University of British Columbia
Math 313, Midterm, February 2016 :
Solutions
Name (please be legible)
Signature
Student number
Instructions Show all your work. Remember to be as
clear as possible and to give reasons for your answers.
You may write on the backs of pages. No memory aids.
The test is worth 100 points and has 8 pages (including
this one).
1
2
3
4
1
2
1 (a) (15 points) Evaluate the Legendre symbol
only Legendre symbols.
105
113
using
Solution : Factoring 105, we have
105
3
5
7
=
113
113
113
113
and so, since 113 ≡ 1 (mod 4), via Quadratic Reciprocity, it
follows that
105
113
113
113
2
3
1
=
=
.
113
3
5
7
3
5
7
We have
2
1
(32 −1)/8
= (−1)
= −1 and
=1
3
7
and so, again appealing to Quadratic Reciprocity,
105
3
5
2
=−
=−
=−
= −(−1) = 1.
113
5
3
3
Math 313, Midterm
1 (b) (15 points) Evaluate the Legendre symbol
treating it as a Jacobi symbol.
3
105
113
by
Solution : Since 113 ≡ 1 (mod 4), using Quadratic Reciprocity
for Jacobi symbols,
3 105
113
8
2
2
=
=
=
=
= 1,
113
105
105
105
105
where the last equality is a consequence of the fact that
105 ≡ 1 (mod 8).
4
2 (a) (15 points) Find the finite simple continued fraction
expansion for 21
.
13
Solution : We have
21 = 1 · 13 + 8, 13 = 1 · 8 + 5, 8 = 1 · 5 + 3, 5 = 1 · 3 + 2
and 3 = 1 · 2 + 1, so that
21
8
1
=1+
=1+
13
13
1+
5
8
=1+
1
1
=1+
,
1
1 + 1+ 3
1 + 1+ 1 1
5
whereby, finally,
21
1
=1+
13
1 + 1+
1
1+
1
1
1+ 1
2
and hence
21
= [1; 1, 1, 1, 1, 2] = [1; 1, 1, 1, 1, 1, 1].
13
1+ 2
3
Math 313, Midterm
5
2 (b) (15 points) Find the rational number having the finite simple continued fraction expansion [1; 2, 3, 4, 5].
Solution : We have
p0 = a0 = 1, q0 = 1, p1 = a0 a1 + 1 = 3, q1 = a1 = 2,
p2 = a2 p1 + p0 = 3 · 3 + 1 = 10, q2 = a2 q1 + q0 = 3 · 2 + 1 = 7,
p3 = a3 p2 + p1 = 4 · 10 + 3 = 43, q3 = a3 q2 + q1 = 4 · 7 + 2 = 30
and
p4 = a4 · p3 + p2 = 5 · 43 + 10 = 225, q4 = a4 q3 + q2 = 5 · 30 + 7 = 157,
so that
[1; 2, 3, 4, 5] =
225
.
157
6
3 (20 points) Prove
that
if p and q are odd primes and
.
q = 2p + 1 then pq = −1
p
Solution : Since p is odd, we may write p = 2k + 1 for some
positive integer k, so that q = 2p + 1 = 4k + 3 is congruent
to 3 modulo 4. We thus have, from Quadratic Reciprocity,
that either
p
q
2p + 1
1
−1
p ≡ 1 (mod 4) =⇒
=
=
=
=1=
,
q
p
p
p
p
or that p ≡ 3 (mod 4), whereby
p
q
2p + 1
1
−1
=−
=−
=−
= −1 =
,
q
p
p
p
p
as desired.
Math 313, Midterm
7
4 (20 points) Prove that if p is an odd prime
and if q is
the least integer such that 0 < q < p and pq = −1, then q
is prime.
Solution : Let us suppose
that q is the least integer such
q
that 0 < q < p and p = −1, and also that q is composite
(and try to deduce a contradiction). Since we suppose
that q its composite, there necessarily exists integers a
and b such that q = a · b and also
1 < a ≤ b < q < p.
We thus have
q
ab
a
b
−1 =
=
=
,
p
p
p
p
and hence exactly one of ap = −1 or pb = −1, contradicting the minimality of q with this property. The resulting contradiction implies that q is prime.
8
Bonus :
Prove that for each k ≥ 1, there are infinitely
many primes p such that p ≡ 1 (mod 2k ).
Solution : See me if interested – this one is tricky!