Maximum amount of divisors of the number $n^m+m^n$

Question

We are given some positive integer $m$. What maximum amount of distinct prime divisors a number $n^m+m^n$ can have, where $n\in\mathbb{Z}_+$?

Edit:

As noted in comments, there is no reason to think that such maximum exists. In that case the question is following: why for every positive integers $m$ and $x$ there exists a positive integer $n$ such that the number $n^m+m^n$ has more than $x$ distinct prime divisors?

Answer 1

Let $\omega(t)$ be the number of distinct prime divisors of $t$ greater than $4$ . We will prove the following theorem which formalizes what you want if I'm not mistaken.

Theorem : for any $m\geq 2,x\geq 2$ there exists an integer $n$ such that: $$\omega(m^n+n^m)\geq x$$

First of all we will prove the following two lemmas:

Lemma 1: For every $m\geq 2,k\geq 1$ there exists an integer $a\geq 1$ such that $$\omega({m^a-ma})\geq k $$

Proof: (Main tools: Fermat's theorem ) let $p_1,....,p_k>m$ be $k$ primes greater than $m$. and let $a=1+\text{lcm}(p_1-1,p_2-1,\dots,p_k-1,p_1\cdots p_k)$ we can see that for every $i$ we have: $$m^{a-1}-a\equiv 0\mod p_i $$ using Fermat's theorem. as we can see $\omega(m^a-ma)\geq k$

Lemma 2: For every $m\geq 2,n\geq 1$ positive integers such that $6\not|n $ we have: $$\omega(1+m^{n})\geq {d(n)} $$

Proof : (Main tools: Zsigmondy's theorem ) A variant of Zsigmondy's theorem states that for every positive integers $a>b>0,n$ such that $\gcd(a,b)=1$ and $n$ odd, then the number $a^n + b^n$ has at least one primitive prime divisor with the exception $n=3,a=2,b=1$. A primitive prime divisor means a prime divisor which does not divide $a^k+b^k$ for all $k< n$ . Let's take $a=m,b=1$ and $d$ a divisor of $n$ this means that there exists a primitive prime divisor of $m^d+1$ let's denote by $p_d$ the smallest such primitive prime divisor.

Now we can argue that $(p_d)_{d|n}$ is a sequence of different primes dividing $a^n+b^n$ (as $a^d+b^d$ divides $a^n+b^n$ for every $d|n$, $n$ is odd of course). Hence we can conclude that: $$\omega(1+m^{n})\geq {d(n)} $$

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Proof of the theorem: The idea of the proof is to take $n=m^a$ and find the appropriate $a$ such that $\omega(1+m^{m^a-ma})\geq x-\omega(m) $ because: $$n^m+m^n=m^{am}(1+m^{m^a-am}) \tag{*} $$ Let's take $k=\lfloor \log_2x\rfloor+1 $,using the lemma $1$ there exists an integer $a$ such that $\omega({m^a-am})\geq k$, so we can write $m^a-am=p_1\cdots p_k \alpha$ where $p_1,\cdots ,p_k>4$ are primes, using the lemma $2$ for $(n,m)$ equals to $(p_1\cdots p_k,m^\alpha)$ we can conclude that $$\omega(1+m^{m^a-am})\geq 2^k$$ This of course using $(*)$ implies that $$\omega(m^n+n^m)\geq 2^k\geq x$$

So we have proved the theorem, which can be formulated in another way saying that for every positive integer $m\geq 1$ : $\limsup\limits_{n\rightarrow\infty}\omega(n^m+m^n)=+\infty$