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CHAPTER 3
Number Theory
1. Factors or not
According to Carl Friedrich Gauss (1777-1855) mathematics is the “queen of sciences” and number theory is the “queen of mathematics”, where “queen” stands for elevated and beautiful. Number theory is mainly the study of the system of integers Z = { 0 , ± 1 , ± 2 ,.. .} and the consequences of the fact that division is not always possible within Z. E.g. 15/ 3 ∈ Z but 15/ 4 ∈/ Z. Let us start with the well-known “long division” that has rather surprising consequences.
EXAMPLE 1.1.
What is really being done and what has been achieved? On the right some zeros are filled in that are not written on the left. We now see that
88323 − 300 · 223 − 90 · 223 − 6 · 223 = 45.
In effect, multiples of 223 were subtracted from 88323 until 223 could not be subtracted anymore without running into negative numbers. The mathematical content of “long divi- sion” is the following theorem and “long division” is just an efficient way of finding q and r in the “Division Theorem”.
1
2 3. NUMBER THEORY
THEOREM 1.2. Division Theorem. Given integers a, b, b > 0 , there exist unique integers q and r such that
a = qb + r , 0 ≤ r < b
The quotient q and the remainder r can be found by repeated subtraction of b. The division algorithm is just an efficient way for computing q and r by repeated subtractions.
EXAMPLE 1.3. Let a = 567923 and b = 735_. Find non-negative integers q and r such that_ 567923 = 735 q + r and 0 ≤ r < 735_._ Answer: By Long Division we find q = 772 and r = 503_._
EXAMPLE 1.4. Let a = 3257397 and b = 7593_. Find non-negative integers q and r such that_ 3257397 = 7593 q + r and 0 ≤ r < 7593_._ Answer: By Long Division we find q = 429 and r = 0_._
DEFINITION 1.5. An integer a is even if a = 2 x for some integer x, i.e., r = 0 in Theorem ??. An integer b is odd if b = 2 x + 1 for some integer x in Z , i.e., r = 1 in Theorem ??.
PROPOSITION 1.6. Every integer is either even or odd. The product of two odd integers is odd.
PROOF. Let a and b be odd integers. According to the definition of “odd integer” there exist integers x and y such that a = 2 x + 1 and b = 2 y + 1. Then
ab = ( 2 x + 1 )( 2 y + 1 ) = ( 2 x + 1 )( 2 y ) + ( 2 x + 1 ) · 1 = 2 ( y ( 2 x + 1 )) + 2 x + 1 = 2 ( y ( 2 x + 1 ) + x ) + 1.
Here z = y ( 2 x + 1 ) + x is an integer by the closure properties of Z, hence ab = 2 z + 1 is an odd number.
THEOREM 1.7.
2 is not rational. PROOF. (Aristotle) By way of contradiction assume that
2 = a / b where either a or b is odd. Then 2 b^2 = a^2 , hence a^2 is even and therefore a is even. This means that a = 2 a ′ and substituting 2 b^2 = 4 a ′^2. Thus b^2 = 2 a ′^2. This says that b is even, a contradiction.
DEFINITION 1.8.
(1) Let a , f be integers. We say that f is a factor of a if a = f · some integer or a = f · a ′^ for some integer a ′. (2) Let a and b be given integers. An integer f is a common factor of a and b if f is a factor of a and a factor of b. (3) The greatest common factor of two integers a and b is the largest among the common factors of a and b. The greatest common factor of a and b is denoted by gcf( a , b ).
4 3. NUMBER THEORY
EXERCISE 1.16.
(1) Rewrite the number 1239584 in the form as 1239584 = 12395 · 100 + 84_. Use Lemma_ ?? and Lemma ?? to show that 4 is a factor of 1239584 because 4 is a factor of 84_._ (2) Rewrite the number 1239582 in the form as 1239582 = 12395 · 100 + 82_. Use Lemma_ ?? and Lemma ?? to show that 4 is NOT a factor of 1239582 because 4 is NOT a factor of 82_._
The Euclidean Algorithm is a very beautiful and efficient way of finding the greatest com- mon factor of two integers. It is bases on the following fact.
