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Solutions to Section 1.3 Homework Problems
S. F. Ellermeyer
1. (a) The prime factorization of 38 is 38 = 2 19 so the set of positive
divisors of 38 is
D (38) = f1; 2; 19; 38g .
The prime factorization of 108 is 108 = 2 2 3 3 3 so the set of
positive divisors of 108 is
D (108) = f1; 2; 3; 4; 6; 9; 12; 18; 27; 36; 54; 108g .
Since D (38) \ D (108) = f1; 2g, we see that gcd (38; 108) = 2.
2.
b. It is clear that jnj is the largest positive divisor of n. That is,
max (D (n)) = jnj. Therefore
gcd (n; n) = max (D (n) \ D (n)) = max (D (n)) = jnj .
3. We want to prove that every common divisor of a and b divides gcd (a; b):
Let d = gcd (a; b) and let let c be a common divisor of a and b. Then
there exist integers m and n such that
d = ma + nb
and there exist integers s and t such that sc = a and tc = b.
This gives us
d = msc + ntc = (ms + nt) c.
Since ms + nt is an integer, we see that c divides d.
4. Given that a and c both divide b and that gcd (a; c) = 1, we want to
prove that ac divides b.
First we give some examples:
Example 1: Observe that a = 2 and c = 3 both divide b = 24 and also
that gcd (a; c) = gcd (2; 3) = 1. As promised by what we are about to
prove, we also observe that ac = 6 divides b = 24.
Example 2: Observe that a = 5 and c = 8 both divide b = 400 and also
that gcd (a; c) = gcd (5; 8) = 1. As promised by what we are about to
prove, we also observe that ac = 40 divides b = 400.
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Now here is the proof: Suppose that a and c both divide b and that
gcd (a; c) = 1. By Proposition 7 (page 22) in the textbook, we know
that ac = lcm (a; c) gcd (a; c). Thus ac = lcm (a; c). Also, since a and
c both divide b, there exist integers s and t such that sa = b and tc = b.
Suppose, for the sake of obtaining a contradiction, that ac does not
divide b. Then by the Division Algorithm, there exist integers q and r
such that b = q (ac) + r and 0 < r < ac. This implies that
r=b
qac,
which implies that
r = sa
qac = (s
qc) a,
which implies that r is a multiple of a.
Likewise
r=b
qac,
which implies that
r = tc
qac = (t
qa) c,
which implies that r is a multiple of c.
However, we have now arrived at a contradiction because we have that
r is a common multiple of a and c and that r > 0 and that r < ac =
lcm (a; c). (It is not possible that some positive common multiple of a
and c is less than lcm (a; c).)
This proves that ac divides b.
5. Suppose that x, y, and m are integers and that gcd (x; m) = 1 and that
gcd (y; m) = 1. We will prove that this implies that gcd (xy; m) = 1.
Since gcd (x; m) = 1, there exist integers s and t such that
sx + tm = 1.
Likewise, since gcd (y; m) = 1, there exist integers w and z such that
wy + zm = 1.
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This gives us (by multiplying each side of the above two equations by
each other)
(sx + tm) (wy + zm) = 1.
By expansion, we obtain
swxy + szxm + twmy + tzmm = 1
which can be written as
(sw) (xy) + (szx + twy + tzm) m = 1.
Since sw and szx + twy + tzm are both integers, we conclude that
gcd (xy; m) = 1.
6. Suppose that a > 0 and that a divides b. We want to prove that
gcd (a; b) = a.
Since a divides b and a > 0, we know that a belongs to the set of
positive divisors of b. (That is, a 2 D (b).) Also, it is clear that a is the
largest member of the set D (a). Thus a 2 D (a) \ D (b) and a must
also be the largest member of this set. Therefore gcd (a; b) = a.
7.
c.
442 = 1 (289)+153
tells us that gcd (442; 289) = gcd (289; 153)
289 = 1 (153)+136
tells us that gcd (289; 153) = gcd (153; 136)
153 = 1 (136) + 17
tells us that gcd (153; 136) = gcd (136; 17)
137 = 8 (17) + 0
tells us that gcd (137; 17) = gcd (17; 0) = 17.
We conclude that gcd (442; 289) = 17.
8. (a)
462 = 1 (442) + 20
442 = 22 (20) + 2
20 = 10 (2) + 0
tells us that gcd (462; 442) = 2.
