# Prove or disprove: There exists a prime p > 3 such that p + 2 and p + 4 are also prime [duplicate]

I'm having a lot of difficulties with this proof. Can someone please solve it and explain to me what's going on at each step? Thank you!

## marked as duplicate by Martin Sleziak, Namaste discrete-mathematics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 27 '17 at 1:31

• Hint: $n, n+2, n+4$ all have different remainders when you divide by $3$. – vadim123 Apr 13 '15 at 22:06
• I proved the theorem for n = 2. Is that not enough? – Adam Apr 13 '15 at 22:09
• @Adam - no... for example, why doesn't 37 work? or 59? or 107? – Joffan Apr 13 '15 at 22:14

Following my hint in the comments, one of the numbers $\{n,n+2, n+4\}$ must always be a multiple of $3$. However the hypotheses of the problem are that all three are prime. Hence one of them must be the specific prime $3$, as that is the only prime number that is also a multiple of $3$. So there are three cases:

1. $n=3$. Excluded, since $p>3$ forbids $p=3$.

2. $n+2=3$. Then $n=1$, which is not prime.

3. $n+4=3$. Then $n=-1$, which is also not prime.

• +1 but why did you use "n" at all? – Kim Jong Un Apr 13 '15 at 22:19
• Because $p$ is always prime, while $n$ may or may not be. – vadim123 Apr 13 '15 at 22:51
• @vadim123 why must one of the numbers be a multiple of 3? – shoestringfries Oct 13 '15 at 1:18
• @shoestringfries, because $n, n+2, n+4$ all give different remainders when you divide by $3$, and there are only three possible remainders. – vadim123 Oct 13 '15 at 1:30

$p$ must be odd and greater than $3$. Thus

$$~~~~~p + 2 \equiv 2 \pm 1 \pmod 3$$ $$\Leftrightarrow p + 4 \equiv 1 \pm 1 \pmod 3$$

If the sign is plus, $p+2 \equiv 3 \equiv 0 \pmod 3$. If the sign is minus, $p+4 \equiv 0 \pmod 3$. In either case, one of these numbers is divisible by $3$, and since they can't be $3$ themselves by assumption, they cannot be prime.

• So the theorem is disproven? – Adam Apr 13 '15 at 22:21
• @Adam Yes, through reductio ad absurdum. – Columbo Apr 13 '15 at 22:22
• Got it. Thanks mate! – Adam Apr 13 '15 at 22:36