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Number Theory Composites

The Chaz

Member
Jan 26, 2012
24
1. There should be a separate (sub)forum for NT. ... and one for abstract algebra, for that matter!

2. Show that there are infinitely many n such that both 6n + 1 and 6n - 1 are composite. Without CRT, if possible.

My work... let n = 6^{2k}.
Then 6n \pm 1 = 6^{2k + 1} \pm 1...
Hmm. Having a hard timing finding the LaTexification button from my iPhone...
To be continued.
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Let n = 35x - 1, for any positive integer x.

6n + 1 = 6(35x - 1) + 1 = 210x - 5 = 5(42x - 1)
6n - 1 = 6(35x - 1) - 1 = 210x - 7 = 7(30x - 1)

QED

PS: there is no LaTeX yet
 

The Chaz

Member
Jan 26, 2012
24
Let n = 35x - 1, for any positive integer x.

6n + 1 = 6(35x - 1) + 1 = 210x - 5 = 5(42x - 1)
6n - 1 = 6(35x - 1) - 1 = 210x - 7 = 7(30x - 1)

QED

PS: there is no LaTeX yet
Slick.
I also like n = 36k^3.
Then 6n \pm 1 is a sum/difference of cubes...