Download Assignments 1-2

Solutions to Assignment 1
A2 Induction
1. Let’s use induction on n 1.
Base case: If n = 1, then n(n+1)
= 1 and thus the base case holds.
2
Inductive case: Assume that 1 + 2 + · · · + n = n(n+1)
. Then
2
1 + 2 + · · · + n + (n + 1) =
n(n + 1)
n(n + 1) + 2(n + 1)
(n + 1)(n + 2)
+ (n + 1) =
=
.
2
2
2
By induction, the result follows for all n 1.
(We used induction to solve this problem, however there is a nice solution by pairing the first and last
numbers from 1 to n.)
3. Again, we use induction on n > 1.
Base case: If n = 2, then 1 + · · · + 2n 1 = 3 = 4 1 = 22 1.
Inductive case: Suppose that 1 + · · · + 2n 1 = 2n 1. Then
1 + · · · + 2n
1
+ 2n = 2n
1 + 2n = 2(2n )
1 = 2n+1
and the result follows for all integers n > 1.
3C Euclid’s Algorithm
35.(ii) The greatest common divisor of 43137 and 21063 is 3 because:
43137 = 21063 · 2 + 1011
21063 = 1011 · 20 + 843
1011 = 843 · 1 + 168
843 = 168 · 5 + 3
168 = 3 · 56 + 0
3D Bezout’s Identity
39.(ii) We use the EEA matrix:
1870
242
242 · 7
176 = 1870 242 · 7
66 = 242 176
66 · 2
44 = 176 66 · 2
22 = 66 44 · 7
44 = 22 · 2
0
1
7
7
8
16
23
31
1
0
0
1
1
2
3
4
1,
Therefore, 22 is the greatest common divisor of 1870 and 242. Moreover, 22 = 242 · 31 + 1870 · ( 4).
55. Suppose c is the least element of J, then we can find integers r, s such that c = ar + bs. Suppose
d = (a, b), then by Bezout’s Identity we can find integers x, y such that d = ax + by. This shows that d
is inside the set J and so c  d. On the other hand, since d divides a and b then d divides any linear
combination of a and b, in particular c. Explicitly, we can find integers m, n such that a = d · m and b = d · n
and so c = ar + bs = d(mr + ns) so d divides c. This implies that d  c and so c = d.
3E Linear Diophantine Equations
57.(i) Using an appropriate method we can find that (114, 270) = 6 (for the details look at the following
solution). By Proposition 12 we know that the solutions to 114x + 270y = 0 are of the form x = 270
6 k = 45k
and y = 114
k
=
19k,
where
k
is
any
integer.
6
57.(ii) We use the EEA matrix:
270
114
114 · 2
42 = 270 114 · 2
42 · 2
30 = 114 42 · 2
12 = 42 30
12 · 2
6 = 30 12 · 2
12 = 6 · 2
0
1
2
2
4
5
7
14
19
1
0
0
1
2
2
3
6
8
Therefore, 6 = 114 · 19 + 270 · ( 8) and so we can find that x0 = 19 · 4 = 76 and y0 = 32 is a solution
for 114x + 270y = 24. By Corollary 13, we have that the solutions of this equation are x = 76 + 45k and
y = 32 19k for every integer k.
58.(i) You can verify that (66, 45) = 3 and so the general solution to 66x + 45y = 0 is x = 45
3 k = 15k and
y = 66
k
=
22k
for
k
any
integer.
The
smallest
possible
x
>
0
is
when
k
=
1
and
this
yields
x = 15 and
3
y = 22.
60.(i) We already verified that 22 = (242, 1870) and that 22 = 242 · 31 + 1870 · ( 4). Therefore a solution
of 242x + 1870y = 66 is x0 = 93 and y0 = 12. Using Corollary 13 we obtain all solutions x = 93 + 85k and
y = 12 11k for any integer k.
62. We start using the EEA matrix to find (133, 203) and the Bezout relation.
203
133
70 = 203 133
63 = 133 70
7 = 70 63
63 = 7 · 9
0
1
1
2
3
1
0
1
1
2
Notice then that 7 = (133, 203) only divides 42 so the only equation having a solution is (iii). By Corollary
13, the solutions are x = 18 + 29k and y = 12 7k for every integer k.
Solutions to Assignment 2
4A
3. One needs to show both sides of the argument.
Suppose there are no prime number dividing both a and b. By the fundamental theorem of arithmetic, let
b
a = pa1 1 . . . pakk be the prime factorization of a with ri > 0, and b = q1b1 . . . qj j be the prime factorization
of b. Note that for any prime number p, p|a , p = pi for some i. Also q|b , q = qi for any prime number
q. Therefore, pi 6= qj for any i or j, i.e. p and q are coprime.
