### Discrete Mathematics

WEN-CHING LIEN Department of Mathematics National Cheng Kung University

2008

### 4.5: The Fundamental Theorem of Arithmetic.

Lemma (4.2)

*IF a,b*∈ Z^{+}*and p is prime, then p|ab*⇒*or p|b.*

proof 4.2.

*If p|a, then we are finished.If not, then because p is prime, it*
*follows that gcd*(p,*a) =*1,and so there exist integer x,y with
*px* +*ay* =1.*Then b*=*p(bx*) + (ab)y,*where p|p and p|ab.So it*
*follows from parts (d) and (e) of Theorem 4.3 that p|b.*

### 4.5: The Fundamental Theorem of Arithmetic.

Lemma (4.2)

*IF a,b*∈ Z^{+}*and p is prime, then p|ab*⇒*or p|b.*

proof 4.2.

*If p|a, then we are finished.If not, then because p is prime, it*
*follows that gcd*(p,*a) =*1,and so there exist integer x,y with
*px* +*ay* =1.*Then b*=*p(bx*) + (ab)y,*where p|p and p|ab.So it*
*follows from parts (d) and (e) of Theorem 4.3 that p|b.*

### 4.5: The Fundamental Theorem of Arithmetic.

Lemma (4.2)

*IF a,b*∈ Z^{+}*and p is prime, then p|ab*⇒*or p|b.*

proof 4.2.

*If p|a, then we are finished.If not, then because p is prime, it*
*follows that gcd*(p,*a) =*1,and so there exist integer x,y with
*px* +*ay* =1.*Then b*=*p(bx*) + (ab)y,*where p|p and p|ab.So it*
*follows from parts (d) and (e) of Theorem 4.3 that p|b.*

### 4.5: The Fundamental Theorem of Arithmetic.

Lemma (4.2)

*IF a,b*∈ Z^{+}*and p is prime, then p|ab*⇒*or p|b.*

proof 4.2.

*If p|a, then we are finished.If not, then because p is prime, it*
*follows that gcd*(p,*a) =*1,and so there exist integer x,y with
*px* +*ay* =1.*Then b*=*p(bx*) + (ab)y,*where p|p and p|ab.So it*
*follows from parts (d) and (e) of Theorem 4.3 that p|b.*

### 4.5: The Fundamental Theorem of Arithmetic.

Lemma (4.2)

*IF a,b*∈ Z^{+}*and p is prime, then p|ab*⇒*or p|b.*

proof 4.2.

*If p|a, then we are finished.If not, then because p is prime, it*
*follows that gcd*(p,*a) =*1,and so there exist integer x,y with
*px* +*ay* =1.*Then b*=*p(bx*) + (ab)y,*where p|p and p|ab.So it*
*follows from parts (d) and (e) of Theorem 4.3 that p|b.*

Lemma (4.3)

*Let a** _{i}* ∈ Z

^{+},

*for all 1*≤

*i*≤

*n.if p is prime and p|a*

_{1}

*a*

_{2}· · ·

*a*

*,*

_{n}*then p*|a

*i*

*for some 1*≤

*i*≤

*n.*

Theorem (4.11)

*Every integer n* >*1 can be written as a product of primes*
*uniquely, up to the order of the primes.(Here a single prime is*
*considered a product of one factor)*

Lemma (4.3)

*Let a** _{i}* ∈ Z

^{+},

*for all 1*≤

*i*≤

*n.if p is prime and p|a*

_{1}

*a*

_{2}· · ·

*a*

*,*

_{n}*then p*|a

*i*

*for some 1*≤

*i*≤

*n.*

Theorem (4.11)

*Every integer n* >*1 can be written as a product of primes*
*uniquely, up to the order of the primes.(Here a single prime is*
*considered a product of one factor)*

proof 4.11 existence.

The proof consists of two parts: The first part covers the existence of a prime factorization, and the second part deals with its uniqueness.

*If the first part is not true, let m*>1 be the smallest integer not
expressible as a product of primes.

*Since m is not a prime, we are able to write m*=*m*_{1}*m*_{2},where
1<*m*_{1}<*m,*1<*m*_{2}<*m.*

*But then m*_{1},*m*_{2}can be written as products of primes, because
*they are less than m. Consequently, with m*=*m*_{1}*m*_{2}we can
obtain a prime factorization of m.

proof 4.11 existence.

