The answer is yes. First, we prove inductively that: $ 3^m|2^{3^m} + 1$ For $ m = 1$ it's clear. Now, we see that: $ 2^{3^{m + 1}} + 1 = (2^{3^m} + 1)(2^{23^{m}} - 2^{3^m} + 1)$ $ 3^m|2^{3^m} + 1$ from the induction hypothesis and the other bracket is divisible by $ 3$, as one can easily verify. So we are done. Now we prove that for $ a \in \mathbb{Z}_ +$ ($ a > 2$) the number $ a^3 + 1$ has a prime divisor which is not prime divisor of $ a + 1$. To prove this, see that $ (\frac {a^3 + 1}{a + 1}, a + 1)|3$ and $ \frac {a^3 + 1}{a + 1}$ is not power of $ 3$ if $ a > 1$ since it is never divisible by $ 9$. Now, we will prove inductively that there exist a number $ n$ such that: $ n = 3^{s}p_1p_2...p_k$ and $ n|2^{n} + 1$ (induction on $ k$). For $ k = 0$ we of course have $ 3|2^3 + 1$. So suppose that for $ k \geq 1$ such number $ n$ exists and see that: $ 2^{3n} + 1 = (2^n + 1)(2^{2n} - 2^n + 1)$. As we proved before, the number $ 2^{3n} + 1$ has a prime divisor which is not prime divisor of $ 2^n + 1$ (so it is not any of the $ p_1, p_2, ..., p_k$). Call it $ p_{k + 1}$. Since $ p_{k + 1}$ is odd it is obvious that for $ i = 1,2,...,k$ we have $ p_i|2^{3np_{k + 1}} + 1$. We have also proved, at the beggining that $ 3^{s + 1}|2^{3^{s + 1}p_1p_2...p_k} + 1$. So $ m = 3^{s + 1}p_1p_2...p_{k + 1}$ also works and we are done with the induction. To end the proof it of course sufficies to take $ k = 1999$.

Zsigmondy's Theorem can be used to quote part of the proof shown in the post above!

Hello, I don't understand why $\frac{a^3 + 1}{a + 1}$ can't add more than one prime divisor?

JayJuly wrote: Hello, I don't understand why $\frac{a^3 + 1}{a + 1}$ can't add more than one prime divisor? Note that it does not have to add just one prime factor (a=10 is an example). However, it adds at least one prime factor, so we just choose any one of them to be $p_{k+1}$

we prove it using the lemma below: lemma: for prime number $p$ that $p|x+y \implies || x^n + y^n ||_p = ||x+y||_p + ||n||_p$ now we use induction, suppose that we have $n$ with $k$ prime factors such as that $n|2^n+1$ and $2^n+1$ has a prime factor like $p$ that $n$ doesn't now to prove for the induction step that there exists some number m with $k+1$ prime factors and the properties above. take $m=pn$, obviously $pn|2^n+1 \implies m|2^m+1$. now we prove there exists a prime number $q$ that $q|2^m+1, (q,m)=1$ assure that $q$ doesn't exist and with the lemma above we see that for all prime prime factor of $n \implies ||2^{pn}+1||_{p_i} = ||2^{n}+1||_{p_i} +||p||_{p_i}= ||2^{n}+1||_{p_i} $ and for $p \implies ||2^{pn}+1||_{p} = ||2^{n}+1||_{p} +||p||_{p}= ||2^{n}+1||_{p} +1 $ which implies that $ 2^{pn}+1= (2^{n}+1)\times p$ which is a contradiction because $p|2^{n}+1 \implies p< 2^{n}+1 \implies 2^{pn} < (2^n+1)^2 < (2^{n+1})^2 $ so there exists some $q$ and the induction step in completed and for the base we can take $n=9$

Zsigmondy theorem