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6
Solutions to Exercises
17.
n
2

n
+
5
=
n(n

1)
+
5. Now either
n

1 or
n
is even, since these integers are consecutive. So
n( n

1) is even. Since the sum of an even integer and the odd integer 5 is odd, the result follows.
18.
[BB]
2X2

4x
+
3
=
2(x
2

2x)
+
3
=
2[(x
_1)2  1]
+
3
=
2(x

1)2
+
1 is the sum of 1 and a
nonnegative number. So it is at least 1 and hence positive.
19. For
a
2

b
2
to be odd, it is necessary and sufficient for one of
a
or
b
to be even while the other is odd.
Here's why.
Case
i:
a, b
even.
In this case,
a
=
2n
and
b
= 2m
for some integers m and
n,
so
a
2

b
2
=
4n
2

4m
2
=
4(n
2

m
2
)
is even.
Case
ii:
a, b
odd.
In this case,
a
=
2n
+
1 and
b
= 2m
+
1 for some integers m and
n,
so
a
2

b
2
=
(4n
2
+
4n
+
1) 
(4m
2
+
4m
+
1)
=
4(n
2
+
n

m
2

m) is even.
Case
iii:
a
even,
b
odd.
In this case,
a
=
2n
and
b
=
2m+1
for some integers
n
and m, so
a
2
_b
2
=
4n
2

(4m2+4m+
1)
=
4(n
2

m
2

m) 
1 is odd.
Case iv:
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory, Integers

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