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MATH 310: Homework 7
Due Thursday, 12/1 in class
Reading: Davenport III.1, III.2, III.3, III.4, III.5
1. Show that x is a root of unity modulo m if and only if (x, m) = 1.
(Hint: Use Euler’s theorem and #4 of HW5)
To say that x is a root of unity means xn ≡ 1 for some n > 0. Hence,
xn = 1 + km for some k, which implies (xn , m) = 1. Using #4 of HW5,
we conclude (x, m) = 1. Conversely, if (x, m) = 1 then by Euler’s
Theorem xφ(m) ≡ 1 (mod m) so that x is a root of unity.
2. Find all primitive roots mod 11.
Using #1 and #2 of HW8 with p = 11:
25 ≡ −1 implies 2 is primitive;
35 = (32 )(33 ) ≡ (−2)(−6) ≡ 1 implies 3 is not;
45 ≡ (25 )2 ≡ 1 implies 4 is not;
55 = (52 )(53 ) ≡ (3)(4) ≡ 1 implies 5 is not;
65 ≡ (25 )(35 ) ≡ −1 implies 6 is primitive;
75 ≡ (−4)5 ≡ −1 implies 7 is primitive;
85 ≡ (−3)5 ≡ −1 implies 8 is primitive;
95 ≡ (−2)5 ≡ 1 implies 9 is not;
102 ≡ 1 implies 10 is not.
Total # should be φ(φ(11)) = 4: 2,6,7,8
3. Find all primitive roots mod 27.
There are φ(φ(27)) = φ(18) = 6 of these. Using the results from the
section More Bonus Stuff of HW8 solutions, we only need to check
that g 9 ≡ −1 and g 3 6≡ −1. Using this, we see that 2 is a primitive
root since 23 = 8 and 29 ≡ −1. We can skip checking any g such that
3|g, since g 3 ≡ 0 (mod 27). We also know that 4 cannot be primitive,
since it is a square. We see that 5 is a primitive since 53 ≡ −10 and
59 ≡ −1.
Next, 73 ≡ −8 and 79 ≡ 1 shows that 7 is not primitive. However, this
calculation shows that −7 ≡ 20 must therefore be primitive. Note also
we can see from this calculation that ±8 ≡ 8, 19 cannot be primitive
(because they are cubes). Similarly, ±10 ≡ 10, 17 cannot be primitive.
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Similarly, we find 11 is primitive because 113 ≡ 8 and 119 ≡ −1, while
13 is not because 133 ≡ 10 and 139 ≡ 1, but that means −13 ≡ 14 is
primitive. Only one possibility remains, namely 23 ≡ −4, which is now
easily seen to be primitive.
Hence, the 6 primitive roots to the modulus 27 are: 2, 5, 11, 14, 20, 23.
4. Find a primitive root to the modulus 73. (Hint: 10 and 2 are solutions
to x8 ≡ 1 and x9 ≡ 1 (mod 73), respectively.)
Note that φ(73) = 72 = 8·9. Recall that the order of the product of two
numbers to any modulus is the product of their orders provided they
are relatively prime. (This was used in the proof of the multiplicative
property of the Euler φ-function.) Hence, it is enough to show that 10
has order 8 and 2 has order 9.
To show that 10 has order 8, it is enough to verify that 104 6≡ 1 (mod 73).
Indeed, 102 = 100 ≡ 27 so that 104 ≡ 729 ≡ −1 6≡ 1. Similarly, to see
that 2 has order 9, we check that 23 6≡ 1 (mod 73), which is clear.
5. Show that if g is a primitive root to the modulus p2 , then it is also a
primitive root to the modulus p. Is the converse true?
Let g be a primitive root to the modulus p2 . This is equivalent to
saying g r 6≡ 1 (mod p2 ) for any 0 < r < φ(p2 ) = p(p − 1). Proceeding
by contradiction, suppose g is not a primitive root to the modulo p.
That means there exists 0 < h < p − 1 such that g h ≡ 1 (mod p).
Hence, g h = 1 + kp for some integer k. Raising both sides to the power
p, we obtain
g ph = (1 + kp)p = 1 + p2 (k + . . . ) ≡ 1 (mod p2 ).
But this is a contradiction, since 0 < ph < φ(p2 ).
The converse holds for p = 3 because 2 is the only primitive root to
the modulus 3 and it happens to be primitive to modulus 9 since it has
φ(9) = 6 distinct powers: 2, 4, 8, 7, 5, 1
The converse also holds for p = 5 because the two primitive roots to
the modulus 5 are ±2 which happen to be primitive to modulus 25
since each has φ(25) = 20 distinct powers:
±2, 4, ±8, −9, ±7, −11, ±3, 6, ±12, −1, ∓2, −4, ∓8, 9, . . . , 1.
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Alternatively, we can use More Bonus Stuff in HW8 solutions and
simply check that (±2)5 ≡ (±7) so that (±2)10 ≡ (±7)2 ≡ −1, which
implies both are primitive to the modulus 25.
When p = 7, we get a counterexample to the converse using either
primitive root g = 3 or 5 to the modulus 7, neither of which is primitive
to modulus 49 since 33 ≡ −22, 34 ≡ −17, 37 ≡ −18 so that 321 ≡ 396 ≡
4 6≡ −1 (mod 49) and 53 ≡ 27, 54 ≡ −12, 57 ≡ (54)(−6) ≡ −30 ≡ 19
so that 5( 21) ≡ (9)(8) 6≡ −1 (mod 49).
