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Question:

Prove or disprove that there are $n$ consecutive odd positive integers that are prime.

If my answer for the question above is correct, then a new question arises.

My Attempt:

Odd numbers consist of multiples of $5$. I think that address the question.

New Question:

Is there at most $3$ consecutive primes? If So how would someone tackle this?

barak manos
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JoeyAndres
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  • There are plenty of odd numbers that are not a multiple of $5$, such as $31$, $47$, $999$. For solution, hint: If we take $3$ consecutive odd numbers, one of them is divisible by $3$. – André Nicolas Jul 12 '14 at 11:48
  • @AndréNicolas I'm just wondering if you are talking about the first question. – JoeyAndres Jul 12 '14 at 11:59
  • If you are not talking about the first question, that means, there are at most 3 odd consecutive integer which are prime, which is 3, 5, 7. For numbers $5$ or more, we will stumble upon an odd integer divisible by 3. Right? – JoeyAndres Jul 12 '14 at 12:02
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    Yes. that's right. The only $3$ consecutive are $3$, $5$, $7$, which does not extend to $4$ consecutive. – André Nicolas Jul 12 '14 at 13:10

2 Answers2

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E.g. $17,19,21,23$ are $4$ consecutive odd numbers, none of which are divisible by $5$. So the observation implies the maximum number of consecutive odd numbers that are primes is at most $4$ (and we should be careful: there is one prime that ends in $5$).

We know $3,5,7$ are three consecutive odd numbers that are primes. So we next have one of two tasks:

  • Find an example of four consecutive odd numbers that are primes; or
  • Prove that no such example exists.

It's natural to consider the numbers $a,a+2,a+4$ modulo $3$ (to see if we find a factor of $3$).

  • If $a \equiv 0 \pmod 3$, then ...?
  • If $a \equiv 1 \pmod 3$, then ...?
  • If $a \equiv 2 \pmod 3$, then ...?
Rebecca J. Stones
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  • AndreNicolas above said that if we take 3 consecutive odd numbers, one of them is divisible by 3. How does 17, 19, 21, 23 implies that there are 4 consecutive odd primes? – JoeyAndres Jul 12 '14 at 12:52
  • They are four consecutive odd numbers that are not multiples of $5$. All I'm saying is that your attempt does not cover this case. (They're not all prime.) – Rebecca J. Stones Jul 12 '14 at 12:55
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Hint: All primes except for $2$ and $3$ are of the form $6n\pm1$.

Lucian
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