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 April 24th, 2019, 06:17 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 Thanks: 2 prime number question Find the smallest number n for which 11 divides 20n^(2)-1. For this question, I have made it so that 20n^2 is congruent to -1mod11; the quadratic residues of 11 are 1 goes to 1, 2 goes to 4, 3 goes to 9, 4 goes to 5, 5 goes to 3. But none of these contain the number -1, so -1 does not belong in the set of squares, so how am I supposed to work out this question? I must have done something wrong in my understanding, so could anyone explain to me what I have done wrong? Thanks! Last edited by skipjack; April 24th, 2019 at 10:11 AM.
 April 24th, 2019, 07:37 AM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 605 Thanks: 88 Try $\displaystyle 20n^2 -1 =11n^2 +9n^2 -1$ , now $\displaystyle 11$ divides $\displaystyle 9n^2 -1 =(3n-1)(3n+1)$. Thanks from Jaket1 Last edited by idontknow; April 24th, 2019 at 07:43 AM.
 April 24th, 2019, 07:57 AM #3 Senior Member   Joined: Mar 2019 From: iran Posts: 318 Thanks: 14 20 n^2 = 1 (mod 11) n = 4 20 × 4^2 = 1 (mod 11) Thanks from idontknow and Jaket1
 April 24th, 2019, 10:07 AM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 605 Thanks: 88 You can solve the problem in many ways but this one doesn’t need complicated theorems of twin primes. 3n-1 and 3n+1 must be both prime and $\displaystyle n=2k$ which leads to the known form of twin primes $\displaystyle (6k-1,6k+1)$. A weak shortcut just to prove n=4 : $\displaystyle 3n-1 \geq 12$ or $\displaystyle n\geq 13/3 > 12/3=4$. Thanks from Jaket1 Last edited by idontknow; April 24th, 2019 at 10:22 AM.

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