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Find all natural numbers $n$ such that $n+1$, $n+5$, $n+7$, $n+11$, $n+13$, $n+17$, $n+23$ are all prime.

So far I've made the following progress on this problem:

a) n must be even. Otherwise some of the $n+x$ numbers would be even (since all $x$s are odd), and therefore not prime (they can't all be $2$).

b) Analyzing $n \mod 3$, it's possible to deduce that $n$ must in fact be divisible with $3$.

If $n ≡ 1 (mod 3)$, then $n + 5$ would be divisible by $3$, and therefore not a prime.

If $n ≡ 2 (mod 3)$, then $n + 1$ would be divisible by $3$, and therefore not a prime.

c) A similar analysis of $n \mod 5$ reveals that the only viable option would be $n ≡ 1 (mod 5)$. For any other option one of the numbers would be divisible by $5$.

Therefore, $n$ could end in either $1$ or $6$, but since it must be even (point a), then it always ends in $6$.

This leaves as candidates all numbers $6$, $36$, $66$, $96$, $126$, and so on, adding $30$ every time.

It's easy to check that $6$ is a solution. And I strongly suspect it's the only one, as I checked with a computer all possibilities up to 100 million.

However, I'm not able to prove that it's the only solution.

Jyrki Lahtonen
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Cristian Lupascu
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1 Answers1

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The numbers $$ n+1, n+5, n+7, n+11, n+13, n+17, n+23 $$ are pairwise non-congruent modulo seven, because $1,5,7,11,13,17,23$ have remainders $1,5,0,4,6,3,2$ respectively.

So one of them is a multiple of seven. The only multiple of seven that is a prime is $7$ itself. Therefore $7$ is one of those numbers leaving $n=6$ and $n=2$ as the alternatives (as well as $n=0$ if that is included into the natural numbers in your book). $n=6$ checks out, but $n=2$ does not. That doubtful candidate $n=0$ won't produce composites, but $n+1=1$ is not a prime.

Jyrki Lahtonen
  • 140,891