The tutor investigates a problem involving composite numbers.

For problem 4b, page 19, of his Elementary Number Theory (second edition), Dudley invites the reader to prove there are infinite n such that both 6n-1 and 6n+1 are composite. (Composite means not prime.)

The first such n I find is 20: 6(20)-1=119=7(17), while 6(20)+1=121=11(11). Now, adding a product of 7 and 11 (such as 77) to 20, we arrive at

6(97)-1=6(20+77)-1=6(20)-1 + 6(77)=119+6(77), which must be divisible by 7, since 119 is. Therefore, it’s composite.

Similarly,

6(97)+1=6(20+77)+1=6(20)+1 + 6(77)=121 + 6(77), which must be divisible by 11, since 121 is.

So, n=20+77k, where k is any integer, gives 6n-1, 6n+1 both composite.

Source:

Dudley, Underwood. Elementary Number Theory, second ed. New York: W H Freeman and Company, 1978.

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