# Conjecture: If $N$ and $N + 10$ are prime, then $N + 20$ is composite.

Conjecture: If $N$ and $N + 10$ are prime, then $N + 20$ is composite.

Here are some examples:

• $19$ and $29$ are prime; $39$ is composite.
• $241$ and $251$ are prime; $261$ is composite.
• $733$ and $743$ are prime; $753$ is composite.
• $1627$ and $1637$ are prime; $1647$ is composite.

I found this list of primes useful.

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What is the smallest prime factor of $N+20$ in all these cases? –  Daniel Fischer Jun 15 '14 at 18:25
But 3, 13 and 23 are primes (even if one of them is clearly divisible by 3). –  Jeppe Stig Nielsen Jun 15 '14 at 18:29
You might find this link helpful: oeis.org/A023203 –  Robert Soupe Jun 15 '14 at 19:06
Conjecture: if $N$ and $N+1000$ is prime then $N+2000$ is composite :) –  mesel Jun 16 '14 at 8:50

## 3 Answers

If $N>3$ is prime then, $N \equiv 1,2 \pmod{3}$. Thus $N+10 \equiv N+1 \equiv 2,0 \pmod{3}$.But $N+10$ is also a prime therefore $N+10 \equiv 2 \pmod{3}$. This means $N \equiv 1 \pmod{3}$. Consequently $N+20 \equiv 0 \pmod{3}$. Hence always composite.

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with the exception $N=3$, as Jeppe pointed out –  Greg Martin Jun 15 '14 at 19:37

Hint $\ 3\,$ divides one of $\,n,\, n\!+\!10,\, n\!+\!20,\,$ since they are $\,\equiv\, n,\, n\!+\!1,\,n\!+\!2\pmod 3$

Thus if $\,n>3\,$ and $\,n\!+\!10\,$ are both prime then $\,3\nmid n\,$ and $\,3\nmid n\!+\!10,\$ so $\ 3\mid n\!+\!20.$

Remark $\$ You may find it instructive to generalize the above proof to show that if prime $\,p < n\,$ then at least one of the $\,p\,$ integers $\ n < n\!+\!a < n\!+\!2a <\ldots < n\!+\!(p\!-\!1)a\$ is composite.

More generally, it is true for any complete system of representatives $> 0$ for the integers mod $\,p\,$ that omits $\,p,\,$ since then the representative for $\,p\equiv 0\,$ is some $\,np > p,\,$ which is composite.

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This is not a conjecture, it's a fact the ancient Greeks knew, only that it's missing the one exception, $N = 3$ (which Jeppe pointed out recently but the ancient Greeks also knew).

You need to look at the sieve of Eratosthenes in the usual 10-column wide rows. Circle 2 and strike out the higher even numbers. There's a very obvious pattern, right? Hold off on 3, though, circle 5 instead and cross out the higher odd multiples of 5 (you don't have to cross out the even multiples of 5 unless you really want to for some reason or other). The pattern is still obvious, right? Now circle 3 and cross off its odd multiples that haven't already been crossed off because of 5.

What I'm getting at is the pattern that repeats every 3 rows. You haven't crossed off 19 because it's not a multiple of 2, 3 or 5; and you haven't crossed off 29 because it's not a multiple of 2, 3 or 5 either. But you have crossed off 39 because it's a multiple of 3. Similarly, you haven't crossed off 49 or 59, but you have crossed off 69. Of course you know 49 is not prime, but it helps make my point.

The language of congruences (e.g., $19 \equiv 1 \mod 3$) will help you out as you venture further to where the patterns are not as graphically obvious on the sieve.

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