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April 24th, 2019, 06:17 AM   #1
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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?
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Last edited by skipjack; April 24th, 2019 at 10:11 AM.
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April 24th, 2019, 07:37 AM   #2
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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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Last edited by idontknow; April 24th, 2019 at 07:43 AM.
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April 24th, 2019, 07:57 AM   #3
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20 n^2 = 1 (mod 11)
n = 4
20 × 4^2 = 1 (mod 11)
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April 24th, 2019, 10:07 AM   #4
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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 $\displaystyle n\geq 13/3 > 12/3=4$.
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Last edited by idontknow; April 24th, 2019 at 10:22 AM.
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