Finish my proof that there are infinitely many pairs of distinct prime numbers $(p, q)$ such that $p \mid (2^{q - 1} - 1)$ and $q \mid (2^{p-1} - 1)$.

Question

This is from the Romania IMO Team Selection Test, 2009, Day 7, Problem 3. Here is my work so far.

Problem: Prove that there are infinitely many pairs of distinct prime numbers $(p, q)$ such that $p \mid (2^{q - 1} - 1)$ and $q \mid (2^{p - 1} - 1)$.

Solution: Let $p$ be an arbitrary prime divisor of the Fermat number $F_n = 2^{2^n} + 1$, and let $q$ be an arbitrary prime divisor of $F_{n + 1} = 2^{2^{n + 1}} + 1$ not equal to $p$.

By Lucas, we know that $2^{n + 2} \mid (p - 1)$, so $\bigl(2^{2^{n + 2}} - 1\bigr) \mid \bigl(2^{p - 1} - 1\bigr)$. Then, since $2^{2^{n + 2}} - 1 = \bigl(2^{2^{n + 1}} - 1\bigr)\bigl(2^{2^{n + 1}} - 1\bigr)$, we know that $F_{n + 1} \mid (2^{p - 1} - 1)$. Since $q \mid F_{n + 1}$ by definition, the first condition is met.

Similarly, by Lucas, we know that $2^{n + 3} \mid (q - 1)$, so $\bigl(2^{2^{n + 3}} - 1\bigr) \mid \bigl(2^{q - 1} - 1\bigr)$. Note that $\bigl(2^{2^{n}} + 1\bigr) \mid \bigl(2^{2^{n + 3}} - 1\bigr)$ since $(x + 1) \mid (x^8 - 1)$, so $F_n \mid \bigl(2^{q - 1} - 1\bigr)$. Since $p \mid F_n$ by definition, the second condition is met.

We will let $n$ vary. To generate infinitely many pairs $(p, q)$, what remains to be shown is that not all Fermat numbers are divisible by a finite set of primes.

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Sadly, this last step seems like such a minutiae that it might even be ignored, but I can’t seem to prove it. Zsigmondy’s Theorem doesn’t work because of the $+1$, not $-1$. Any help?

Answer 1

How I would solve:

1. If there are infinitely many positive integers $n$ such that both $2^{2^{n}}+1$ and $2^{2^{n+1}}+1$ are both prime: Then let $n$ be any sufficiently large positive integer such that $p=2^{2^n} +1$ and $q=2^{2^{n+1}}+1$ are both prime. Then $2^{p-1} \equiv_q 1$ and $2^{q-1} \equiv_p 1$ and the result follows for Case 1. [Indeed, $p-1 = 2^{2^n}$ [namely $2^{n+2}$ multiplied by many more factors of $2$ [as $2^n >> n+2$ for $n$ large enough]] and $q-1=2^{2^{n+1}}$. Furthermore, $2^{2^{n+2}} \equiv_q 1$ gives a desired $2^{p-1} \equiv_q 1$ [as $p-1$ is $2^{n+2}$ times many more factors of $2$, so $2^{p-1}$ is $2^{2^{n+2}}$ squared many times in particular an integral power of $2^{2^{n+2}}$]. And likewise, $2^{2^{n+1}} \equiv_p 1$ gives $2^{q-1} \equiv_p 1$.]

2. Otherwise there are infinitely many positive integers $n$ such that $2^{2^n}+1$ is composite. So for each such $n$, let $p_n$ and $q_n$ be distinct primes that both divide $2^{2^n}+1$. Then both equations $2^{p_n-1} \equiv_{q_n} 1$ and $2^{q_n-1} \equiv_{p_n} 1$ hold. [Indeed, as $p_n$ and $q_n$ are primes dividing $2^{2^n}+1$, Claim 0 [below] implies the following: $2^{n+1}$ must divide both $(p_n-1)$ and $(q_n-1)$. This however gives $2^{p_n-1}$ an integral power of $2^{2^{n+1}}$ [because $2^{n+1}$ divides $p_n-1$] and as the equation $2^{2^{n+1}} \equiv_{q_n} 1$ holds, it follows that the equation $2^{p_n-1} \equiv_{q_n} 1$ also holds. Likewise, $2^{q_n-1}$ an integral power of $2^{2^{n+1}}$ and so the equation $2^{q_n-1} \equiv_{p_n} 1$ also holds.] Finish this case 2. by noting that the Fermat numbers are relatively prime to each other, so the pairs $(p_n, q_n)$; $n$ a positive integer satisfying $2^{2^n}+1$ composite; $p_n,q_n$ distinct primes dividing $2^{2^n}+1$; are distinct pairs and are in fact disjoint pairs.

