Chapter 6

Student ID: U10011024 Name: Kuan-Lun Wang

75 Theorem. Eevery integer greater than 1 is a product
of prime numbers. (P.34)
We use proof by contradiction. Assume that some positive
integer cannot be written as the product of primes. Let n be
the smallest such integer (such an integer must exist, from
the well-ordering property). If n is prime, it is obviously the
product of set a set of primes, namely the one prime n. So n
must be composite. Let n = ab, with 1<a<n and 1<b<n.
But because a and b are smaller than n, they must be the
product of primes. Then, because n = ab, we conclude that
n is also a product of primes. This contradiction shows that
every positive integer can be written as the product of primes.
Pratice 73. If p | a1a2 · · · an, then p | ai for some 1 ≤ i ≤
n. (P.34)
Lemma 3.5. If p divides a1a2 · · · an, where p is a prime and
a1, a2, . . . , an are positive integers, then there is an integer
i with 1 ≤ i ≤ n such that p divides ai. (P.113)
We prove this result by induction. The case where n = 1
is trivial. Assume that the results is true for n. Consider
a product of n + 1 integers a1a2 · · · an+1 that is divisible by
the integer p. We know that either (p, a1a2 · · · an) = 1 or
(p, a1a2 · · · an) = p. If (p, a1a2 · · · an) = 1, then by Lemma
3.4, p | an+1. On the other hand, if p | a1a2 · · · an, using the
induction hypothesis, there is an integer i where 1 ≤ i ≤ n
such that p | ai. Consequently, p | ai for some i with 1 ≤
i ≤ n + 1. This prove the result.

CHAPTER 6 1


Lemma 3.4. If a, b, and c are positive integers such that
(a, b) = 1 and a | bc, then a | c. (P.34)
76 Theorem (Fundamental Theorem of Arithmetic). Ev-
ery integer greater than 1 can be represented as a product of
primes in only one way, apart from the order of the factors.
We arrange the prime factorization of n as follows:
a
n = p a1 p a2 · · · p k k ,
1 2

where a1>0, a2>0, . . . , ak >0 and p1<p2< · · · <pk . This
factorization of n is called the canonical factorization of n.
For example 23 · 32 · 5 · 72 is the canonical factorization of
17640. (P.34)
Theorem 3.15. The fundamental Theorem of Arith-
metic. Every positive integer greater then 1 can be written
uniquely as product of primes, with the prime factors in the
product written in nondecreasing order. (P.112)
We use proof by contradiction. Assume that some positive
integer cannot be written as the product of primes. Let n be
the smallest such integer (such an integer must exist, from
the well-ordering property). If n is prime, it is obviously the
product of set a set of primes, namely the one prime n. So n
must be composite. Let n = ab, with 1<a<n and 1<b<n.
But because a and b are smaller than n, they must be the
product of primes. Then, because n = ab, we conclude that
n is also a product of primes. This contradiction shows that
every positive integer can be written as the product of primes.
We now ﬁnish the proof of the fundamental theorem of
arithmetic by showing that the factorization is unique. Sup-
CHAPTER 6 2


pose that there is an integer n that has two different factor-
izations into primes:
n = p 1 p2 · · · ps = q 1 q 2 · · · q t ,
where p1, p2, . . ., ps, and q1, q2, . . ., qt are all primes, with
p1 ≤ p2 ≤ · · · ≤ ps and q1 ≤ q2 ≤ · · · ≤ qt.
Remove all common primes from the two factorizations to
obtain
pi1 pi2 · · · piu = qj1 qj2 · · · qjv ,
where the primes on the left-hand side of this equation differ
from those on the right-hand side, u ≥ 1, and v ≥ 1 (be-
cause the two original factorizations were presumed to differ).
However, this leads to a contradiction of Lemma 3.5; by this
lemma, pi must divide qjk for some k, which is impossible,
because each qjk is prime and is different from pi1. Hence,
the prime factorization of a positive integer n is unique.
Lemma 3.5. If p divides a1a2 · · · an, where p is a prime
and a1, a2, . . . , an are positive integers, then there is an
integer i with 1 ≤ i ≤ n such that p divides ai. (P.113)
78 Theorem. For a, b ∈ Z+, a · b = (a, b) · [a, b]. (P.35)
Theorem 3.16. If a and b are positive integers, then
[a, b] = ab/(a, b), where [a, b] and (a, b) are the least common
multiple and greater common divisor of a and b, respectively.
(P.116)
Let a and b have prime-power factorizations a = pa1 pa2 · · · pan
1 2 n
b1 b2 bn
and b = p1 p2 · · · pn , where the exponents are nonnega-
tive integers and all primes occurring in either factorization
CHAPTER 6 3


