MATH 135
Fall 2008
Assignment #1 Solutions
HandIn Problems
1.
Counterexample
If
n
= 20, then
n
2
+ (
n
+ 1)
2
= 20
2
+ 21
2
= 400 + 441 = 841 which is a perfect square, so the
given statement is FALSE.
How did we find this?
We could find this by trying positive integer values starting at
n
= 4 until we obtained a value
for
n
2
+ (
n
+ 1)
2
that was a perfect square. Alternatively, we might remember that (20
,
21
,
29)
forms a Pythagorean triple, because 20
2
+ 21
2
= 29
2
.
2.
(a) Treating the given equation as a quadratic equation in
x
with coefficients in terms of
y
,
4
x
2
+
x
(4
y

4) + (2
y
2

2
y
+ 1) = 0
Using the quadratic formula, we get
x
=

(4
y

4)
±
(4
y

4)
2

4(4)(2
y
2

2
y
+ 1)
8
=

(4
y

4)
±

16
y
2
8
=

(
y

1)
±

y
2
2
(b)
Counterexample
If (
x, y
) =
(
1
2
,
0
)
, 4
x
2
+4
xy
+2
y
2

4
x

2
y
+1 = 4
(
1
2
)
2
+4
·
1
2
·
0

2
·
0
2

4
·
1
2

2
·
0+1 = 0,
so the given statement is FALSE.
How did we find this?
From (a), for a real value for
x
to exist, we must have

y
2
≥
0, so
y
= 0.
If
y
= 0, then
x
=
1
2
, so (
x, y
) =
(
1
2
,
0
)
makes the expression equal to 0.
3.
LS
=
sin
4
x
+ 2 cos
2
x

cos
4
x
=
sin
4
x
+ 2(1

sin
2
x
)

(1

sin
2
x
)
2
=
sin
4
x
+ 2

2 sin
2
x

(1

2 sin
2
x
+ sin
4
x
)
=
1
=
RS
as required.
4. Consider (
x
2

3
x
)

10 =
x
2

3
x

10 = (
x

5)(
x
+ 2).
Since 2
≤
x
≤
5, then
x

5
≤
0 and
x
+ 2
≥
0, so (
x

5)(
x
+ 2)
≤
0.
Therefore,
x
2

3
x

10
≤
0 or
x
2

3
x
≤
10.
If
x
≥
2, then
x

1
≥
1 and
x

2
≥
0, so
x
2

3
x
= (
x
2

3
x
+ 2)

2 = (
x

1)(
x

2)

2
≥
1(0)

2 =

2
Thus, if 2
≤
x
≤
5 then

2
≤
x
2

3
x
≤
10.
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5. Suppose that
n
has units digit 5.
Then
n
= 10
k
+ 5 for some nonnegative integer
k
.
Thus,
n
2
= (10
k
+ 5)
2
= 100
k
2
+ 100
k
+ 25 = 100(
k
2
+
k
) + 25.
In other words,
n
2
is a multiple of 100 plus 25, so its last two digits are 25.
6. First, we factor to obtain
n
3

n
=
n
(
n
2

1) =
n
(
n

1)(
n
+ 1).
This tells us that
n
3

n
is the product of three consecutive positive integers, namely
n

1,
n
and
n
+ 1.
Not all three of these integers can be odd, so at least one is even, so
n
3

n
is even (ie. is a
multple of 2).
Also, at least one of
n

1,
n
,
n
+ 1 must be a multiple of 3, since they are three consecutive
integers.
Thus,
n
3

n
is a multiple of 3.
Therefore,
n
3

n
is a multiple of 2 and of 3, so is a multiple of 2(3) = 6, as required.
(We will see later that it is important that 2 and 3 have no common factors larger than 1.)
7.
(a) This “if and only if” statement is equivalent to
“If
x
2
+
y
2

2
y
=

1, then
x
= 0 and
y
= 1.”
and
“If
x
= 0 and
y
= 1, then
x
2
+
y
2

2
y
=

1.”
(b) “
⇐
”
Suppose that
x
= 0 and
y
= 1.
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 Fall '08
 ANDREWCHILDS
 Math, Algebra, Pythagorean Theorem, Prime number

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