Arithmetic progression and properties of prime numbers

Question

If three prime numbers, all greater than 3, are in AP, then their common difference must be divisible by both 2 and three.

I tried to prove it experimentally for first 10 prime numbers and it worked but was not satisfactory.Can it be proved algebracilly.

Thanking you all in advancs :-)

Yes the prime numbers for an AP. All the terms are greater than 3 — AkashdeepKar, Jun 07 '17 at 12:29
@Arnaldo I think the OP must mean consecutive, else we could consider all the natural numbers (an arithmetic progression with common difference $1$). — lulu, Jun 07 '17 at 12:29

score 2 · Accepted Answer · edited Jun 07 '17 at 12:36

2

Suppose $p,p+n,$ and $p+2n$ are prime, with $p>3$ . Then $p$ is odd. If $n$ is odd, then $p+n$ is even and therefore not prime, a contradiction. If $n$ is not divisible by $3$, then one of the three terms are divisible by three, and so that term is not prime, a contradiction.

edited Jun 07 '17 at 12:36

Jaideep Khare

19,293

answered Jun 07 '17 at 12:31

florence

12,819

score 1 · Answer 2 · answered Jun 07 '17 at 12:30

1

So you have prime numbers $p,p+a,p+2a$ and want to show that $a$ is divisible both by two and three. For that, look at the problem mod 2 and mod 3. What can your number $p$ be mod $2$, what possible values are there mod $3$? Then every case where $a$ is not divisible by two or three will yield a contradiction, as in these cases at least one of your numbers is itself divisible by two or three, and therefore would not be prime.

answered Jun 07 '17 at 12:30

Dirk

6,359

Thank you Florence and Bemte, your explanations helped a lot..... – AkashdeepKar Jun 07 '17 at 12:37

Arithmetic progression and properties of prime numbers

2 Answers2