6. Find all sets of two decimal digits which can occur as the last two digits
of a perfect square.
Let n be a square. The problem asks for the possible values of n (mod 100).
In other words, we are asked to find all squares to the modulus 100.
Since n is a square to the modulus 100, it is also a square to the modulus 4. Hence, n (mod 4) = 0, 1. Similarly, n (mod 5) = 0, 1, 4. From
this, it follows according to the 3 possibilities that n (mod 25) = 0
or 1, 6, 11, 16, 21 or 4, 9, 14, 19, 24. By Chinese Remainder Theorem,
n (mod 100) = 0, 25 or o6, e1 or e4, e9 (where ‘e’ and ‘o’ stand for
any even or odd digit, respectively) giving a total of 22 squares to the
modulus 100.
19
33
−26
,
and
.
7. Calculate the Legendre symbols
73
73
73
−1
2
13
73
8
2
−26
=
=
=
=
= −1
73
73
73
73
13
13
13
2
73
16
4
19
=
=
=
=1
73
19
19
19
33
3
11
73
7
11
4
=
=
=
=−
=−
= −1
73
73
73
11
11
7
7
8. Determine which of the following congruences are soluble:
(a) x2 ≡ 125 (mod 1016)
Since 1016 = 23 × 127, the congruence has a solution if and only if it
has a solution to the modulus 8 as well as to the modulus 127. Since
125
−2
−1
2
=
=
= (−1)(+1) = −1
127
127
127
127
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it follows that 125 is not a square to the modulus 127 (in other words,
not a quadratic residue). Hence, (a) is insoluble.
(b) x2 ≡ 129 (mod 1016)
Since 129 ≡ 1 (mod 8), 129 is a square to the modulus 8. That is, there
exists a residue a to the modulus 8 such that a2 ≡ 129. (It’s obvious
that a = ±1, but the purpose
of introducing
the a will become clear
129
2
shortly.) Similarly, since
=
= 1, 129 is also a square
127
127
to the modulus 127. Let b be a residue to the modulus 127 such that
b2 ≡ 129 (mod 127). (Here, it is not as obvious how to determine b,
but doing that is not the point.) By the Chinese Remainder Theorem,
there is a residue c to the modulus 1016 such that c ≡ a (mod 8) and
c ≡ b (mod 127). (The fact that c is uniquely determined is irrelevant
for this problem.) Moreover, c2 ≡ 129 (mod m) holds for m = 8 and
m = 127. Since (8, 127) = 1, it follows that (b) is soluble.
(c) x2 ≡ 41 (mod 79)
This is insoluble since
79
−3
−1
3
41
2
41
=
=
=
=
=
= −1
79
41
41
41
41
3
3
(d) 41x2 ≡ 43 (mod 79)
Method 1. First, observe that the quadratic character of a residue, i.e.
whether or not it is a square, is the same as that of its multiplicative
inverse. In other words, if xy ≡ 1 (mod p) then x is a square to the
modulus p if and only if y is a square
the modulus
p. The quickest
to
y
xy
x
to conclude that
way to see this is to use the fact p = p
p
the RHS of this identity is one. Alternatively, one can argue that the
sum of the indices of x and y (relative to any primitive root of unity)
vanishes so that one is even if and only if the other is.1
Second, determine the quadratic character of 43 by the calculation
43
79
−1
7
43
1
=−
=−
=−
=−
= −1
79
43
43
43
7
7
1
The latter approach uses the existence of primitive roots of unity, which we also used
to establish the homomorphism property of the Legendre symbol.
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or more quickly via
43
79
=
−36
79
=
−1
79
= −1.
Now, let k be the solution to 41k ≡ 1 (mod 79). (In other words, let k
be the inverse of 41 in the group (Z/79Z)× .) Note that the fact that k
is exists (and is uniquely determined) is a consequence of the fact that
gcd(41, 79) = 1. Then (d) is soluble if and only if x2 ≡ 43k (mod 79)
is soluble, which it is because by the previous calculation and the fact
that (c) is insoluble, we have
43
k
43k
=
= (−1)(−1) = 1
79
79
79
so that (d) is soluble.
Method 2. First, multiply (d) by 2 to get an equivalent congruence
3x2 ≡ 7 (mod 79), then multiply by -26 to get x2 ≡ −182 (mod 79),
then calculate
55
5
11
4
2
−182
=
=
=
×−
= 1.
79
79
79
79
5
11
(e) 43x2 ≡ 47 (mod 79)
Follow Method 1 and compute
47
−32
−1
2
=
=
= (−1)(+1) = −1
79
79
79
79
to conclude that (e) is soluble.
Or, follow Method 2 and first multiply by 2 to get 7x2 ≡ 15 (mod 79),
then multiply by 11 to get −2x2 ≡ 7 (mod 79), then multiply by 39 to
get x2 ≡ 273 (mod 79), then compute
273
36
=
=1
79
79
(f) x2 ≡ 151 (mod 840)
Since 8 divides 840 and x2 ≡ −1 (mod 8) is insoluble, (f) is insoluble.
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9. (BONUS) Show that a primitive root of unity modulo a prime p cannot
be a quadratic residue modulo p.
Let g be a primitive root of unity modulo p. For any x 6≡ 0 (mod p),
x is a quadratic residue modulo p if and only if its index relative to g
is even. Since g has index one relative to itself, it is not a quadratic
residue.