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This is a question from the textbook 'Cambridge Senior Mathematics for Queensland', Exercise 4B.

(25) Show that there is no infinite arithmetic sequence whose terms are all prime numbers.

While the fact that there is no known pattern in prime gaps is enough to prove that an arbitrarily long arithmetic sequence consisting only of primes could exist, I am not sure that it is enough to prove that an infinite arithmetic sequence. Conversely, I can find no proven evidence that there does not exist an infinitely long arithmetic sequence whose terms are all prime numbers (though, perhaps I am looking in the wrong places).

Joseph Chung
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  • I don't know how I missed that question. Nevertheless, I would like to keep this question open as the answer given by @D_S is much more elementary than the answers given in that question. – Joseph Chung Feb 02 '22 at 09:40

2 Answers2

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Given any arithmetic sequence $$n, n+k, n+2k, ...$$ you will eventually get to the term $n + nk$, which is not prime.

Edit: my answer does not cover the case $n =1$. In this case, I still claim there is a composite number in the sequence

$$1, 1+k, 1+2k, ...$$

Let $p$ be any prime number which is bigger than $k$. Since $k$ and $p$ are relatively prime, there exist infinitely many positive integers $a$ such that $ak \equiv -1 \pmod{p}$. Thus infinitely many terms $1+ak$ in the arithmetic sequence are properly divisible by $p$ and hence not prime.

D_S
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  • ... Yeah, that should have been obvious. Thanks for the answer! – Joseph Chung Feb 01 '22 at 23:48
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    This answer would do well with a bit of polishing. What if $n=1$? (Not that I think it isn't obvious how to plug this gap, but I feel it should be done for completeness.) –  Feb 01 '22 at 23:49
  • What if $k = 0$ and $n$ is prime? – Joseph Chung Feb 01 '22 at 23:57
  • @JosephChung in that case, you do not have an arithmetic sequence at all, only a single number. – D_S Feb 02 '22 at 04:39
  • @StinkingBishop good point, I have edited. – D_S Feb 02 '22 at 04:39
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    It is a trivial arithmetic sequence, but it is still an arithmetic sequence. A teacher should probably thank the student the student for the contribution and ask that the question be answered for non-constant sequences. – Peter Feb 02 '22 at 08:08
  • Not a teacher, just a guy who likes answering interesting questions. And everyone who isn't this guy (https://imgur.com/a/P10zWP4 ) thinks of arithmetic sequences as having infinitely many terms. – D_S Feb 02 '22 at 15:27
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    An alternate way to deal with the $n=1$ case is just to consider $1+(k+2)k = (k+1)^2$. One way to see how one could derive similar expressions is to rewrite the $n$th term as $1+k+k(n-1)$. This gives rise to the general case of $n+(n+k+1)k = (n+k)(k+1)$. – Alexander51413 Feb 02 '22 at 17:12
  • @D_S yes, but do the infinitely many terms have to be unique? – Joseph Chung Feb 02 '22 at 23:46
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Not exactly what you're asking for, but Dirichlet's theorem says that if $(a,b)=1$ then the arithmetic progression $\{an+b\}, n\in\mathbb{N}$ contains infinitely many primes.

Luis Alexandher
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