guessed from real analysis, but remember these are all p-adic series. For our uses, Hensel's lemma will look like this,
$$|F_t(s)|_p < |F_t'(s)|_p^2$$
(Roughly speaking as a quick refresher, when $x$ is a rational number you can interchange these two statements $x \equiv 0 \mod p^m \iff |x|_p \le \frac{1}{p^m}$.)
This means we know from the original problem we have some $n=t+(p-1)s$ which gives us $|f(n)|_p \le \frac 1 p$ so we have $|F_t(s)|_p \le \frac 1 p$. Now looking at the derivative,
$$|F_t'(s)|_p = |a^t(a^{p-1})^s \log(a^{p-1}) + b^t(b^{p-1})^s \log(b^{p-1}) + c^t(c^{p-1})^s \log(c^{p-1})|_p$$
For the terms individually we have $|x^t(x^{p-1})^s|_p=1$ and $|\log(x)|_p\le \frac 1 p$ and the ultrametric inequality will give us,
$$|F_t'(s)|_p \le \max(|a^t(a^{p-1})^s\log(a^{p-1})|_p, |b^t(b^{p-1})^s \log(b^{p-1})|_p, |c^t(c^{p-1})^s \log(c^{p-1})|_p)$$
$$|F_t'(s)|_p \le \frac 1 p$$
Since we care about when $|F_t(s)|_p < |F_t'(s)|_p^2$, we're left with $|F_t(s)|_p \le \frac 1 p$ and $|F_t'(s)|_p^2 \le \frac{1}{p^2}$. In other words, the assumption that $f(n)$ is divisible by $p$ is not strong enough to tell us that we will have a solution that's divisible by $p^k$. However, you might expect that if $f(n)$ is divisible by $p^3$ then it's more likely you will have a solution that's divisible by $p^k$ for all $k$.
This might be able to be strengthened with some work, to see when it's certain that it's divisible by $p^3$ is not just necessary, but also sufficient. If that is the case, and I'm not mistaken then that means by continuity we can safely check finitely many values of $s$ so that we exhaust the possible values that satisfy the Hensel's lemma inequality. I wish I had more time to polish this and check for errors before sending it, but that's life!