Find all $n$ such that $n,n+2,n+4$ are all prime?

Asked Sep 02 '20 at 00:55

Active Sep 02 '20 at 00:55

Viewed 59 times

I am trying to solve the following problem:

Find all $n$ such that $n,n+2,n+4$ are all prime.

I know $n=3$ works but no other $n$ works. I suspect the answer is as following: If $p$ and $q$ are different primes, then $p\not\equiv 0 \pmod q$, otherwise $p$ would be a multiple of $q$. Then we have:

$$n\equiv n \pmod 3\\ n+2\equiv n+2 \pmod 3\\ n+4\equiv n+1 \pmod 3 $$

So, testing for $n=3$, we see that no other possibility exists because at least one of them must be multiple of $3$ greater than $3$. Is that correct?

asked Sep 02 '20 at 00:55

Red Banana

23,956
20
91
192

2

Yes, that's right. – jjagmath Sep 02 '20 at 00:58
1

If n>3 and is prime then must be congruent with 1 or 2 mod3 (other case will be divisible by 3). Case1: if n=1(mod3), then n+2=0(mod 3) absurd, and Case2: if n=2(mod 3), then n+4=6=0(mod3), also absurd. – MathTc Sep 02 '20 at 01:09
3

This depends, of course, on the modern convention that $n=1$ is not a prime, otherwise $n=1$ is a solution. Also there seems to be a tacit assumption that only positive primes are allowed, otherwise $n=-7$ is a solution. – bof Sep 02 '20 at 01:19
1

Yes, mod $3$ they're congruent to $,n,n+1,n+2,$ and one of $3$ consecutive integers is a multiple of $3$ – Bill Dubuque Sep 02 '20 at 03:19

Find all $n$ such that $n,n+2,n+4$ are all prime?

0 Answers0