LEMMA 1.17. Let a and b be integers and a = bq + r for some integers q and r. Then the common factors of a and b are exactly the same as those of b and r. In particular,
gcf( a , b ) = gcf( b , r ).
EXAMPLE 1.18. Let a = 56371 and b = 3476. Then a = 16 · b + 755, hence
gcf( 56371 , 3476 ) = gcf( 3476 , 755 ).
Further, 3476 = 4 · 755 + 456, so
gcf( 56371 , 3476 ) = gcf( 3476 , 755 ) = gcf( 755 , 456 ).
Further, 755 = 1 · 456 + 299, 456 = 1 · 299 + 157, 299 = 1 · 157 + 122, 157 = 1 · 122 + 35, hence
gcf( 56371 , 3476 ) = gcf( 755 , 456 ) = gcf( 456 , 299 ) = gcf( 299 , 157 ) = gcf( 157 , 122 ) = gcf( 122 , 35 ).
It is clear that 35 has the positive factors 1, 5 , 7 , 35 and of these only 1 is a factor of 122. Hence gcf( 56371 , 3476 ) = 1.
THEOREM 1.19. Let a and b be (positive) integers. Then there exist integers u and v such that
gcf( a , b ) = ua + vb.
Consequently, every common factor of a and b is a factor of gcf( a , b ). The integers u, v and gcf( a , b ) can be found efficiently using the Euclidean Algorithm described below.
EXAMPLE 1.20. Let a = 569321 and b = 347_. Then_ gcf( 69321 , 347 ) = gcf( a , b ) = ua + vb where gcf( a , b ) = 1 , u = 36 and v = − 59065_._
PROOF.
The long divisions used:
The same process can be done in a short form that only lists the essential data.
- 1. FACTORS OR NOT - ( 1 ) 569321 = 1 · 569321 + 0 · - ( 2 ) 347 = 0 · 569321 + 1 ·
- ( 3 ) = ( 1 ) − 1640 · ( 2 ) 241 = 1 · 569321 − 1640 ·
- ( 4 ) = ( 2 ) − 1 · ( 3 ) 106 = − 1 · 569321 + 1641 ·
- ( 5 ) = ( 3 ) − 2 · ( 4 ) 29 = 3 · 569321 − 4922 ·
- ( 6 ) = ( 4 ) − 3 · ( 5 ) 19 = − 10 · 569321 + 16407 ·
- ( 7 ) = ( 5 ) − ( 6 ) 10 = 13 · 569321 − 21329 ·
- ( 8 ) = ( 6 ) − ( 7 ) 9 = − 23 · 569321 + 37736 ·
- ( 9 ) = ( 7 ) − ( 8 ) 1 = 36 · 569321 − 59065 · - 569321 = 1640 · 347 + - 347 = 1 · 241 + - 241 = 2 · 106 + - 106 = 3 · 29 + - 29 = 1 · 19 + - 19 = 1 · 10 + - 10 = 1 · 9 + - (1) action a b - (2)
- ( 3 ) = ( 1 ) − 1640 · ( 2 ) 241 1 -
- ( 4 ) = ( 2 ) − 1 · ( 3 ) 106 -1 +
- ( 5 ) = ( 3 ) − 2 · ( 4 ) 29 3 -
- ( 6 ) = ( 4 ) − 3 · ( 5 ) 19 -10 + - (7)=(5)-(6) 10 13 - - (8)=(6)-(7) 9 -23 + - (9)=(7)-(8) 1 36 -
- EXAMPLE 1.21. Let a = 3675 and b = 791 Then gcf( a , b ) = 7 = 48 · 3675 − 223 ·
- THE FUNDAMENTAL THEOREM OF ARITHMETIC 7
EXAMPLE 1.27. 1 240 +^
1 330 =^
11 2640 +^
6 2640 =^
17
2. The Fundamental Theorem of Arithmetic
Given a positive integer, say 113, it can always be factored as 113 = 1 · 113, in general a = 1 · a = a · 1. This is an uninteresting “trivial” factorization.
