prime number question

Jan 2016
104
2
Blackpool
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!
 
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Dec 2015
1,069
166
Earth
Try \(\displaystyle 20n^2 -1 =11n^2 +9n^2 -1 \) , now \(\displaystyle 11\) divides \(\displaystyle 9n^2 -1 =(3n-1)(3n+1)\).
 
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Dec 2015
1,069
166
Earth
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 \(\displaystyle n\geq 13/3 > 12/3=4\).:giggle:
 
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