The proof consists of two parts: The first part covers the existence of a prime factorization, and the second part deals with its uniqueness.

*If the first part is not true, let m*>1 be the smallest integer not
expressible as a product of primes.

*Since m is not a prime, we are able to write m*=*m*_{1}*m*_{2},where
1<*m*_{1}<*m,*1<*m*_{2}<*m.*

*But then m*_{1},*m*_{2}can be written as products of primes, because
*they are less than m. Consequently, with m*=*m*_{1}*m*_{2}we can
obtain a prime factorization of m.

proof 4.11 existence.

The proof consists of two parts: The first part covers the existence of a prime factorization, and the second part deals with its uniqueness.

*If the first part is not true, let m*>1 be the smallest integer not
expressible as a product of primes.

*Since m is not a prime, we are able to write m*=*m*_{1}*m*_{2},where
1<*m*_{1}<*m,*1<*m*_{2}<*m.*

*But then m*_{1},*m*_{2}can be written as products of primes, because
*they are less than m. Consequently, with m*=*m*_{1}*m*_{2}we can
obtain a prime factorization of m.

proof 4.11 existence.

*If the first part is not true, let m*>1 be the smallest integer not
expressible as a product of primes.

*Since m is not a prime, we are able to write m*=*m*_{1}*m*_{2},where
1<*m*_{1}<*m,*1<*m*_{2}<*m.*

*But then m*_{1},*m*_{2}can be written as products of primes, because
*they are less than m. Consequently, with m*=*m*_{1}*m*_{2}we can
obtain a prime factorization of m.

proof 4.11 uniqueness.

In order to establish the uniqueness of a prime factorization, we shall use the alternative form of the Principle of Mathematical Induction.

For the integer 2, we have a unique prime factorization, and
assuming uniqueness of representation for 3,4,5, . . . ,*n*−1,we
*suppose that n* =*p*_{1}^{s}^{1}*p*_{2}^{s}^{2}· · ·*p*_{k}^{s}* ^{k}* =

*q*

_{1}

^{t}^{1}

*q*

_{2}

^{t}^{2}· · ·

*q*

_{r}

^{t}*,where each*

^{r}*p*

*,1≤*

_{i}*i*≤

*k*,

*and each q*

*,1≤*

_{j}*j*≤

*r*,is a prime.Also

*p*1<*p*2< · · · <*p*_{k}*and q*1<*q*2< · · · <*q**r*,*and s** _{i}* ≥0 for all
1≤

*i*≤

*k*,

*t*

*>0 for all 1≤*

_{i}*j*≤

*r*.

proof 4.11 uniqueness.

In order to establish the uniqueness of a prime factorization, we shall use the alternative form of the Principle of Mathematical Induction.

For the integer 2, we have a unique prime factorization, and
assuming uniqueness of representation for 3,4,5, . . . ,*n*−1,we
*suppose that n* =*p*_{1}^{s}^{1}*p*_{2}^{s}^{2}· · ·*p*_{k}^{s}* ^{k}* =

*q*

_{1}

^{t}^{1}

*q*

_{2}

^{t}^{2}· · ·

*q*

_{r}

^{t}*,where each*

^{r}*p*

*,1≤*

_{i}*i*≤

*k*,

*and each q*

*,1≤*

_{j}*j*≤

*r*,is a prime.Also

*p*1<*p*2< · · · <*p*_{k}*and q*1<*q*2< · · · <*q**r*,*and s** _{i}* ≥0 for all
1≤

*i*≤

*k*,

*t*

*>0 for all 1≤*

_{i}*j*≤

*r*.

proof 4.11 uniqueness.

In order to establish the uniqueness of a prime factorization, we shall use the alternative form of the Principle of Mathematical Induction.

