# prime number question

#### Jaket1

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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#### idontknow

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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#### youngmath

20 n^2 = 1 (mod 11)
n = 4
20 Ã— 4^2 = 1 (mod 11)

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#### idontknow

You can solve the problem in many ways but this one doesnâ€™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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