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MT 430 Intro to Number Theory
MIDTERM 2 PRACTICE
1. Material covered
Midterm 2 is comprehensive but will focus on the material of all the lectures
from February 19 up to April 4. Please review the following topics covered before
Midterm 1:
(1) Euler’s φ-function.
(2) Algebraic congruences. Hensel’s lemma.
(3) Euler’s criterion.
New topics for Midterm 2 are:
(1) Primitive roots of prime numbers.
(2) Solving algebraic congruences xa ≡ b (mod p).
(3) Legendre’s symbol.
(4) Gauss’ lemma.
(5) Quadratic reciprocity laws.
(6) Continued fractions.
(7) Representation of rational numbers by simple continued fraction using Euclidean algorithm.
(8) Infinite continued fractions.
(9) Representation of quadratic irrationals by periodic infinite continued fractions.
(10) Solving Pell’s equation using continued fractions.
To prepare for the midterm, review your lecture notes and redo Problems Sets
4, 5, 6 (also read and work through the posted solutions).
Here are a few extra practice problems:
2. Practice problems
Problem 1. What are all possible orders of elements in F∗19 ?
Solution: The order of any element in F∗19 divides 19 − 1 = 18. It follows that the
orders can be only 1, 2, 3, 6, 9, 18. To see that these orders are possible, consider
any primitive root ζ. Then ζ has order 18, ζ 2 has order 9, ζ 3 has order 6, ζ 6 has
order 3, ζ 9 has order 2, and of course ζ 0 = 1 has order 1.
Problem 2. Find all primitive roots of 19.
Solution: A primitive root of 19 is an element ζ ∈ F∗19 of order 18. By Problem 1,
an element of F∗19 is a primitive root as long as it does not satisfy equation x6 = 1
or x9 = 1.
We start checking non-zero residues one by one. Begin with 2: clearly, 26 =
64 ≡ 7 6≡ 1 (mod 19) and 29 = 512 ≡ −1 6≡ (mod 19). It follows that 2 is a
primitive root of 19. Every primitive roots of 19 can now be written as 2k , where
k ∈ {1, 2, . . . , 18} is coprime to 18:
21 , 25 , 27 , 211 , 213 , 217 .
1
2
MIDTERM 2 PRACTICE
Problem 3. Find all solutions of x12 ≡ 7 (mod 19) and x12 ≡ 6 (mod 19).
Solution: We use the fact that 2 is a primitive root of 19 from the previous problem.
Write x = 2k . Then because 7 ≡ 64 = 26 (mod 19) the first equation can be
rewritten as 212k ≡ 26 (mod 19). This is equivalent to 18 | 12k − 6. The solution
to 12k − 6 ≡ 0 (mod 18) are k ≡ 2 (mod 3) or k ≡ 2, 5, 8, 11, 14, 17 (mod 18). It
follows that the solutions of x12 ≡ 7 (mod 19) are
x = 22 = 4
(mod 19)
5
x = 2 = 32 ≡ 13
x = 28 ≡ 9
(mod 19)
11
≡ 15
14
≡6
17
≡ 10
x=2
x=2
x=2
(mod 19)
(mod 19)
(mod 19)
(mod 19)
Since we’ve discovered that 6 ≡ 214 (mod 19), the second equation can now be
rewritten as 212k ≡ 214 (mod 19) or 12k ≡ 14 (mod 18). This equation clearly has
no solutions because 6 - 14.
Problem 4. Is 8 a square modulo 31? Does equation x2 + 6x + 1 ≡ 0 (mod 31)
have a solution?
Solution: Complete the square to re-write the given equation as
(x + 3)2 ≡ 8
(mod 31).
The question is now whether 8 is a quadratic residue modulo 31. Clearly
3 2
2
8
=
=
.
31
31
31
Finally, because 31 ≡ 7 (mod 8), we conclude that 2 is a square modulo 31. (Indeed
2 ≡ 64 (mod 31).) It follows that 8 is also a square modulo 31.
The answer to both questions is thus YES.
15
15
and
.
Problem 5. Compute Legendre’s symbols
17
31
Solution: First,
15
17
=
3
17
5
17
.
By quadratic reciprocity,
17
17
2
3
= (−1)(3−1)(17−1)/4
=
=
= −1.
17
3
3
3
Also,
5
17
It follows that
= (−1)
(5−1)(17−1)/4
15
17
=
3
17
17
5
5
17
2
2
=
= (−1)(5 −1)/8 = −1.
