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Show that if there are 3 primes (>3) in arithmetic progression then d (the common difference) is a multiple of 6
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\[\text{If a,b and c are three numbers( prime ).}\] \[\text{then, b - a = c - b = d .....(I can't think anything good from here)}\]Now the first A.P which fits is 5, 11, 17 with d=6
d must be even, right?
Yes because all primes > 3 are odd.
How about is d divisible by 3?
ah I can't prove it.
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Then another A.P 7,13 ,19
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.
That wake u up:-)
No medals, Joe might look to look when he comes back.....
Okay : ) lol
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