Assignment 5 – Solutions
1. Prove that any two consecutive integers are coprime.
Solution:
Consider the consecutive integers
n
and
n
+ 1. Because
n
(

1) + (
n
+ 1)(1) = 1, by
Proposition 2.27(i), gcd(
n, n
+ 1) = 1 and the consecutive integers are coprime.
2. Prove that the Diophantine equation
ax
2
+
by
2
=
c
does not have any integer solutions
unless gcd(
a, b
)

c
. If gcd(
a, b
)

c
, does the equation always have an integer solution?
Solution:
If
a
=
b
= 0 then the equation only has solutions if
c
= 0, and then any pair
(
x, y
)
∈
Z
will work. Otherwise suppose that the equation has a solution
x
0
and
y
0
.
Then
ax
2
0
+
by
2
0
=
c
. Let
d
= gcd(
a, b
)
6
= 0, so
d

a
and
d

b
. By Proposition 2.11(ii),
d

ax
2
0
+
by
2
0
, so
d

c
.
The converse of this statement is not true, to see this let
a
=
b
= 1 and
c
= 3. Then
gcd(1
,
1) = 1, which does divide 3. However, if
x
2
+
y
2
= 3, then
x
2
≤
3, so

x
 ≤
1
and

y
 ≤
1. None of these integer values give solutions.
3. (a) Prove that if
n
≥
3, then
n
! + 3 is not prime.
Solution:
For
n
≥
3,
n
! =
n
×
(
n

1)
×· · ·×
3
×
2
×
1, so 3

n
! and
n
!
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 Winter '08
 ANDREWCHILDS
 Math, Integers, Prime number, prime Factorization, consecutive integers

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