Structure of Examination Paper:
• There are 3 pages including this page.
• There are TWO sections.
• There are 7 questions in total.
• There are no appendices
• The maximum mark for the examination paper is 100 and the mark obtainable for a question
of part of a question is shown in brackets alongside the question.
Instructions for completing the examination:
• Answer ALL questions from Section A and Section B.
• Show all your workings.
1MA2011/21
Section A
1. (a) Use Fermat’s Method to factorize 1147. [5]
(b) Euclid’s algorithm applied to two numbers a and b computed the quotients q1 = 3,
q2 = 3, q3 = 3, q4 = 3, q5 = 3 (in this order) and their greatest common divisor
gcd(a, b) = 3. Compute a and b. [5]
2. Using the extended version of Euclid’s algorithm, find a solution to the following Dio-
phantine linear equation.
77x + 91y + 143z = 2 [10]
3. Find all solutions for the following pair of simultaneous congruences.
262x ≡ 3 mod 807
3x ≡ 2 mod 5 [10]
4. Show that the equation
2x3 + 7y3 = 4
has no solution in integers. [10]
5. (a) Derive the continued fraction of √7. [5]
(b) Find the value of β, given its continued fraction expression β = [1, ̄7], i.e., a0 = 1
and ai = 7 for all i ∈ {1, 2, . . .}. [5]
– 2 –MA2011/21
Section B
6. The Euler’s function φ(m) counts the number of integers a with 0 ≤ a < m and
gcd(a, m) = 1.
(a) Let m = pa where p is prime. Show that φ(m) = m(1 − 1/p). [5]
(b) Let m = pa and n = qb where p and q are distinct primes. Show that
φ(mn) = φ(m)φ(n). [5]
(c) Compute φ(17) and φ(77). [4]
(d) Show that if gcd(a, m) = 1, then
aφ(m) ≡ 1 mod m. [11]
7. (a) Show that there exists a constant c > 0 such that
|b√7 − a| ≥ 1
cb ,
for all natural numbers a, b (b 6 = 0). [7]
(b) Compute two distinct, positive integer solutions to x2 − 17y2 = 1. [7]
(c) Let α > 0 be a real number and let pn/qn denote the corresponding n-th conver-
gent.
Show that ∣∣∣∣pn qn
− α
∣
∣
∣
∣ ≥ 1
2q2
n
, and
∣
∣
∣
∣
pn+1
qn+1
− α
∣
∣
∣
∣ ≥ 1
2q2
n+1
,
implies (qn+1 − qn)2 ≤ 0.
Conclude that either
∣
∣
∣ pn
qn − α
∣
∣
∣ < 1
2q2
n
or
∣
∣
∣ pn+1
qn+1 − α
∣
∣
∣ < 1
2q2
n+1
. [11]
