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 $\displaystyle 9n^2 1 =(3n1)(3n+1)$.
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) 
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. 3n1 and 3n+1 must be both prime and $\displaystyle n=2k$ which leads to the known form of twin primes $\displaystyle (6k1,6k+1)$. A weak shortcut just to prove n=4 : $\displaystyle 3n1 \geq 12 $ or $\displaystyle n\geq 13/3 > 12/3=4$. Last edited by idontknow; April 24th, 2019 at 10:22 AM. 

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