Practice Midterm Solutions, Math 313
1 For which positiveintegers n that are relatively prime to 15 does
equal 1?
the Jacobi symbol 15
n
Solution : First note that for the Jacobi symbol to be defined, we
require that n is odd. Suppose first that n ≡ 1 (mod 4). Then
15
3
5
n
n
=
=
n
n
n
3
5
= 1, we require that either
and so to have 15
n
n n
n n
=
= 1 or
=
= −1.
3
5
3
5
The first of these occurs when
n≡1
(mod 3) and n ≡ 1, 4
(mod 5),
while the second corresponds to
n≡2
(mod 3) and n ≡ 2, 3
(mod 5).
Similarly, if n ≡ 3 (mod 4), we have
n n
15
3
5
=
=−
n
n
n
3
5
= 1, we require that either
and so to have 15
n
n
n
n
n
= 1,
= −1 or
= −1,
= 1.
3
5
3
5
The first of these occurs when
n≡1
(mod 3) and n ≡ 2, 3
(mod 5),
while the second corresponds to
n≡2
(mod 3) and n ≡ 1, 4
(mod 5).
In summary, we find, via the Chinese Remainder Theorem, that there
are precisely 8 residue classes for n modulo 60 for which 15
= 1.
n
They are (after a little work) given by
n ≡ 1, 7, 11, 17, 43, 49, 53, 59
(mod 60).
2 Show that there exist infinitely many primes p ≡ 1 (mod 4). Hint
: recall Assignment 1; you may assume without proof, if p is an odd
prime, that there exists an integer x such that x2 ≡ −1 (mod p) iff
p ≡ 1 (mod 4)
1
2
Solution : There are several ways to approach this problem. Let us
suppose that there are only finitely many primes congruent to 1 modulo
4, say
p1 < p2 < · · · < pk .
Set
N = (pk !)2 + 1.
Then we have that N is odd and (arguing as in Euclid’s proof) not
divisible by any of the primes p1 , p2 , . . . , pk . But N must have an odd
prime divisor, say q, so that
(pk !)2 ≡ −1
and so
(mod q)
−1
= 1.
q
It follows that q ≡ 1 (mod 4), contradicting our initial assumption.
We conclude that there exist infinitely many primes congruent to 1
modulo 4.
3 We calculate the continued fraction expansion of e as
e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, ...].
Find the first 5 convergents to e.
Solution : We have
1
= 3,
1
2
8
1
= ,
C2 = [2; 1, 2] = 2 +
1 = 2+
3
3
1+ 2
1
3
11
C3 = [2; 1, 2, 1] = 2 +
=2+ =
1
4
4
1 + 2+ 1
C0 = [2] = 2, C1 = [2; 1] = 2 +
1
and
C4 = [2; 1, 2, 1, 1] = 2 +
1
1+
1
2+
1
1
1+ 1
=2+
5
19
= .
7
7
4 Suppose that
α = [a0 ; a1 , . . . , an ],
where ai ∈ Z for each i and we have that ai ≥ 2 for i ≥ 1. If we write
Ck = pk /qk for the kth convergent, prove that qk ≥ 2k .
Solution : We proceed by induction upon k. We have that
q0 = 1 = 20 and q1 = a1 ≥ 2 = 21 .
For k ≥ 2, we have that
qk = ak qk−1 + qk−2 .
3
Given k ≥ 2, suppose that we have, for each i ≤ k, qi ≥ 2i . It follows
that
qk+1 = ak+1 qk + qk−1 ≥ 2 · qk + qk−1 ≥ 2 · 2k + 2k−1 > 2k+1 ,
which, by the principle of mathematical induction, implies the desired
claim.
5 Prove that if p ≥ 7 is prime, then there are always at least two
consecutive quadratic residues, say a and a + 1, modulo p.
Solution : We been by observing that
2 10
10
10
2
5
10
1=
=
.
=
p
p
p
p
p
p
It follows that we cannot have
10
2
5
=
=
= −1
p
p
p
and so
a
a+1
=
=1
p
p
for one of a ∈ {1, 4, 9}.