Dr. Marques Sophie
Office 519
Number theory
Fall Semester 2014
marques@cims.nyu.edu
Quiz # 3
(10 points for each question).
Problems:
1. Determine if the congruence x2 ≡ 3(mod53) has a solution.
Solution:
3
The question is whether the Legendre symbol 53
is 1 or −1. Since 53 ≡ 1 (mod 4)
and since 53 ≡ 2 (mod 3), we have by Quadratic Reciprocity
3
53 2
=
=
53
3
3
Again by Quadratic Reciprocity, the last Legendre symbol is −1 since 3 ≡ 3(mod 8)
(or obverse that 2 ≡ −1 (mod 3) and use Quadratic reciprocity). This the original
quadratic congruence has no solutions.
21
;
2. Compute 53
Solution:
21
53
7
53
53
2
4
1·26
3·26
=
=
(−1)
(−1)
=
53
3
7
3
7
2
2
2
= (−1)
= (−1)(−1)(7 −1)/4 = (−1)(−1)12 = −1.
7
3
53
3. Give the number of element in U9 . Give a primitive root mod 9. Describe explicitly all the element of U9 in term of this primitive root mod 9.
Solution:
Number of elements: φ(9) = 6.
2i 6≡ 1mod9, for i ∈ {1, 2, ..., 6} and 26 ≡ 1 mod 9.
U9 = {2, 22 , 23 , 24 , 25 , 26 }.
4. Give explicitly all the elements of U2n where n ∈ N.
Solution:
(−1)s 5t where s ∈ {0, 1} and 0 ≤ t ≤ 2n−2 .
5. List all a ∈ (Z/35Z)∗ such that a is not a quadratic residue (mod 35).
Solution:
12 ≡ 1 mod35, 22 ≡ 4 mod35, 32 ≡ 9 mod35, 42 ≡ 16 mod35, 52 ≡ 25 ≡ −10 mod35
1
62 ≡ 1 mod35, 72 ≡ 39 ≡ 4 mod35, 82 ≡ 64 ≡ −6 mod35, 92 ≡ 81 ≡ 11 mod35
102 ≡ 100 ≡ −5 mod35, 112 ≡ 121 ≡ 16 mod35, 122 ≡ 22 62 ≡ 4 mod35
132 ≡ 169 ≡ −6 mod35, 142 ≡ 22 72 ≡ 4 × 4 ≡ 16 mod35
152 ≡ 32 ×52 ≡ 9×(−10) ≡ 15 mod35, 162 ≡ 22 82 ≡ (−6)×4 ≡ 11 mod35, 172 ≡ 289 ≡ 9 mod35
−10, −6, −5, 1, 4, 9, 11, 15, 16 are quadratic residue
and then −17, −16, −15, −14, −13, −12, −11, −9, −8, −7, −4, −3, −2, −1
2, 3, 5, 6, 7, 8, 10, 12, 13, 14, 17 are the non residue.
2