5
= (−1)(−1) = 1.
MIDTERM 2 PRACTICE
3
Next,
15
31
=
3
31
5
31
.
By quadratic reciprocity,
31
1
3
(3−1)(31−1)/4
= (−1)
=−
= −1.
31
3
3
Also,
5
31
(5−1)(31−1)/4
= (−1)
31
5
1
=
= 1.
5
It follows that
15
31
=
3
31
5
31
= (−1)(1) = −1.
Problem 6. For which primes p is 6 a square modulo p?
Solution: Clearly, 6 is a square modulo p = 2 and modulo p = 3. Suppose now
p ≥ 5. Then
2
3
6
=
.
p
p
p
Recall that
(
1
2
=
p
−1
if p ≡ 1, 7 (mod 8)
.
if p ≡ 3, 5 (mod 8)
Next,
p
p
3
= (−1)(3−1)(p−1)/4
= (−1)(p−1)/2
.
p
3
3
p
Clearly, (−1)(p−1)/2 depends on the residue of p modulo 4 and
depends
3
2
3
on the residue of p modulo 3. Summarizing, both
and
depend only on
p
p
2
= 1 and
the residue of p modulo 24. For example, if p ≡ 1 (mod 24), then
p
p p
3
1
6
= (−1)(3−1)(p−1)/4
=
=
= 1, so that
= 1.
p
3
3
3
p
Checking the reduced
residue system modulo 24, we conclude that if p ≡ 1, 5, 19, 23
6
(mod 24), then
= 1 and so 6 is a square modulo p ≥ 5. If p ≡ 7, 11, 13, 17
p
(mod 24), then 6 is not a square modulo p ≥ 5.
Problem 7. Calculate the continued fraction expansion of
Solution: Apply the Euclidean algorithm:
676
.
107
4
MIDTERM 2 PRACTICE
676 = 6 ∗ 107 + 34
107 = 3 ∗ 34 + 5
34 = 6 ∗ 5 + 4
5=1∗4+1
4=4∗1
It follows that
676
= h6, 3, 6, 1, 4i.
107
Problem 8. Find the infinite continued fraction representation of
√
20.
Solution: First, we compute
√ the infinite continued fraction expansion of the reduced
quadratic irrational 4 + 20. It turns out to be h8, 2i. Indeed, we start with
a0 = 8, m0 = 4, q0 = 1. Then
m1 = a0 q0 − m0 = 4,
q1 = (20 − m21 )/1 = 4,
√
a1 = b(4 + 20)/4c = 2
and
m2 = a1 q1 − m1 = 8 − 4 = 4,
q2 = (20 − m22 )/4 = 1,
√
a2 = b4 + 20c = 8
From now on the values
√ of ai repeat themselves with period 8, 2.
It follows that 4 + 20 = h8, 2i = h8, 2, 8, 2, . . . i and so
√
20 = h4, 2, 8, 2, 8, . . . i = h4, 2, 8i.
Problem 9. Find the irrational number having continued fraction expansion h4, 4, 8i.
√
Solution: 18.
hn
hn+1
and
are two successive convergents of a
kn
kn+1
a
in the open interval
number x. Show that for any reduced rational number
b
(hn /kn , hn+1 /kn+1 ) one necessarily has b ≥ kn + kn+1 .
Conclude that hn /kn is the best possible approximation of x among all rational
numbers a/b satisfying b ≤ kn .
Problem 10. Suppose that
Solution: Without loss of generality, we assume that n is even, so that
Then
hn
hn+1
<
.
kn
kn+1
hn+1
hn
hn+1
a
a hn
−
=
−
+
−
.
kn+1
kn
kn+1
b
b
kn
Clearing the denominators and using the fact that hn+1 kn − hn kn+1 = (−1)n = 1
we obtain
1
(hn+1 b − akn+1 ) (akn − bhn )
=
+
.
kn kn+1
kn+1 b
bkn
MIDTERM 2 PRACTICE
5
Let S = (hn+1 b − akn+1 ) and T = (akn − bhn ). Then because hn /kn < a/b <
hn+1 /kn+1 , we must have S, T > 0. Because S and T are actually integers, we
obtain S, T ≥ 1.
We now have
1
S
T
=
+
.
kn kn+1
kn+1 b bkn
Multiplying by kn kn+1 b, we obtain
b = Skn + T kn+1 ≥ kn + kn+1 .