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Claim 0: Let $n$ be a positive integer, and let $p_n$ be a prime that divides $2^{2^n}+1$. Then $2^{n+1}$ divides $p_n-1$.

Indeed, let $b_n$ be the smallest positive integer such that the equation $2^{b_n} \equiv_{p_n} 1$. Then it follows that $b_n$ must divide $(p_n-1)$. However, we claim that $b_n$ must infact be $2^{n+1}$. [Indeed, for any other positive integer $y$, the equation $2^y \equiv_{p_n} 1$ holds iff $b_n$ divides $y$. Now from this $b_n$ does not divide $y=2^n$ [as $p_n$ divides $2^y+1=2^{2^n}+1$ so $2^y =2^{2^n} \equiv_{p_n} -1 \not = 1$], but $b_n$ does divide $2^{n+1}$ [as $2^{2^{n+1}} = (2^y)^2 \equiv_{p_n} (-1)^2 = 1$]. So $b_n$ must be a power of $2$ [as it divides another power of $2$ namely $2^{n+1}$] and must infact be $2^{n+1}$.[Indeed, as $2^{2^n} \equiv_{p_n} -1 \not = 1$ it follows that $b_n$ does not divide $2^n$. This together with $b_n$ being a power of $2$ and $b_n$ dividing $2^{n+1}$, leaves $b_n =2^{n+1}$ as the only possibility.]] So from this it follows from the above that, as $b_n$ divides $p_n-1$ and $b_n$ is $2^{n+1}$, that indeed, $b_n=2^{n+1}$ divides $p_n-1$. And so Claim 0 follows. $\surd$

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I am sorry I could not fix/follow what you had above in the OP, I was not seeing how your idea could be made to work. In particular, going by the notation in your OP, I'm still not seeing why $2^{n+2}$ divides $p-1$. [Yes, $2^{2^{n+2}} \equiv_p 1$ but that does not imply $2^{n+2}$ divides $p-1$.]

Answer 2

Never mind, I think figured out how to clear my small doubt. Note that the Fermat numbers are pairwise relatively prime. This is because $$F_{n + 1} = F_n F_{n - 1} \cdots F_0 + 2.$$ To prove this, we will use induction. Note that $5 = F_1 = F_0 + 2$, so the base case is clear. Then, assume that $$F_k = F_{k - 1} F_{k - 2} \cdots F_0 + 2$$ for some integer $k$. Subtracting $1$ from both sides and then multiplying both sides by $F_k$, we get that $$(F_k - 1)^2 = F_kF_{k - 1} \cdots F_0 + 1.$$ However, note that $F_{k + 1} - 1 = (F_k - 1)^2$, so $$F_{k + 1} = F_kF_{k - 1} \cdots F_0 + 2$$ as desired.

Then, assume for the sake of contradiction that $p \mid F_m$ and $p \mid F_n$ for $m < n$. The Fermat numbers are odd, so $p$ is odd. Then, $$0 \equiv F_n \equiv F_{n - 1} F_{n - 2} \cdots F_0 + 2 \equiv 2 \pmod{p},$$ since $0 \le m \le n - 1$. This is a contradiction, so the Fermat numbers are relatively prime.

Once we have shown that the Fermat numbers are relatively prime, then it is clear that infinitely many primes $(p, q)$ can be generated, because each $F_{k + 1}$ contains a prime factor that none of the previous Fermat numbers had.