For the integer 2, we have a unique prime factorization, and
assuming uniqueness of representation for 3,4,5, . . . ,*n*−1,we
*suppose that n* =*p*_{1}^{s}^{1}*p*_{2}^{s}^{2}· · ·*p*_{k}^{s}* ^{k}* =

*q*

_{1}

^{t}^{1}

*q*

_{2}

^{t}^{2}· · ·

*q*

_{r}

^{t}*,where each*

^{r}*p*

*,1≤*

_{i}*i*≤

*k*,

*and each q*

*,1≤*

_{j}*j*≤

*r*,is a prime.Also

*p*1<*p*2< · · · <*p*_{k}*and q*1<*q*2< · · · <*q**r*,*and s** _{i}* ≥0 for all
1≤

*i*≤

*k*,

*t*

*>0 for all 1≤*

_{i}*j*≤

*r*.

4.11 continued.

*Since p*_{1}|n,*we have p*_{1}|q_{1}^{t}^{1}*q*_{2}^{t}^{2}· · ·*q**r**t**r*.*By lemma 4.3, p*_{1}|q* _{j}* for
some 1≤

*j*≤

*r*.

*Because p*1*and q*_{j}*are primes, we have p*1=*q** _{j}*.

*In fact j=1, for otherwise q*_{1}|n⇒*q*_{1}=*p** _{e}*for some 1<

*e*≤

*l and*

*p*

_{1}<

*p*

*=*

_{e}*q*

_{1}<

*q*

*=*

_{j}*p*

_{1}.

*With p*

_{1}=

*q*

_{1}, we find that

*n*_{1}=*n/p*_{1}=*p*_{1}^{s}^{1}^{−1}*p*_{2}^{s}^{2}· · ·*p*_{k}^{s}* ^{k}* =

*q*

_{1}

^{t}^{1}

^{−1}

*q*

_{2}

^{t}^{2}· · ·

*q*

*r*

*t*

*r*.

*Since n*

_{1}<

*n,*by the induction hypothesis it follows that

*k*=

*r*,

*p*

*=*

_{i}*q*

*for 1≤*

_{i}*i*≤

*k*,

*s*

_{1}−1=

*t*

_{1}−

*1 so (s*

_{1}=

*t*

_{1}), and

*s*

*=*

_{i}*t*

*for 2≤*

_{i}*i*≤

*k*.

Hence the prime factorization of n is unique.

4.11 continued.

*Since p*_{1}|n,*we have p*_{1}|q_{1}^{t}^{1}*q*_{2}^{t}^{2}· · ·*q**r**t**r*.*By lemma 4.3, p*_{1}|q* _{j}* for
some 1≤

*j*≤

*r*.

*Because p*1*and q*_{j}*are primes, we have p*1=*q** _{j}*.

*In fact j=1, for otherwise q*_{1}|n⇒*q*_{1}=*p** _{e}*for some 1<

*e*≤

*l and*

*p*

_{1}<

*p*

*=*

_{e}*q*

_{1}<

*q*

*=*

_{j}*p*

_{1}.

*With p*

_{1}=

*q*

_{1}, we find that

*n*_{1}=*n/p*_{1}=*p*_{1}^{s}^{1}^{−1}*p*_{2}^{s}^{2}· · ·*p*_{k}^{s}* ^{k}* =

*q*

_{1}

^{t}^{1}

^{−1}

*q*

_{2}

^{t}^{2}· · ·

*q*

*r*

*t*

*r*.

*Since n*

_{1}<

*n,*by the induction hypothesis it follows that

*k*=

*r*,

*p*

*=*

_{i}*q*

*for 1≤*

_{i}*i*≤

*k*,

*s*

_{1}−1=

*t*

_{1}−

*1 so (s*

_{1}=

*t*

_{1}), and

*s*

*=*

_{i}*t*

*for 2≤*

_{i}*i*≤

*k*.

Hence the prime factorization of n is unique.

4.11 continued.

*Since p*_{1}|n,*we have p*_{1}|q_{1}^{t}^{1}*q*_{2}^{t}^{2}· · ·*q**r**t**r*.*By lemma 4.3, p*_{1}|q* _{j}* for
some 1≤

*j*≤

*r*.

*Because p*1*and q*_{j}*are primes, we have p*1=*q** _{j}*.

*In fact j=1, for otherwise q*_{1}|n⇒*q*_{1}=*p** _{e}*for some 1<

*e*≤

*l and*

*p*

_{1}<

*p*

*=*

_{e}*q*

_{1}<

*q*

*=*

_{j}*p*

_{1}.

