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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.


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

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same solution slightly different | Comment 4 of 5 |

Another slightly different way of seeing this:
All primes, "p", above 3 are odd, and none are divisible by 3
p^2 - 1 = (p+1)*(p-1) of these factors, one must be divisible by 3 and both must be divisible by 2, so (p^2 - 1) is a multiple of:
3*2*2 = 12, so mod(p^2,12)=1 for all such primes

  Posted by Larry on 2010-10-02 00:18:37
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