To determine the least natural number ( k ) such that there are only finitely many consecutive primes ( p_n ) and ( p_{n+1} ) with ( k \mid (p_n + p_{n+1}) ), we analyze the divisibility of sums of consecutive primes for increasing values of ( k ).

Key Observations:

1. Small ( k ) Values (k=1,2,3,4,6): The user already showed these have infinitely many such pairs. For example:...

Short answer: There is no such k. For every positive integer k, there are infinitely many pairs of consecutive primes p, q with k | (p + q).

Explanation and references:

• Your proofs for k = 2, 3, 4, 6 can be streamlined. For example, modulo 3 (or 4, or 6), the primes > 3 occupy exactly two reduced residue classes. If only finitely many consecutive pairs have different residues, then beyond some ...

Here is what one can say in complete generality.

1. What you have already proved.
   Whenever the multiplicative group of units modulo (k) has at most two elements, one can run exactly the same “two‐class” Dirichlet argument you gave for (3,4,6). In fact:

   – if (\phi(k)\le1), then (k=1) or (2), and obviously (P_n+P_{n+1}\equiv0\pmod k) for infinitely many (n) (ev...

Let

p1 < p2 < p3 < …

denote the sequence of primes (so pi and pi+1 are consecutive).
For a fixed positive integer k put

Si = pi + pi+1 . (1)

The question is:

For which k is the set
Ak = { i : k | Si } (2)

finite, and what is the smallest such k ?

1. A quick remark about parity

Except for the first pair (2,3) the two primes in (1) are b...

An excellent and insightful question! You've made great progress on it. Let's break down your reasoning, see where it leads, and address the core of your question.

Summary of Your Proof Method

Your proofs for the cases k = 3, 4, 6 are perfectly correct. The underlying logic can be generalized:

1. Assumption: Assume there are only a finite number of pairs of consecutive primes $(P_n, ...

<think> I need to find the smallest natural number \( k \) such that only finitely many pairs of consecutive primes sum to a multiple of \( k \). The user has already shown that for \( k = 1, 2, 3, 4, 6 \), there are infinitely many such pairs, so \( k \) must be larger than 6.

The key point is that for these small ( k ), the residues modulo ( k ) are such that there are primes in every resid...

Write the primes in increasing order

  P1 = 2 , P2 = 3 , P3 = 5 , … , Pn , Pn+1 , …

and put

  Sn = Pn + Pn+1 .

The problem asks for the least natural number k such that only finitely many Sn are divisible by k.

1. What we can prove at present
   • Every Sn with n ≥ 2 is even, so k = 1 or 2 certainly will not work.
   • Using the elementary argument you gave (it can be rewritte...

Finding the Smallest Such That Finitely Many Consecutive Prime Pairs Sum to a Multiple of

This is an interesting number theory problem about the divisibility properties of consecutive prime sums. Let me work through it step by step.

Understanding the Problem

We need to find the smallest natural number such that there are only finitely many pairs of consecutive primes $(p_n, p_{...

This is an interesting question about the divisibility properties of sums of consecutive primes. Let me analyze this problem systematically.

First, let's understand what we're looking for: the smallest such that only finitely many pairs of consecutive primes satisfy .

Analysis of Small Values

You've correctly shown that cannot be 1, 2, 3, 4, o...