Question

Show that if there are 3 primes (>3) in arithmetic progression then d (the common difference) is a multiple of 6

Answer 1

\[\text{If a,b and c are three numbers( prime ).}\]
\[\text{then, b - a = c - b = d .....(I can't think anything good from here)}\]Now the first A.P which fits is 5, 11, 17 with d=6

Answer 2

d must be even, right?

Answer 3

Yes because all primes > 3 are odd.

Answer 4

How about is d divisible by 3?

Answer 5

ah I can't prove it.

Answer 6

Then another A.P 7,13 ,19

Answer 7

Prime p >3 must be of form 3k+1 or 3k+2 and each d of form 3r+1 or 3r +2 and if u just work it out for p+d or p+2d, 3 divides.

So with 2 and 3 as divisors, d is a multiple of 6.

Answer 8

That wake u up:-)

Answer 9

No medals, Joe might look to look when he comes back.....

Answer 10

Okay : )
lol