occur in both, perhaps with 0 exponents. Now let Mj =
max(aj , bj ) and mj = min(aj , bj ). Then we have
(a, b) · [a, b] = pM1 pM2 · · · pMn · pm1 pm2 · · · pmn
1 2 n 1 2 n
= pM1+m1 p2 2+m2 · · · pMn+mn
1
M
n
= pa1+b1 p22+b2 · · · pan+bn
1
a
n
= p a1 p a2 · · · p an · p b 1 p b 2 · · · p b n
1 2 n 1 2 n
= a · b,
because Mj + mj = max(aj , bj ) + min(aj , bj ) = aj + bj by
Lemma 3.6.
Lemma 3.6. If x and y are real numbers, then max(x, y)+
min(x, y) = x + y.
√
79 Proposition. Prove that √ is irrational. (P.35)√
2
Example 3.20. Suppose that 2 is rational. Then 2 =
a/b, where a and b are relatively prime integers with b = 0.
It follows that 2 = a2/b2, so that 2b2 = a2. Because 2 | a2, it
follows (see Exercise 40 at the end of this section) that 2 | a.
Let a = 2c, so that b2 = 2c2. Hence, 2 | b2, and by Exercise
40, 2 also divides b. However, because (a, b) = 1, we know
that 2 cannot divide both a and b. This contradiction shows
√
that 2 is irrational. (P.119)
Exercise 40. Show that if a, b, and c are integers with
c | ab, then c | (a, c)(b, c). (P.123)
80 Theorem. Let α be a real root of the polynomial xn +
cn−1xn−1 + · · · + c1x + c0, where the coeﬃcients c0, c1, . . .,
cn−1 are integers. Then α is either an integer or an irrational
CHAPTER 6 4


number. (P.36)
Theorem 3.18. Let α be a real number that is a root of
the polynomial xn + cn−1xn−1 + · · · + c1x + c0, where the
coeﬃcients c0, c1, . . ., cn−1 are integers. Then α is either an
integer or an irrational number. (P.119)
Suppose that α is rational. Then we can write α = a/b,
where a and b are relatively prime integers with b = 0. Be-
cause α is a root of xn + cn−1xn−1 + · · · + c1x + c0, we have
(a/b)n + cn−1(a/b)n−1 + · · · + c1(a/b) + c0 = 0.
Multiplying by bn, we ﬁnd that
an + cn−1an−1b + · · · + c1abn−1 + c0bn = 0.
Because
an = b(−cn−1an−1 − · · · − c1abn−2 − c0bn−1),
we see that b | an. Assume that b = ±1. Then b has a prime
divisor p. Because p | b and b | an, we know that p | an.
Hence, by Exercise 41, we see that p | a. However, because
(a, b) = 1, this is a contradiction, which shows that b = ±1.
Consequently, if α is rational then α = ±a, so that α must
be an integer.
Exercise 41. (P.123)
a) Show that if a and b are positive integers with (a, b) = 1,
than (an, bn) = 1 for all positive integers n.
b) Use part (a) to proof that if a and b are integers such that
an | bn, where n is a positive integer, then a | b.

