For any consideration of partial credit all work must be shown in the space provided (you can hit enter if more room is needed) and the answer spaces completely filled in. Chapter 1: Integers Find n!...

· For any consideration of partial credit all work must be shown in the space provided (you can hit enter if more room is needed) and the answer spaces completely filled in.
I. Chapter 1: Integers
1. Find n! For n equal to each of the first ten positive integers. 1. ___________

Using the rule ! (1)!, we find that
1! 1, 2! 2, 3! 6, 4!24, 5!120, 6!720,
7! 5040,
8! 40320, 9! 362880, 10! 3628800.
nnn
and
=×-
=======
===
2. Use mathematical induction to prove that x – y is a factor of xn – yn, where x and y are variables.

1) Base Case: Let 1. Then
()(), and () is obviously
divisible by ().
Therefore, the formula holds true for
1.
2) Induction Hypothesis: Assume the form
ula holds true
nn
n
xyxyxyxy
n
=
-=---
=
11
11
for .
() divides ()
( We want to prove that () divides ()

Now subtract and add . After all, subtr
acting and t
kk
kk
kkkk
kk
nk
xyxy
xyxy
xyxxyy
xyxy
++
++
=
--
--
-=-
11
11
hen adding the
same term is the same as adding zero.
= +
Now, factor the first two terms and the
last two terms.
= ()()
(
kkkkkk
kkkkk
k
xyxxxyxyyy
xyxxyyxy
yxy
++
++
---
--+-
-
) is obviously divisible by ().
By our induction hypothesis, we assumed
that () is divisible by
(). Therefore, () is divisible by ().
The sum of two terms both divisible by (
) is
kk
kk
xy
xy
xyxxyxy
xy
-
-
---
-
also divisible by().
Therefore, what we started with, ()() i
s divisible by
().
Therefore, by the principle of mathemati
cal induction, () is divisible
by () for all natural n
kkk
nn
xy
xxyyxy
xy
xy
xy
-
-+-
-
-
-
umbers n.
3. Prove that fn - 2 + fn + 2 = 3f​n whenever n is an integer with n ≥ 2.
(Recall that f0 = 0.)
042
-22
By using the principle of mathematical i
nduction,
(1) Base case, 2,
033.
(2) Induction hypothesis, ,
3 for 2, 3, . .
. ,.
T
kkk
n
fff
nk
fffkn
+
=
+=+=
=
+==
132321
2231
hen by the definition of the Fibonacci s
equence,

( )(
)
3
nnnnnn
nnnn
n
ffffff
ffff
f
----++
-+-+
+=+++
=+++
=
11
22
3 3 .
Therefore 3 whenever n is an integer
with 2.
nn
nnn
ff
fffn
-+
-+
+=
+=³
4. Show that if a is an integer, then 3 divides a3 – a
32
3
Let . Then (1)(1)(1)(1)(1)
(1)(1).
Now (1)(1) is a product of three consecu
tive integers 1,
; 1 and 6 divides (1)(1). Now 36
aZaaaaaaaaaa
aaaaa
aaaa
aaaaa
Î-=-=+-=-+
Þ-=-+
-+-
+-+
3
and 6(1)(1).
If and , then , 3(1)(1). Hence, 3 div
ides .
aaa
abbcacaaaaa
-+
-+-
II. Chapter 3: Primes and Greatest Common Divisor
5. Find the smallest prime between n2 and (n + 1)2 for all positive integers n with n ≤ 10.
5:___________________________________
(
)
(
)
22
22
1 1
1 1 4
2
2 4 9
5
3
nnnsmallestprimebetweennandn
++
9 16
11
4 16 25
17
5 25 36
29
6 36
49
37
7 49 64
53
8 64 81
67
9 81 100
83
10 100 121
101
6. Let a be a positive integer. What is the greatest common divisor of a and a2?
6: ________________________________
2(1)(1)(2), we know that (, 2) is eithe
r 2 or 1. The GCD
of 2 integers is the least positive line
ar combination of those 2 integers.
Since 2 is a linear combination of and
2, the leas
aaaa
aa
=-+++
+
t positive linear
combination can only be either 1 or 2. I
f a is even, it is 2. If is odd, then
we know 2 is not a common divisor of a
nd 2, so the GCD cannot
be 2. It must be 1.
a
aa
+
7. Use the fact that every prime divisor of

1
2
5
2
5
+
=
F
is of the form 27k + 1 = 128k + 1
to demonstrate that the prime factorization of F5 is F5 = 641*6700417.
5
27
5
First we have to find the prime factoriz
ation of
214294967297 is of
the form 211281.
Now put 1, 2, 3, 4, 5
12811129 is not prime,
Fkk
k
=+=+=+
=
×+=
12821257 is prime but not divided the
number 4294967297,
12831385 is not prime,
12841513 is not prime,
and 12851 641 is prime
and divide 4
×+=
×+=
×+=
×+=
294967297 with 42949672976416700417. No
w we can
see that 6700417 is also a factor of 429
4967297. But 6700417 is not divided
by 641. All integers are less than 670
0417, the integers 769, 897, 1025,
=×

1153, 1281, 1409, 1537, 1665, 1793, 1921
, 2049, 2177, 2305, 2433, and
2561, but out of these only 769, 1153,
and 1409 are prime, and none of them
divide 6700417. Hence 6700417 is prime a
nd the prime
5
factorization of is
6416700417.
F
×
8. The Indian astronomer and mathematician Mahavira, who lived in the ninth century, posed this puzzle:
A band of 23 weary travelers entered a lush forest where they found 63 piles each...