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Let p be the largest prime with 2010 digits. What is the smallest positive integer k such that p^2-k is divisible by 12?

 Jul 18, 2019
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I think that no matter how big the prime number is, you will always get the same answer:

(p^2 - k) mod 12 = 0

k =12n + 1, where n =0, 1, 2, 3.....etc.

So, the smallest poisitive integer k =12*0 + 1 = 1. This is because every prime number > 5 when squared - 1, has at least 2^2 x 3 = 12 as factors. Example: 109^2 - 1 =11880 = 2^3 * 3^3 * 5 * 11. Another example:1000133^2 - 1 =1000266017688 = 2^3 * 3^3 * 7 * 23 * 1553 * 18521. And this applies to all primes => 5.

 Jul 19, 2019

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