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86
Solutions to Exercises
(c) If
n
is not a power of 2, we can write
n
=
ab
with
b
odd. Then
2
ab
+
1
=
(2
a
+
1)(2
a
(bl) 
2
a
(b2)
+
... + 1). Note the last tenn in the second factor is "+"1 because
b
is odd.
20. (a)
If
P
=
2k
+ 1 and
q
=
2l + 1, then
P
+
q
=
2(k +.e
+ 1), so 2
1
(p
+
q).
Also
P
+
q
>
2 since
P
>
2 and
q
>
2. Thus,
P
+
q
>
2 is divisible by 2 and so not prime.
(b) No. Consider
2
+
3 =
5.
21. [BB] Suppose, to the contrary, that
P
and
q
are consecutive primes (in that order) such that
P
+
q
=
2r,
where
r
is a prime. Then
P
=
P
+
P
<
P
+
q
=
r
<
q
+
q
=
qj
2
2
2
that is,
P
<
r
<
q.
This contradicts the fact that
P
and
q
are consecutive primes.
22. Suppose there exist integers
x
and
y
such that
x
2

y2
=
2n.
Then 21
(x
+
y)(x

y)
so 2 1
(x
+
y)
or 21
(x

y).
But
x
+y
=
(x

y)
+
2y
shows that
x
+
y
and
x

y
are both even or both odd. Thus,
2 divides both
x
+
y
and
x

y
so 41
(x
2

y2),
412n and 21
n,
a contradiction.
23. [BB] This is true, and the proof has given Andrew Wiles a place in history. This is Fennat's Last
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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

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