# Primes number $n,n+2,n+6,n+8,n+12,n+14$

Find all natural number $$n$$ such that all the following numbers are primes : $$n,\;\; n+2,\;\;n+6,\;\;n+8,\;\;n+12,\;\;n+14$$ are all prime numbers

• Consider these three: $$n, n+6, n+12$$ What do you notice about those three? – Zubin Mukerjee May 15 at 17:06

As $$n$$ is a prime, $$n\ge 2$$.

When $$n=2$$, $$n+2=4$$ is not prime.

When $$n=3$$, $$n+6=9$$ is not prime.

When $$n=5$$, all of them are prime.

$$(n+6)-n\equiv 1$$ (mod $$5$$)

$$(n+2)-n\equiv 2$$ (mod $$5$$)

$$(n+8)-n\equiv 3$$ (mod $$5$$)

$$(n+14)-n\equiv 4$$ (mod $$5$$)

At least one of these integers is a multiple of $$5$$.

When $$n>5$$, at least one of the numbers is not prime.

The only possible answer is $$5$$.

• Thank you very much Sir – user674299 May 15 at 17:34

Hint: $$n\equiv_5 n+0$$ $$n+2\equiv_5 n+2$$

$$n+6\equiv_5 n+1$$ $$n+8\equiv_5 n+3$$ $$n+14\equiv_5 n+4$$

So among $$n,n+2,n+6,n+8, n+14$$ exactly one is divisible by $$5$$ so one of them is $$5$$...

Note that 2,6,8,14 are all different mod 5. Furthermore, for each $$x=2,6,8,14$$, note that if $$n+x \equiv_5 0$$ then $$n+x$$ is a proper multiple of 5 i.e., $$5|n+x$$ but $$5 \not = n+x$$ so if $$5|n+x$$ then $$n+x$$ isn't prime. Thus the only way that $$n+x$$ can be prime for all $$x \in \{2,6,8,14\}$$ is if $$n \equiv_5 0$$. But the only way $$n \equiv_5 0$$ and $$n$$ still also be prime is if $$n$$ is 5 itself. So $$n=5$$ is the only possibility.

If you allow negative numbers for $$n$$ then $$n=-19$$ will work, and [if you consider $$\pm 1$$ to be prime] so will $$n=-1,-7,-13$$. But by the same reasoning as in the above pragraph $$|n| < 14+5+1$$ [as at least $$x \in \{0,2,6,8,14\}$$ one must satisfy $$|n+x| \le 5$$] so those are it.