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 Three forever (Posted on 2010-09-30)
Choose a prime number greater than 3.
Multiply it by itself and add 14.
Divide by 12 and write down the remainder.
It will always be 3.

WHY?

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 solution | Comment 1 of 5
All prime numbers greater than 3 have an odd remainder themselves, when divided by 12. Furthermore, they cannot be congruent to 3 or 9 mod 12, as that value would always be a multiple of 3. Therefore, all prime numbers greater than 3 are either 1, 5, 7, or 11 mod 12.

The square of each of these remainders comes out to be a multiple of 12, plus 1. By adding 14, we are essentially adding 2 to the remainder, leading to a remainder of 3 for all primes greater than 3. That is:

1 * 1 + 14 = 1 + 14 = 15 = 1 * 12 + 3
5 * 5 + 14 = 25 + 14 = 39 =3 * 12 + 3
7 * 7 + 14 = 49 + 14 = 63 = 5 * 12 + 3
11 * 11 + 14 = 121 + 14 = 135 = 11 * 12 + 3

This trick can be repeated by adding any constant, c, and dividing (p^2 + c) by 2, 3, 4, 6, 8, or 24. In all of these cases, for p > 3, p^2 will become 1 mod x. At this point, adding any constant, c, to all p^2 values will result in a consistent remainder.

 Posted by Justin on 2010-09-30 15:27:31

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