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Let $(p_{n})_{n≥1}$ be the sequence of prime numbers.

Then my question is: How one can proves that there is infinitely many indices $n$ such that $p_{n+1}$ is not of the form $p_{n}+2d$. Already we know a finite set of those indices. Here $d≥1$. The case when $d=1$ is known to be true from this question: Prove that there is infinitely many indices $n$ such that $p_{n+1}$ is not of the form $p_{n}+2$

Safwane
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2 Answers2

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Think about congruences modulo $2d+1$. If there were $2d+1$ consecutive primes $>2d+1$ with difference $2d$, then they would cover all congruence classes modulo $2d+1$, and so one would be a multiple of $2d+1$, contradiction. So in any sequence of $2d+1$ consecutive primes each $>2d+1$ there are a pair not differing by $2d$.

Angina Seng
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For each $d$, note that every integer $k$ from $(2d+1)!+2$ to $(2d+1)!+2d+1$ is composite.

Let $p_n$ be the greatest prime such that $p_n\le (2d+1)!+1$. Then

  • If $p_n+2d\le (2d+1)!+1$, then $p_{n+1}>p_n+2d$ by the definition of $p_n$.
  • Otherwise $(2d+1)!+2\le p_n+2d\le (2d+1)!+2d+1$ and $p_n+2d$ is composite.
ajotatxe
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