Math 480
HOMEWORK solutions
#
3
W1.
Find all integer solutions of the equation 2
x
+ 3
y
= 11
Answer.
(1 + 3
t,
3

2
t
)
, t
∈
Z
.
W2.
(a) For which
n
is it possible to simplify the fraction
39
n
+8
65
n
+13
?
Solution.
The fraction
39
n
+8
65
n
+13
is reducible if and only if GCD(39
n
+ 8
,
65
n
+ 13)
>
1.
To find the GCD, we employ Euclidean algorithm.
65
n
+ 13 = (39
n
+ 8) + (26
n
+ 5)
39
n
+ 8 = (26
n
+ 5) + (13
n
+ 3)
26
n
+ 5 = (13
n
+ 3) + (13
n
+ 2)
13
n
+ 3 = (13
n
+ 2) + 1
Hence, the GCD=1; and the fraction is already reduced.
(b) Solve in integers:
x
3
+ 21
y
2
+ 5 = 0.
Solution.
Consider the equation mod 7.
We have
x
3
≡ 
5
≡
2 (mod 7); hence,
x
6
≡
4 (mod 7). This is impossible by the Fermat’s Little theorem (which says that
either
x
≡
0 or
x
6
≡
1 (mod 7)).
W3.
Recall that the sequence of Fibonacci numbers is defined by
F
1
=
F
2
= 1,
F
n
+1
=
F
n
+
F
n

1
if
n
≥
2
.
Show that if
F
n
is divisible by
p
for some
n
≥
0, then there are infinitely
n
such that
F
n
is divisible by
p
.
Solution.
Fix a prime
p
such that there exists
n
0
with
p

F
n
0
.
Consider now pairs of
consecutive Fibonacci numbers (
F
0
, F
1
)
,
(
F
2
, F
3
)
, . . .
. By the pigeonhole principle, there must
be two pairs (
F
i
, F
i
+1
) and (
F
j
, F
j
+1
),
i < j
, such that the remainders mod
p
are pairwise
the same, that is
F
j
≡
F
i
(mod
p
) and
F
j
+1
≡
F
i
+1
(mod
p
).
Let
d
=
j

i
, and let
a
n
=
F
n
+
d

F
n
,
n
≥
0. We have
a
n
+2
=
F
n
+2+
d

F
n
+2
=
F
n
+1+
d
+
F
n
+
d

F
n
+1

F
n
=
a
n
+1
+
a
n
By the choice of
d
, we also have
a
i
≡
a
i
+1
≡
0(mod
p
)
The recurrence relation now implies that
a
n
≡
0(mod
p
), for any
n
≥
0. Hence,
F
n
+
d
≡
F
n
(mod
p
) for any
n
≥
0. By assumption, there exists
n
0
such that
p

F
n
0
. Hence,
p

F
n
0
+
dk
for any
k
≥
0.