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Let $ a_0 = a> 1$ be an integer, and for $ n \ge 0$ , define $ a_{n + 1} = 2 ^ {a_n}-1$ . Show that the set of prime divisors of the terms of the sequence $ a_n$ is infinite.

This is a problem from 38th Brazilian Mathematics Olympiad.

Pablo A. Barros
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    I don't know the intended method, but an overkill method would be to use Zsigmondy's Theorem or similar. See https://math.stackexchange.com/questions/2251/for-any-n-is-there-a-prime-factor-of-2n-1-which-is-not-a-factor-of-2m-1 – Mark S. Jun 04 '17 at 18:29
  • Tip: Many primes are found by using $2^n-1$ for different values of n – Xii Jun 04 '17 at 19:04
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    If $p \mid a_n$, then $(2^p - 1) \mid a_{n+1}$. What do you know about the primes dividing $2^p-1$? – Daniel Fischer Jun 04 '17 at 19:28
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    (By the way, Daniel's comment gives a very neat proof that there are infinitely many primes) – Maxime Ramzi Jun 04 '17 at 20:10

2 Answers2

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In base $10$, a repeated decimal of any form can be represented as $\frac{x}{10^n-1}$ where $x$ is the repeated digits and $n$ is how many digits $x$ has. Since any repeated fraction can be represented this way, a repeated sequence of $9$s implied by $10^n-1$ should be divisible by any number which has divisors other than $2$ and $5$ if you put in enough $9$s.

You can apply this same logic to base $2$, see if you get the answer...

u8y7541
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  • It's indeed true that every odd number divides some number of the form $2^n - 1$, but we're not considering all of the numbers of the form $2^n - 1$, we're only considering those that are in the given sequence. – Dylan Jun 04 '17 at 20:49
  • @Dylan Hm... Maybe the same logic applies to the bigger set, but we'd have to prove it... – u8y7541 Jun 05 '17 at 00:40
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Assume to the contrary that the set of primes is finite. Then there must exist a prime $p$ such that $p$ divides infinitely many elements of the sequence. Pick such a prime.

Suppose that $p\mid a_n=2^{a_{n-1}}-1$ with $n>>p$. Let $\ell_1$ be order of $2$ modulo $p$. Thus we must have $\ell_1\mid a_{n-1}=2^{a_{n-2}}-1$. Let $\ell_2$ be the order of $2$ modulo $\ell_1$. Then we also must have $\ell_2\mid a_{n-2}$. Continuing this we must have, $\ell_k\mid a_{n-k}$. Observe that $\ell_i\ne 1$, otherwise we will have $\ell_{i-1}\mid 1$, a contradiction. Thus this process will continue until we have $\ell_i=2$. But we also should have $2=\ell_i\mid a_{n-i}$. But $a_{n-i}$ is odd, contradiction! $\square$

Takamoto Yuji
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