$\displaystyle 3,5,7$ are primes.

And:

$\displaystyle 7-5=2=5-3$

Question:

Are there any triples of primes so the difference between any two of sequential number equals 2? (Except the above...)

Results 1 to 10 of 10

- Aug 6th 2010, 01:19 PM #1

- Aug 6th 2010, 01:56 PM #2

- Aug 6th 2010, 03:04 PM #3

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 12,028
- Thanks
- 848

Hello, Also sprach Zarathustra!

$\displaystyle 3,5,7$ are primes.

And: .$\displaystyle 7-5\:=\:2\:=\:5-3$

Are there any other triples of primes that differ by 2?

Since $\displaystyle 2$ is the only even prime,

. . any such triple of primes must be consecutivenumbers.*odd*

It can be shown that, in a set of three consecutive odd numbers,

. . one of them will be divisible by 3.

Crank out a list of three consecutive odd numbers:

. . $\displaystyle \begin{array}{c}\rlap{/}1-3-5 \\ 3-5-7 \\ 5-7-\rlap{/}9 \\ 7-\rlap{/}9-11 \\ \rlap{/}9-11-13 \\ 11-13-\rlap{//}15 \\ 13-\rlap{//}15-17 \\ \vdots \end{array}$

And we see that $\displaystyle 3-5-7$ is thesuch triple of primes.*only*

- Aug 6th 2010, 03:05 PM #4

- Aug 6th 2010, 03:17 PM #5

- Aug 7th 2010, 06:30 AM #6

- Joined
- Oct 2009
- Posts
- 769

- Aug 7th 2010, 09:17 AM #7

- Joined
- Dec 2007
- From
- Ottawa, Canada
- Posts
- 3,184
- Thanks
- 80

- Aug 7th 2010, 10:55 AM #8
Along those lines,

Green-Tao Theorem

- Aug 7th 2010, 12:36 PM #9

- Joined
- Dec 2007
- From
- Ottawa, Canada
- Posts
- 3,184
- Thanks
- 80

- Aug 7th 2010, 07:14 PM #10
More nutjob info here

Primes in arithmetic progression - Wikipedia, the free encyclopedia