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Three forever (Posted on 2010-09-30) Difficulty: 2 of 5
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?

No Solution Yet Submitted by Ady TZIDON    
Rating: 2.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
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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