DEFINITION 2.1. An (positive) integer a is composite if it can be factored as a = b · c where neither b nor c is equal to one. In other words, the (positive) integer a is composite if it is the product of two positive integers that are both smaller than a. A positive integer that is not composite - can only be factored trivially - is a prime number or simply a prime.
LEMMA 2.2. Let p be a prime. Then p has exactly two positive factors, namely 1 and p. Let a be any other integer. Then gcf( a , p ) = 1 or gcf( a , p ) = p depending on whether p is a factor of a or not.
EXAMPLE 2.3. The numbers 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 91 , 97 are all prime numbers. The number 123 is composite because 123 = 3 · 41_. The number_ 3127 is composite because 3127 = 53 · 59_. The number_ 1111 is composite because 1111 = 11 · 101_._
PROPOSITION 2.4. Let n be a positive integer and suppose that
n = a · b
where a and b are again positive integers. If b ≥
n, then a ≤
n. PROOF. If b ≥
n and also a >
n then a · b >
n ·
n = n while a · b = n by hypoth- esis.
COROLLARY 2.5. If the positive integer p has no factors that are less than or equal to
p then p is a prime number.
PROOF. If p had a factor larger than
n then it would have a factor ≤
n and these have all been eliminated.
COROLLARY 2.6. If the positive integer p has no prime factors which are less than or equal to
p then p is a prime number.
PROOF. If p has no prime factors ≤
n then there is no composite factor ≤ n either, since such a factor would in turn contain a prime factor of n.
THEOREM 2.7. There exist infinitely many primes.
PROOF. This is an example of a proof by contradiction. We make an assumption, then derive an impossibility from it and conclude that the assumption was false. Assume that there are only finitely many primes. List them in increasing order:
(2.8) 2 , 3 , 5 , 7 , 11 , 13 , 17 ,.. ., P ,
8 3. NUMBER THEORY
so P is the largest prime. Study the number n obtained by multiplying all the primes and adding 1:
n = 2 · 3 · 5 · 7 · 11 · 13 · 17 · · · p · · · P + 1.
Now 2 is not a factor of n since 2 is not a factor of 1; 3 is not a factor of n since 3 is not a factor of 1; 5 is not a factor of n since 5 is not a factor of 1; in general, p is not a factor or n because p is not a factor of 1. So n is either a prime itself different from any one in the list ( ?? ) or it is a product of primes, each of which is not contained in the list ( ?? ). This means that the list ( ?? ) did not contain all the primes after all, so the assumption that there are only finitely many primes is false. Therefore, there are infinitely many primes.
THEOREM 2.9. Fundamental Theorem of Arithmetic. Every integer a > 1 can be fac- tored uniquely as
a = pn 11 pn 22 · · · pn k k ,
ni ≥ 1 , p 1 < p 2 < · · · < pk , pi primes.
We will indicate a proof later. For the moment we observe that every integer a > 1 is a product of primes for the following reason. If a is itself a prime it is considered a “product of primes”. If a is composite, then a = a 1 · a 2 and both a 1 and a 2 are smaller than a. Now look at a 1 and a 2. Factor them if they are composite, and keep going until you arrive at primes. The process have to come to a halt because integers cannot become smaller and smaller forever.
EXAMPLE 2.10. 155771 = 539 · 289 = ( 11 · 49 ) · ( 17 · 17 ) = 11 · ( 7 · 7 ) · 17 · 17 = 11 · 7 · 7 · 17 · 17 = 72 · 11 · 172_._
Before getting into the uniqueness proof, let us see what it can do for us. The uniqueness says that however we arrive at a factorization into a product of primes, the result is always the same.
EXAMPLE 2.11. Let a = 23 · 52 · 7_. Now suppose that b is a factor of a and a_ = b · c. Let b = pn 11 · · · pn kk and c = pm 1 1 · · · pm k k, where p 1 < p 2 < · · · < pk, and 0 ≤ ni , mi, be the prime
factorizations of b and c as in Theorem ??. Now a = b · c = pn 11 + m^1 · · · pn k k^ + mk. By the uniqueness of the prime factorization we must have
a = 23 · 52 · 7 = pn 11 + m^1 · · · pn k k^ + mk.
