**Proposition:**

Prove (or disprove) that for every integer \(n \ge 4\), there exists at least one ordered triplet \( (p_1 , n , p_2) \) where \(p_1\), \(n\) and \(p_2\) are in arithmetic progression and \(p_1\) and \(p_2\) are distinct primes.

This problem came to me upon pondering over numbers. I do not know if this idea has already been discussed by an individual or any mathematical community.

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TopNewestSince, \(p_{1}, n, p_{2}\) are in an AP;

Therefore;

\(2n=p_{1} + p_{2}\)

Now, if Goldbach's conjecture is true; our proof will be complete.

There exists an AP for n=2 and 3 also because there is no restriction that tells that \(p_{1}, p_{2}\) are distinct; or the AP is non-constant.

Is there any conjecture for every even integer greater than 6 as sum of 2 distinct primes? – Yatin Khanna · 5 months ago

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This is the real deal. For now, even I have only observed this result but cannot prove it. – Tapas Mazumdar · 5 months ago

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– Spandan Senapati · 5 months ago

Hey try googling the prime distributions and something on Green-Tao Theorem( on arithmetic progressions of primes).See if the first is relevant to the propositionLog in to reply