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I am learning about arithmetic progressions and I came across the question: "Prove that there is no infinite arithmetic sequence whose terms are all prime numbers."

I can see that there is no constant difference between all the prime numbers and hence an arithmetic sequence consisting only of prime numbers can't exist. However, I am unsure about how to prove this mathematically.

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  • $\begingroup$ Terms in an arithmetic progression have the form $na+b$ for constants $a$ and $b$. $a$ and $b$ have a common factor, or they are coprime. Consider these two cases separately. $\endgroup$
    – MJD
    Commented Jan 30, 2017 at 11:27
  • $\begingroup$ math.stackexchange.com/questions/251258/… $\endgroup$
    – lab bhattacharjee
    Commented Jan 30, 2017 at 11:44
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    $\begingroup$ You don't need all the primes to be in it. $\endgroup$
    – user645636
    Commented Jan 1, 2020 at 2:32

2 Answers 2

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Let us suppose this arithmetic sequence exists. Then, let the first term be p, and the common difference is d. Then, the (p+1)th term is p+pd, but it is a multiple of p, so it is composite. But, we assumed all terms are prime. Therefore, no arithmetic sequence has all terms as primes.

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  • $\begingroup$ and the $p+nd+n+1$ th term is divisible by the $n+1$ th in general. $\endgroup$
    – user645636
    Commented Jan 1, 2020 at 2:36
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Here’s another proof. Suppose otherwise, and take an arithmetic progression with only primes. Let its first term be $n$, and its common difference be $d$. Take a prime $p>d$. There exists an integer $a$ such that $$p\mid da+n,$$ namely, $n$ times the modular inverse of $-d$. Therefore, $$d(p+a)+n$$ will be a multiple of $p$ bigger than $p$ in the progression, and we’re done.

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