-4
$\begingroup$

Given a prime $p$ Let $$\phi(p):=\sum_{p_i <p} v_{p_i}(p-1) e_i$$ where $e_i$ is the $i$-th standard basis vector, $v_p(n)$ is the valuation of $n$ for the prime $p$ and $p_i$ is the $i$-th prime number. A prime $p$ is called linearly independent, if its phi-vector does not lie in the $\mathbb{Q}$ span of phi-vectors of previous primes. Can you prove that there are $\infty$ many linearly independent primes without using Dirichlets theorem?

Edit: Here is the motivation for asking this question: et $(q_j)_{j \ge 1}$ be the sequence of linearly independent (LI) primes (with $q_1 = 3$), defined as the primes $p_k$ whose $p$-adic valuation vector $\phi(p_k) = (v_{p_i}(p_k-1))_{i<k}$ is linearly independent from the vectors of all preceding primes. Let $m_j = q_j - 1$.

The conjecture posits a rigid "parent-child" structure governing the prime factorizations of $m_j$.

The Arithmetic Conjecture

For every index $j \ge 2$, there exist:

  1. An index $p(j)$ with $1 \le p(j) < j$ (the parent of $q_j$),
  2. An integer $\alpha_j \ge 0$,
  3. And an odd prime $\ell_j$,

such that the following multiplicative factorization holds:

$$m_j = q_j - 1 = 2^{\alpha_j} \cdot m_{p(j)} \cdot \ell_j$$ $$q_j - 1 = 2^{\alpha_j} (q_{p(j)}-1) \ell_j$$

This structure must satisfy two key conditions on the prime factors:

  • (New Odd Prime): The prime $\ell_j$ is new at step $j$. It does not divide $m_i$ for any $i < j$, and $v_{\ell_j}(m_j) = 1$.
  • (Copied Odd Primes): For every other odd prime $\ell \neq \ell_j$, the $\ell$-adic valuation is simply copied from the parent: $$v_\ell(m_j) = v_\ell(m_{p(j)})$$

(The $2$-adic valuation is $v_2(m_j) = v_2(m_{p(j)}) + \alpha_j$).

Informally: Each new LI prime $q_j$ is generated from a previous one $q_{p(j)}$ by simply multiplying $(q_{p(j)}-1)$ by a power of $2$ and exactly one new odd prime $\ell_j$ that has never appeared before in the sequence.

Second edit: Here is a different proof from Wojowus which does not use Dirichlet: In https://oeis.org/A061303 it is linked that Amarnath Murthy, On the divisors of Smarandache Unary Sequence. Smarandache Notions Journal, Vol. 11, No. 1-2-3, Spring 2000. proves the existence of a prime $q \equiv 1 \mod(p)$ given any prime $p$. Now given any prime $p$ choose the smallest such prime $q \equiv 1 \mod(p)$. Then this prime $q$ must be linearly independent. Hence by repeatadly applying Murthy's result to find a minimal such prime, we get a sequence of primes $p_0,p_1,p_2,\cdots$ such that $p_1,p_2,p_3,\cdots$ are linearly independent, hence proving $\infty$ of such primes.

Improve this question
$\endgroup$
15
  • 4
    $\begingroup$ That's a convoluted way of asking to prove that the values of the Euler totient function $\varphi(p)=p-1$ are not contained in a finitely generated subgroup of $\mathbb{Q}^{\times}$ as $p$ varies in the primes. Or simply that there are infinitely many different prime divisors occurring in the sequence $\varphi(p)$. $\endgroup$ Commented Nov 14 at 11:33
  • 2
    $\begingroup$ If I understand this correctly, if this were not the case, then there would be a finite set of primes $P$ such that for all primes $p$, all prime divisors of $p-1$ lie in $P$. But such numbers is exponentially sparse, and primes are not exponentially sparse. $\endgroup$ Commented Nov 14 at 12:58
  • 4
    $\begingroup$ An initial version of this question was posted and deleted yesterday after receiving negative feedback: mathoverflow.net/questions/503831/… $\endgroup$ Commented Nov 14 at 13:21
  • 1
    $\begingroup$ PNT is an overkill, we don't need anything remotely this strong. It follows for instance from divergence of $\sum_p\frac{1}{p}$. $\endgroup$ Commented Nov 14 at 14:04
  • 2
    $\begingroup$ I wouldn't say this new proof "doesn't use Dirichlet" when it uses a special case of Dirichlet which can be proven in an elementary way. You can prove Dirichlet for primes $1\pmod n$ for any $n$ very elementarily using cyclotomic polynomials. For $n$ a prime, the standard proof simplifies a lot, I have no idea how it compares to the proof in the article you reference $\endgroup$ Commented Nov 17 at 17:22

1 Answer 1

Reset to default
3
$\begingroup$

Suppose there were only finitely many LI primes. Denoting them by $q_i$, this means that for all primes $p$, $\phi(p)$ is a linear combination of $\phi(q_i)$. In particular, this means that there is a finite set of primes $P$ such that, for every prime $p$, $v_q(p-1)=0$ for all primes outside of $P$, that is, $p-1$ is only divisible by primes in $P$. But this is impossible, since the set $S$ of numbers with prime factors contained in $P$ is exponentially sparse, unlike the set of all primes. A very elementary proof uses the fact that the sum of reciprocals of primes diverges, while $$\sum_{n\in S}\frac{1}{n}=\prod_{q\in P}\sum_{k=0}^\infty\frac{1}{q^k}$$ is finite.

While at it, let me explain why your arithmetic conjecture is probably false - Dickson's conjecture implies take a prime $p$ such that $kp+1$ is not prime for $k<14$ and $q=14p+1$ is prime. Then $q$ is obviously LI (since by assumption it is the first prime with $q-1$ divisible by $p$). But the only odd factors of $q-1$ are $7$ and $p$, and neither $(q-1)/7=2p$ nor $(q-1)/p=14$ is a power of two times $q'-1$ for another prime $q'$. I tried finding such $p$ numerically but the least such is probably very large.

Edit: I have made a silly error in my code which caused me to miss small solutions. The least counterexample arising this way is for $p=167$, meaning $q=14p+1=2339$ is a counterexample to your conjecture.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ Thanks for your detailled answer. I did not know about Dickson's conjecture. $\endgroup$ Commented Nov 14 at 14:41
  • $\begingroup$ I have reasons to believe that this AC is true. $\endgroup$ Commented Nov 14 at 14:44
  • 1
    $\begingroup$ I would be curious what those reasons are, given your conjecture contradicts the (widely believed) Dickson's conjecture. $\endgroup$ Commented Nov 14 at 15:14
  • $\begingroup$ I have formulated some conclusions based on this conjecture and one of them is that the lattices generated by linear independent primes are isomorphic to the cubic lattices, which is the simplest possible. If you do not mind, I would use your nice proof in my writup: orges-leka.de/linear_independent_prime_numbers.pdf $\endgroup$ Commented Nov 14 at 15:21
  • 3
    $\begingroup$ @mathoverflowUser Please see my edit, turns out there exist rather small counterexamples. $\endgroup$ Commented Nov 14 at 15:40

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.