*With p*

_{1}=

*q*

_{1}, we find that

*n*_{1}=*n/p*_{1}=*p*_{1}^{s}^{1}^{−1}*p*_{2}^{s}^{2}· · ·*p*_{k}^{s}* ^{k}* =

*q*

_{1}

^{t}^{1}

^{−1}

*q*

_{2}

^{t}^{2}· · ·

*q*

*r*

*t*

*r*.

*Since n*

_{1}<

*n,*by the induction hypothesis it follows that

*k*=

*r*,

*p*

*=*

_{i}*q*

*for 1≤*

_{i}*i*≤

*k*,

*s*

_{1}−1=

*t*

_{1}−

*1 so (s*

_{1}=

*t*

_{1}), and

*s*

*=*

_{i}*t*

*for 2≤*

_{i}*i*≤

*k*.

Hence the prime factorization of n is unique.

4.11 continued.

*Since p*_{1}|n,*we have p*_{1}|q_{1}^{t}^{1}*q*_{2}^{t}^{2}· · ·*q**r**t**r*.*By lemma 4.3, p*_{1}|q* _{j}* for
some 1≤

*j*≤

*r*.

*Because p*1*and q*_{j}*are primes, we have p*1=*q** _{j}*.

*In fact j=1, for otherwise q*_{1}|n⇒*q*_{1}=*p** _{e}*for some 1<

*e*≤

*l and*

*p*

_{1}<

*p*

*=*

_{e}*q*

_{1}<

*q*

*=*

_{j}*p*

_{1}.

*With p*

_{1}=

*q*

_{1}, we find that

*n*_{1}=*n/p*_{1}=*p*_{1}^{s}^{1}^{−1}*p*_{2}^{s}^{2}· · ·*p*_{k}^{s}* ^{k}* =

*q*

_{1}

^{t}^{1}

^{−1}

*q*

_{2}

^{t}^{2}· · ·

*q*

*r*

*t*

*r*.

*Since n*

_{1}<

*n,*by the induction hypothesis it follows that

*k*=

*r*,

*p*

*=*

_{i}*q*

*for 1≤*

_{i}*i*≤

*k*,

*s*

_{1}−1=

*t*

_{1}−

*1 so (s*

_{1}=

*t*

_{1}), and

*s*

*=*

_{i}*t*

*for 2≤*

_{i}*i*≤

*k*.

Hence the prime factorization of n is unique.

4.11 continued.

*Since p*_{1}|n,*we have p*_{1}|q_{1}^{t}^{1}*q*_{2}^{t}^{2}· · ·*q**r**t**r*.*By lemma 4.3, p*_{1}|q* _{j}* for
some 1≤

*j*≤

*r*.

*Because p*1*and q*_{j}*are primes, we have p*1=*q** _{j}*.

*In fact j=1, for otherwise q*_{1}|n⇒*q*_{1}=*p** _{e}*for some 1<

*e*≤

*l and*

*p*

_{1}<

*p*

*=*

_{e}*q*

_{1}<

*q*

*=*

_{j}*p*

_{1}.

*With p*

_{1}=

*q*

_{1}, we find that

*n*_{1}=*n/p*_{1}=*p*_{1}^{s}^{1}^{−1}*p*_{2}^{s}^{2}· · ·*p*_{k}^{s}* ^{k}* =

*q*

_{1}

^{t}^{1}

^{−1}

*q*

_{2}

^{t}^{2}· · ·

*q*

*r*

*t*

*r*.

*Since n*

_{1}<

*n,*by the induction hypothesis it follows that

*k*=

*r*,

*p*

*=*

_{i}*q*

*for 1≤*

_{i}*i*≤

*k*,

*s*

_{1}−1=

*t*

_{1}−

*1 so (s*

_{1}=

*t*

_{1}), and

*s*

*=*

_{i}*t*

*for 2≤*

_{i}*i*≤

*k*.

Hence the prime factorization of n is unique.

Example (4.46)

Can we find three consecutive positive integers whose product
*is a perfect square— that is, do there exist m,n*∈ Z^{+}with
*m(m*+1)(m+2) =*n*^{2}?