Thus
(462) (442)
lcm (462; 442) =
= 102; 102.
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9. We want to prove that if n is an odd integer, then gcd (3n; 3n + 2) = 1.
Examples of this are gcd (3; 5) = 1, gcd (15; 17) = 1, and gcd (33; 35) =
1.
First observe that if n is an odd integer, then n = 2k + 1 where k is
some integer. This means that 3n = 3 (2k + 1) = 6k + 3. Since 6k is
an even integer, then 6k + 3 must be odd. Thus 3n is odd. This means
that 3n + 2 is also odd. Applying the Division Algorithm to 3n + 2 and
3n, we obtain
3n + 2 = 1 3n + 2
and we know (by Proposition 6 on page 18) that gcd (3n + 2; 3n) =
gcd (3n; 2). Since the only positive divisors of 2 are 1 and 2 and since 2
is not a divisor of 3n (because 3n is odd), we see that gcd (3n; 2) = 1.
Therefore gcd (3n + 2; 3n) = 1.
10. Let us …nd all of the solutions of the Diophantine equation 35x + 25y =
15: If we divide both sides of this equation by 5 we obtain 7x + 5y = 3.
Any (integer pair) solution of this latter equation is also a solution of
the former equation and vice versa.
Since gcd (7; 5) = 1 and since 3 is a multiple of 1, we know that our
equation does have solutions. We are going to …nd a solution of the
equation 7x + 5y = 1 (and then multiply our solution by 3 to obtain a
solution of 7x + 5y = 3). To …nd a solution of 7x + 5y = 1, we use the
Euclidean Algorithm:
7 = 1 (5) + 2
5 = 2 (2) + 1
gives us
1=
2=
2 (2) + 5
1 (5) + 7
which gives us
1 = 2 ( 1 (5) + 7) + 5
= 2 (5) 2 (7) + 5
= 2 (7) + 3 (5) .
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Thus we see that ( 2; 3) is a solution of the equation 7x + 5y = 1. This
means that (x0 ; y0 ) = ( 6; 9) is a solution of the equation 7x + 5y = 3.
Furthermore, all of the solutions of the equation 7x + 5y = 3 are
(x; y) =
x0 +
b
t; y0
gcd (a; b)
a
t
gcd (a; b)
= ( 6 + 5t; 9
7t)
where t can be any integer. Thus, for example, another solution of
7x + 5y = 3 (taking t = 2) is (x; y) = ( 16; 23).
11. To solve the Diophantine equation 18x 5y = 15, we note that gcd (18; 5) =
1 and that, since 15 is a multiple of 1, this equation does have solutions.
We will …rst solve 18x 5y = 1. We will actually solve 18x + 5y = 1 (in
order for the Division Algorithm to apply) and then note that if (x; y)
is a solution of 18x + 5y = 1, then (x; y) is a solution of 18x 5y = 1.
18 = 3 (5) + 3
5 = 1 (3) + 2
3 = 1 (2) + 1
gives
1=
2=
3=
1 (2) + 3
1 (3) + 5
3 (5) + 18
which gives
1 = 1 (2) + 3
= 1 ( 1 (3) + 5) + 3
= 1 (3) 1 (5) + 3
= 2 (3) 1 (5)
= 2 ( 3 (5) + 18) 1 (5)
= 6 (5) + 2 (18) 1 (5)
= 2 (18) 7 (5) .
We now see that (2; 7) is a solution of the equation 18x+5y = 1. This
means that (2; 7) is a solution of the equation 18x 5y = 1. Therefore
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a solution of the equation 18x 5y = 15 is (x0 ; y0 ) = (30; 105). All of
the solutions of the equation 18x 5y = 15 are
(x; y) = (30
5t; 105
18t)
where t can be any integer. For example, taking t = 9, we can verify
that ( 15; 57) is also a solution.
We want to determine which of these solutions are positive integer
solutions. Thus we need to determine the values of t for which both
30 5t > 0 and 105 18t > 0. The …rst inequality tells us that we
must have t < 6 and the second inequality tells us that we must have
t < 105=18 5:8. Thus, to obtain positive solutions, me must choose
t 5. Note that for t = 6, we obtain the solution (x; y) = (0; 3) and
that for t = 5 we obtain the solution (x; y) = (5; 15).
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