On the other hand, suppose there is a prime number p dividing both a and b, then by definition of gcd,
p|gcd(a, b) and a, b cannot be coprime.
4B
10. Suppose
by b2 , we get
p
p = ab , with a and b being natural numbers. By squaring and multiplying both sides
a2 = pb2
Consider prime factorization of a and b. Suppose the power of p appearing in the factorization of a is
r, and the power of p appearing in the factorization of b is s. Then comparing the powers of p in the
equation above, we get
2r = 1 + 2s
which gives a contradiction, since LHS is even while RHS is odd.
3rk
1
12.(i) Suppose the exponent of each prime factor of a is a multiple of 3, i.e. a = p3r
1 . . . pk . Then
r
r
a = x3 , where x = p11 . . . pkk .
b
On the other hand, if a is a cube, then a = x3 for some natural number x. Write x = q1b1 . . . qj j , then
3b
a = x3 = q13b1 . . . qj j . Therefore the exponent of each prime factor is a multiple of 3.
(ii) If not, suppose a1/3 = xy , with x and y being natural numbers. By taking cube powers and multiplying,
we get
y 3 = ax3
3f
3f1
3ek
j
1
Consider prime factorization of x and y, i.e. x = p3e
1 . . . pk , y = q1 . . . qj . By comparing the exponent of each prime on both sides of the equation above, the exponent of each prime factor of a must be
a multiple of 3. By (i), a must be a cube. Contradiction.
27. If a|bc and (a, b)|c. Write the prime factorization of a, b, c as
a = pa1 1 . . . pai i
b = pb11 . . . pbi i
c = pc11 . . . pci i
where we allow the some of the exponents to be zero. The first condition says aj  bj + cj for all j, and
the second condition says min{aj , bj }  cj for all j.
If min{aj , bj } = aj , then the second equation says aj  cj  2cj for all j. And hence we are done.
If min{aj , bj } = bj , then the second equation says bj  cj for all j, and the first equation says
aj  bj + cj  cj + cj = 2cj . And hence we are done.
1
37. 62 73 84 95 106 = 22+12+6 32+10 56 73 , 65 76 82 93 104 = 25+6+4 35+6 54 76 . So the lcm of the two numbers
is
220 312 56 76
5A
5. We just need to find one number in the range, and ± that number by a multiple of m so that the
number still lies within the range.
(i) 1912, 1925, 1938, 1951, 1964, 1977, 1990
(ii) 1776 ⌘ 1(mod 25). 1901, 1926, 1951, 1976.
(iii) 1914 ⌘ 24(mod 27). Actually we just need to add multiples of 27 to 1914 in this case. 1914, 1941, 1968, 1995.
6. We will learn Chinese remainder theorem in the future to solve questions like this. But for now,
list the numbers satisfying a ⌘ 5(mod 8), i.e. 5, 13, 21, 29, 37, 45, 53, 61, . . . and see which of them satisfies
a ⌘ 3(mod 7). Note that 45 works in our case. Check that 45 + 56n also works for all natural number n.
m
7. Note that if m > 4 is not prime, then m = p · m
p for some prime number p and p is an integer. If
m
m
m
p 6= p , then both p and p  (m 1) and hence m = p · p |(m 1)!.
2
Suppose p = m
1) < p2 = m,
p , then m = p > 4 is a square number, with p > 2. Therefore, p, 2p  (m
and hence
m|p · (2p)|(m 1)!
5B
15. 71 6 = (72 )8 = 498 ⌘ ( 2)8 (mod 17), since
2 ⌘ 49(mod 17). Now
( 2)8 = 256 ⌘ 1(mod 17)
7546 = 734·16+2 = (716 )34 · 72 ⌘ 134 · 49 ⌘ 15(mod 17).
16. (i) 518 = 259 ⌘ 39 = 273 ⌘ 53 = 125 ⌘ 4(mod 11).
(iii) 447 = 43 · 444 = 43 · 1622 ⌘ 43 · 422 = 43 · 1611 ⌘ 43 · 411 = 414 = 167 ⌘ 47 = (16)(16)(16)(4) ⌘
(4)(4)(4)(4) = (16)(16) ⌘ (4)(4) = 16 ⌘ 4(mod 12)
17. 16 = 1 ⌘ 1(mod 7).
26 = 8 · 8 = 64 ⌘ 1(mod 7).
36 = 27 · 27 ⌘ ( 1) · ( 1) = 1(mod 7).
46 = 26 · 26 ⌘ (1) · (1) = 1(mod 7).
56 ⌘ ( 2)6 = 26 = 1(mod 7).
66 ⌘ ( 1)6 = 1(mod 7)
2