CHAPTER 6 5


n
84 Deﬁnition. The integers Fn = 22 + 1 are called the
Fermat numbers.
For example,
F0 = 3, F1 = 5, F2 = 17, F3 = 257, andF4 = 65537.
These numbers Fi, 0 ≤ i ≤ 4, are all primes. (P.37)
The Fermat Number
n
The integers Fn = 22 + 1 are called the Fermat numbers.
Fermat conjectured that these integers are all primes. Indeed,
the ﬁrst few are primes, namely, F0 = 3, F1 = 5, F2 = 17,
5
F3 = 257, and F4 = 65, 537. Unfortunately, F5 = 22 + 1 is
composite, as we will now demonstrate. (P.131)
5
85 Proposition. The Fermat number F5 = 22 + 1 is
divisible by 641. (P.37)
5
Example 3.24. The Fermat number F5 = 22 + 1 is di-
visible by 641. We can show that 641 | F5 without actually
performing the division, using several not-so-obvious obser-
vations. Note that
641 = 5 · 27 + 1 = 24 + 54.
Hence,
5
22 + 1 = 232 + 1 = 24 · 228 + 1 = (641 − 54)228 + 1
641 · 228 − (5 · 27)4 + 1 = 641 · 228 − (641 − 1)4 + 1
641(228 − 6413 + 4 · 6412 − 6 · 641 + 4).
Therefore, we see that 641 | F5. (P.131-132)
86 Theorem. For all positive integers n, we have
F0F1F2 · · · Fn−1 = Fn − 2.
CHAPTER 6 6


(P.37)
k
Lemma 3.10. Let Fk = 22 + 1 denote the kth Fermat
number, where k is a nonnegative integer. Then for all posi-
tive integers n, we have
F0F1F2 · · · Fn−1 = Fn − 2.
(P.133-134)
We will prove the lemma using mathematical induction.
For n = 1, the identity reads
F0 = F1 − 2.
This is obviously true, because F0 = 3 and F1 = 5. Now, let
us assume that the identity holds for the positive integer n,
so that
F0F1F2 · · · Fn−1 = Fn − 2.
With this assumption, we can easily show that the identity
holds for the integer n + 1, because
F0F1F2 · · · Fn−1Fn = (F0F1F2 · · · Fn−1)Fn
n n
= (Fn − 2)Fn = (22 − 1)(22 + 1)
n
= (22 )2 − 1 = 22n+1 − 1 = Fn+1 − 2.
This leads to the following theorem.
Pratice 81. Prove that the least digit in the decimal ex-
n
pansion of Fn = 22 + 1 is 7 if n ≥ 2. (P.37)
Exercise 17. Show that the last digit in the decimal expan-
n
sion of Fn = 22 +1 is 7 if n ≥ 2. (Hint: Using mathematical
n
induction, show that the last decimal digit of 22 is 6.) (P.136,
622)
CHAPTER 6 7


We can prove that the last digit in the decimal expansion of
Fn is 7 for n ≥ 2 by proving that the last digit in the decimal
n
expansion of 22 is 6 for n ≥ 2. This can be done using
2
mathematical induction. We have 22 = 16, so the result is
true for n = 2. Now assume that the last decimal digit of
n n n+1
22 is 6, that is, 22 ≡ 6 (mod 10). It follows that 22 =
n+1 −2n
2n 22 n+1 −2n
(2 ) ≡ 62 ≡ 6 (mod 10). This completes the
proof.
87 Theorem. Let m and n be distinct nonnegative inte-
gers. Then (Fm, Fn) = 1. (P.38)
Theorem 3.21. Let m and n be distinct nonnegative in-
tegers. Then the Fermat numbers Fm and Fn are relatively
prime. (P.134)
Let us assume that m<n. By Lemma 3.10, we know that
F0F1F2 · · · Fn−1 = Fn − 2.
Assume that d is a common divisor of Fm and Fn. Then,
Theorem 1.8 tells us that
d | (Fn − F0F1F2 · · · Fm · · · Fn−1) = 2.
Hence, either d = 1 or d = 2. However, because Fm and Fn
are odd, d cannot be 2.
k
Lemma 3.10. Let Fk = 22 + 1 denote the kth Fer-
mat number, where k is a nonnegative integer. Then for all
positive integers n, we have
F0F1F2 · · · Fn−1 = Fn − 2.

CHAPTER 6 8


(P.133-134)
Theorem 1.8. If a, b, and c are integers with a | b and
b | c, then a | c. (P.37)

CHAPTER 6 9

Chapter 6

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Chapter 6