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Solutions to Math 332 Homework 9
2. Direct computation using (5.10) – (5.13) and (5.18) yields
311−1
14
2
7
311
3
× 7−1
2
2
=
= (−1)
=−
= −(−1) = 1
311
311 311
7
7
313−1
313
165
3
5
11
5
× 11−1
2
2
=
= 1 × (−1) × (−1)
=−
= −1
313
313 313 313
11
11
2
1891
−108
−2 × 33
−3
=
=
=
=1
1999
1999
1999
1999
6. (a) (37/43) ≡ 37
43−1
2
≡ 3721 ≡ (373 )7 ≡ (−1)7 ≡ −1 (mod 43). So (37/43) = −1.
(b) Note (19 − 1)/2 = 9. Now 13 × 1, 13 × 2, . . . , 13 × 9 ≡ 13 , 7, 1, 14 , 8, 2, 15 , 9, 3 (mod 19). Since
13, 14, 15 are the only numbers greater than 9, we have N = 3 and (13/19) = (−1)3 = −1.
(c) As in Problem 2,
323
−30
−2
3
5
=
=
= 1 × (−1) × (−1) = 1.
353
353
353 353 353
25. (a) Note that 175 = 52 × 7. x2 ≡ 6 ≡ −1 (mod 7) has no solution since (−1/7) = −1 by (5.11). So
x2 ≡ 6 (mod 175) has no solution.
(b) Note that 361 = 192 , so x2 ≡ 361 (mod 32 × 7 × 11) is solvable and thus has 23 = 8 solutions by
(5.5).
(c) x2 ≡ 41 (mod 26 ) has exactly 4 solutions by (4.14)(iii) since 41 ≡ 1 (mod 8).
x2 ≡ 41 (mod 53 ) has exactly 2 solutions by (4.13) since (41/5) = (1/5) = 1.
x2 ≡ 41 (mod 372 ) has exactly 2 solutions by (4.13) since (41/37) = (4/37) = 1.
x2 ≡ 41 (mod 733 ) has exactly 2 solutions by (4.13) since (41/73) = (73/41) = (−9/41) =
(−1/41) = 1.
Therefore number of solutions of x2 ≡ 41 (mod 26 × 53 × 372 × 733 ) is 4 × 2 × 2 × 2 = 32.
29. (a) x2 ≡ 17 (mod 25 ) has exactly 4 solutions by (4.14)(iii) since 17 ≡ 1 (mod 8).
x2 ≡ 17 (mod 132 ) has exactly 2 solutions by (4.13) since (17/13) = (4/13) = 1.
x2 ≡ 17 (mod 19) has exactly 2 solutions by (4.13) since (17/19) = (−2/19) = 1.
Therefore number of solutions of x2 ≡ 17 (mod 25 × 132 × 19) is 4 × 2 × 2 = 16.
(b) Note that 9 = 32 , so x2 ≡ 9 (mod 24 × 53 × 72 ) is solvable and thus has 23+2 = 16 solutions by
(5.5).
(c) x2 ≡ 57 (mod 27 ) has exactly 4 solutions by (4.14)(iii) since 57 ≡ 1 (mod 8).
x2 ≡ 57 (mod 75 ) has exactly 2 solutions by (4.13) since (57/7) = (1/7) = 1.
x2 ≡ 57 (mod 592 ) has exactly 2 solutions by (4.13) since (57/59) = (−2/59) = 1.
Therefore number of solutions of x2 ≡ 57 (mod 27 × 75 × 592 ) is 4 × 2 × 2 = 16.
35. Check that 79 is prime. (−3/79) = 1 by (5.13)(iii), so x2 ≡ −3 (mod 79) is solvable. Since 79 =
4 × 19 + 3, the solutions are x ≡ ±(−3)19+1 ≡ ±(34 )5 ≡ ±25 ≡ ±32 (mod 79) by 5-3.
38. (a) Check that 263 is prime. (2/263) = 1 by (5.12), so x2 ≡ 2 (mod 263) is solvable. Since 263 =
4 × 65 + 3, the solutions are x ≡ ±265+1 ≡ ±(28 )8 × 22 ≡ ±(−7)8 × 22 ≡ ±110 (mod 263) by 5-3.
(b) Check that 83 is prime. (−53/83) = (30/83) = (2/83)(3/83)(5/83) = −1 × 1 × (−1) = 1 by
(5.12) and (5.13), so x2 ≡ −53 (mod 83) is solvable. Since 83 = 4 × 20 + 3, the solutions are
x ≡ ±(−53)20+1 ≡ ±3021 ≡ ±(303 )7 ≡ ±257 ≡ ±514 ≡ ±(53 )4 ×52 ≡ ±424 ×52 ≡ ±(422 )2 ×52 ≡
±212 × 52 ≡ ±26 × 25 ≡ ±14 (mod 83) by 5-3.
