# How do I prove that there is no strictly monotonically increasing arithmetic sequence in which all elements are primes?

Does anyone have an idea on how to prove this?

"There is no strictly monotonically increasing arithmetic sequence in which all elements are primes."

Any help appreciated! Thanks:)

• Well, write what it means to be a monotonically increasing arithmetic sequence. – Thomas Andrews Sep 17 '13 at 17:16

If $p$ is a polynomial with integer coefficients, and $k$ divides $p(n)$, then $k$ divides $p(mk+n)$ for any $m$.

In particular, this applies to $p$ a linear polynomial. Now note that terms in arithmetic progression are the consecutive values of a linear polynomial with integer coefficients: $a,a+b,a+2b,\dots$ are $p(0),p(1),\dots$ for $p(x)=a+xb$.

• That sounds good, but how do I know that there is an element in the sequence that's equal to p(mk + n)? – nitrogenhurricane Sep 17 '13 at 17:21
• The $(mk+n)$-th member of the sequence is is $p(mk+n)$. – Andrés E. Caicedo Sep 17 '13 at 17:22
• In case you are confused about this: For any $N$, no matter how large, there is a strictly increasing arithmetic progression of length $N$ consisting solely of primes. The question is about infinite sequences, so no matter how large $k,n,m$ are, the sequence has an $(mk+n)$-th term. – Andrés E. Caicedo Sep 17 '13 at 17:25
• Okay, it's clear now thanks, I just couldn't put the existence of element and that k divides it together:) Thanks! – nitrogenhurricane Sep 17 '13 at 17:26

Suppose the first term in the sequence is prime $p$. Then the $(p+1)$th term in the sequence is $p + pn$ for some positive integer $n$, since by assumption your sequence must be all integer-valued if it's all prime. Can this term $p + pn$ be prime?

• @nitrogenhurricane, in this answer the difference between two elements is $n$, not $p$. – Antonio Vargas Sep 17 '13 at 17:54

Consider the arithmetic sequence $a,a+d,a+2d,a+3d,\dots$. The sequence is increasing, so for some $k$ we have $a+kd\gt 1$.

Now consider $(a+kd)+ (a+kd)d$. This is in the sequence, for $$(a+kd)+(a+kd)d= a+(k+a+kd)d.$$ But $(a+kd)+(a+kd)d$ is not prime, since it is clearly divisible by $a+kd$, and greater than $a+kd$.

• Thanks for your answer, but as Andres was faster I think, so it's fair to accept his. Please anyone correct me if I'm wrong since I'm a newbie here. – nitrogenhurricane Sep 17 '13 at 17:27
• In my opinion the solution by user2566092 is much cleaner and better. Mine is essentially the same, but hides the simplicity of the idea. So I think you should not accept mine. – André Nicolas Sep 17 '13 at 17:31
• You should accept whichever answer you think best answers the question. It's okay for this to be a late answer (it's even okay to change your mind) – Ben Millwood Sep 17 '13 at 17:32