MATH 135, Solutions to Challenge Problem Set 1
1:
Determine whether there exists a positive integer
n
such that
n
2
+
n
+ 1 is a square.
Solution: No such integer
n
exists, as we shall prove. Suppose (for a contradiction) that such an integer does
exist, say
n
2
+
n
+ 1 =
m
2
, where
n >
0 and
m
≥
0. Then we have
m
2
=
n
2
+
n
+ 1
> n
2
, so
m > n
, and
hence
m
≥
n
+ 1. But then
m
2
≥
(
n
+ 1)
2
=
n
2
+ 2
n
+ 1
> n
2
+
n
+ 1 and so
m
2
=
n
2
+
n
+ 1 giving the
desired contradiction.
2:
Let
a
0
=
c
and for
n
≥
1 let
a
n
=
pa
n

1
+
q
where
p
= 0. Find a formula for
a
n
in terms of
c
,
p
and
q
.
Solution: We have
a
0
=
c
a
1
=
pc
+
q
a
2
=
p
(
pc
+
q
) +
q
=
p
2
c
+
pq
+
q
a
3
=
p
(
p
2
c
+
pq
+
q
) +
q
=
p
3
c
+
p
2
q
+
pq
+
q
a
4
=
p
(
p
3
c
+
p
2
q
+
pq
+
q
) +
q
=
p
4
c
+
p
3
q
+
p
2
q
+
pq
+
q
and in general
a
n
=
p
n
c
+
p
n

1
q
+
p
n

2
q
+
· · ·
+
p
2
q
+
pq
+
q
=
p
n
c
+ (
p
n

1
+
p
n

2
+
· · ·
p
2
+
p
+ 1)
q .
We can obtain a (nonrecursive) formula for the geometric sum
p
n

1
+
p
n

2
+
· · ·
+
p
2
+
p
+ 1 as follows.
Let
S
=
p
n

1
+
p
n

2
+
· · ·
+
p
2
+
p
+ 1 (1). Note that
pS
=
p
n
+
p
n

1
+
p
n

2
+
· · ·
+
p
2
+
p
(2). Subtracting
(1) from (2) gives
S
(
p

1) =
p
n

1 and so
S
=
p
n

1
p

1
. Thus we have
a
n
=
p
n
c
+
p
n

1
p

1
q .
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3:
Let
{
F
n
}
be the Fibonacci sequence (so
F
1
= 1,
F
2
= 1, and for
n
≥
3,
F
n
=
F
n

1
+
F
n

2
). Prove that
F
n
2
+
F
2
n
+1
=
F
2
n
+1
for all
n
≥
1.
Solution: We begin by trying (and failing) to use induction to prove that
F
n
2
+
F
2
n
+1
=
F
2
n
+1
for all
n
≥
1.
When
n
= 1, we have
LS
=
F
1
2
+
F
2
2
= 1
2
+ 1
2
= 2 and
RS
=
F
3
=
F
2
+
F
1
= 1 + 1 = 2 =
LS
, so the
equality holds. Let
k
≥
1 and suppose (inductively) that
F
k
2
+
F
k
+1
2
=
F
2
k
+1
. Then when
n
=
k
+ 1 we
have
LS
=
F
k
+1
2
+
F
k
+2
2
=
F
k
+1
2
+ (
F
k
+1
+
F
k
)
2
=
F
k
+1
2
+
F
k
+1
2
+ 2
F
k
F
k
+1
+
F
k
2
=
(
F
k
+1
2
+ 2
F
k
F
k
+1
)
+
(
F
k
2
+
F
k
+1
2
)
=
(
F
k
+1
2
+ 2
F
k
F
k
+1
)
+
F
2
k
+1
(where the last inequality follows from the induction hypothesis), and we have
RS
=
F
2
k
+3
=
F
2
k
+2
+
F
2
k
+1
.
If we could show that
(
F
k
+1
2
+ 2
F
k
F
k
+1
)
=
F
2
k
+2
then we would have
LS
=
RS
and our induction proof
would work. We shall modify this abortive proof by proving two equalities at once.
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 Fall '08
 ANDREWCHILDS
 Math, Mathematical Induction, Recursion, Natural number, line segments, Structural induction, Peano axioms

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