(c) Check that 79 is prime. (20/79) = (22 × 5/79) = (5/79) = 1 by (5.13)(iv), so x2 ≡ 20 (mod 79)
is solvable. Since 79 = 4 × 19 + 3, the solutions are x ≡ ±2019+1 ≡ ±2020 ≡ ±(202 )10 ≡ ±510 ≡
±(53 )3 × 5 ≡ ±353 × 5 ≡ ±62 × 33 × 5 ≡ ±40 (mod 79) by 5-3.
42. (a) 4x2 − 12x + 5 ≡ 0 (mod 77) is equivalent to
(
y 2 ≡ 64 (mod 77)
8x ≡ y + 12 (mod 77)
by (5.3). y 2 ≡ 64 (mod 77) is in turn equivalent to
(
y 2 ≡ 64 (mod 7)
y 2 ≡ 64 (mod 11).
Since 64 = (±8)2 , y ≡ ±8 ≡ ±1 (mod 7) and y ≡ ±8 ≡ ±3 (mod 11) are easily seen to be
solutions of the respective congruences.
Solving 11b1 ≡ 1 (mod 7) and 7b2 ≡ 1 (mod 11), we get b1 ≡ 2 (mod 7) and b2 ≡ −3 (mod 11).
By the Chinese Remainder Theorem, the solutions for y are
y ≡ 11 × 2 × 1 + 7 × (−3) × 3, 11 × 2 × 1 + 7 × (−3) × (−3),
11 × 2 × (−1) + 7 × (−3) × 3, 11 × 2 × (−1) + 7 × (−3) × (−3)
≡ ±8, ±36 (mod 77).
(mod 77)
Multiply 8x ≡ y + 12 (mod 77) by suitable numbers to reduce coefficient of x to 1:
10 × 8x ≡ 10 × (y + 12)
(mod 77)
3x ≡ 10y − 34 (mod 77)
26 × 3x ≡ 26 × (10y − 34) (mod 77)
x ≡ 29y + 40 (mod 77).
Hence we get
x ≡ −36, −38, 6, −3
(mod 77)
upon substituting the values of y found earlier.
(b) 2x2 − x + 7 ≡ 0 (mod 91) is equivalent to
(
y 2 ≡ 36 (mod 91)
4x ≡ y + 1 (mod 91)
by (5.3). y 2 ≡ 36 (mod 91) is in turn equivalent to
(
y 2 ≡ 36 (mod 7)
y 2 ≡ 36 (mod 13).
Since 36 = (±6)2 , y ≡ ±6 ≡ ±1 (mod 7) and y ≡ ±6 (mod 13) are easily seen to be solutions of
the respective congruences.
Solving 13b1 ≡ 1 (mod 7) and 7b2 ≡ 1 (mod 13), we get b1 ≡ −1 (mod 7) and b2 ≡ 2 (mod 13).
By the Chinese Remainder Theorem, the solutions for y are
y ≡ 13 × (−1) × 1 + 7 × 2 × 6, 13 × (−1) × 1 + 7 × 2 × (−6),
13 × (−1) × (−1) + 7 × 2 × 6, 13 × (−1) × (−1) + 7 × 2 × (−6)
≡ ±6, ±20 (mod 91).
(mod 91)
Multiply 4x ≡ y + 1 (mod 91) by suitable numbers to reduce coefficient of x to 1:
23 × 4x ≡ 23 × (y + 1)
(mod 91)
x ≡ 23 × (y + 1)
(mod 91)
Hence we get
x ≡ −21, −24, 28, 18
(mod 91)
upon substituting the values of y found earlier.
43. 7x2 − x + 24 ≡ 0 (mod 36) is equivalent to
(
7x2 − x + 24 ≡ 0
7x2 − x + 24 ≡ 0
(mod 4)
(mod 9).
Upon reducing the coefficients, we get
(
x2 + x ≡ 0 (mod 4)
2x2 + x + 3 ≡ 0 (mod 9).
Observe that
=⇒
x ≡ 0, 1, 2, 3
(mod 4)
x + x ≡ 0, 2, 2, 0
(mod 4).
2
Hence the solutions to x2 + x ≡ 0 (mod 4) are x ≡ 0, 3 (mod 4).
Observe that
=⇒
x ≡ 0, 1, 2, 3, 4, 5, 6, 7, 8
(mod 9)
2x + x + 3 ≡ 3, 6, 4, 6, 3, 4, 0, 0, 4
(mod 9).
2
Hence the solutions to 2x2 + x + 3 ≡ 0 (mod 9) are x ≡ 6, 7 (mod 9).
Solving 9b1 ≡ 1 (mod 4) and 4b2 ≡ 1 (mod 9), we get b1 ≡ 1 (mod 4) and b2 ≡ −2 (mod 9). By the
Chinese Remainder Theorem, the solutions to 7x2 − x + 24 ≡ 0 (mod 36) are
y ≡ 9 × 1 × 0 + 4 × (−2) × 6,
9 × 1 × 0 + 4 × (−2) × 7,
9 × 1 × 3 + 4 × (−2) × 6,
≡ −12, 16, 15, 7 (mod 36).
9 × 1 × 3 + 4 × (−2) × 7
(mod 36)