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OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

d must be even, right?

OpenStudy (anonymous):

Yes because all primes > 3 are odd.

OpenStudy (anonymous):

How about is d divisible by 3?

OpenStudy (anonymous):

ah I can't prove it.

OpenStudy (anonymous):

Then another A.P 7,13 ,19

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

That wake u up:-)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Okay : ) lol

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