Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find all numbers $p$ such that all six numbers $p$, $p+2$, $p+6$, $p+8$, $p+12$, $p+14$ are primes

I know that $p=5$ works, but I don't know how to find all values for $p$, if any?

share|improve this question

1 Answer 1

Hint:

Consider these numbers modulo $5$ $$p,p+2,p+1,p+3,p+2,p+4$$ This is a complete set of residues. Hence, one of these number must be divisible by $5$.

This means that $p\leq 5$, otherwise all these numbers would be greater than $5$ and one of them will be a multiple of $5$. Hence, one of them is composite. (contradicting the fact that all of them are primes.)

share|improve this answer
    
So it doesn't matter that there are two p+2 elements in the set of residues modulo 5? Also, does this mean there exists only one number p? –  flamingohats Dec 27 '12 at 14:42
    
@flamingohats I edited my answer to include an answer for the question in your comment –  Amr Dec 27 '12 at 16:06
    
(+1) @flamingohats: Since $5$ must divide one of the numbers and all of them are prime, then one of the numbers must be $5$. It does matter, then, that there are two $p+2$ elements in the set of residues modulo $5$. In particular, it means that we cannot allow $5$ to divide $p+2$ (for then $p+2=5$, so $p+12=15$, which is not prime). Under the assumption that $p>0$, the only option left is $p=5$. –  Cameron Buie Dec 27 '12 at 16:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.