This means that p 1 = 2 , n 1 + m 1 = 3 , p 2 = 5 , n 2 + m 2 = 2 , p 3 = 7 , and n 3 + m 3 = 1_. We conclude that k_ = 3 , n 1 ≤ 3 , n 2 ≤ 2 and n 3 ≤ 1_._
10 3. NUMBER THEORY
COROLLARY 2.16. Let p , p 1 , p 2 primes. If p is a factor of p 1 p 2 , then p = p 1 or p = p 2_._
COROLLARY 2.17. Let p , p 1 ,... , pn be primes. If p is a factor of p 1 · p 2 · · · pn, then p = p 1 or p = p 2 , or ... or p = pn.
Corollary ?? is the result that is needed to prove the uniqueness part of the Fundamental Theorem of Arithmetic. We illustrate the formal proof by an example.
EXAMPLE 2.18. We find that
10500 = 2 · 2 · 3 · 5 · 5 · 5 · 7
Somebody else come up with a “mysterious” factorization
(2.19) 10500 = pn 11 pn 22 · · · pn kk , p 1 < p 2 < · · · < pk.
Therefore we have that 2 · 2 · 3 · 5 · 5 · 5 · 7 = pn 11 pn 22 · · · pn kk
Take a factor pi of the right hand side which is then also a factor of the left hand side. By Corollary ?? pi = 2 or pi = 3 or pi = 5 or pi = 7. Hence ( ?? ) must look like
10500 = 2 n^1 3 n^2 5 n^3 7 n^4
and we have 2 · 2 · 3 · 5 · 5 · 5 · 7 = 2 n^1 3 n^2 5 n^3 7 n^4
Now 2 is a factor of the left hand side so it is a factor of the right hand side by Corollary ??. This means that n 1 ≥ 1 and we can cancel 2 to get
2 · 3 · 5 · 5 · 5 · 7 = 2 n^1 −^13 n^2 5 n^3 7 n^4
The prime 2 is still a factor of the left hand side, so of the right hand side, and we must have n 1 − 1 ≥ 1. This means we can cancel another 2 and obtain
3 · 5 · 5 · 5 · 7 = 2 n^1 −^23 n^2 5 n^3 7 n^4.
Now there is no 2 on the left so there cannot be a 2 on the right and we conclude that n 1 − 2 = 0 or n 1 = 2. We next look at the prime 3 that appears on the left. It must appear on the right also, so n 2 ≥ 1. Canceling the 3 we have
5 · 5 · 5 · 7 = 3 n^2 −^15 n^3 7 n^4.
We conclude that n 2 = 1 because no 3 appears on the left and have now
5 · 5 · 5 · 7 = 5 n^3 7 n^4.
We have a five on the left, so n 3 ≥ 1 enabling us to cancel a 5 and get
5 · 5 · 7 = 5 n^3 −^17 n^4.
We still have a five on the left, so n 3 − 1 ≥ 1 enabling us to cancel another 5 to get
5 · 7 = 5 n^3 −^27 n^4.
- THE FUNDAMENTAL THEOREM OF ARITHMETIC 11
We still have a five on the left, so n 3 − 2 ≥ 1 enabling us to cancel another 5 to get
7 = 5 n^3 −^37 n^4.
We finally get n 3 = 3 and n 4 = 1 showing that the “mysterious” factorization is identical with ours.
PROPOSITION 2.20. Common Factors. Suppose
a = pn 11 pn 22 pn 33... pn kk
is the factorization of the integer a into a product of primes, and
b = pm 1 1 pm 2 2 pm 3 3... pm kk
is the factorization of the integer b into a product of primes, then the common factors of a and b are all integers
f = pE 1 1 pE 2 2 pE 3 3... pE kk
where Ei is less than or equal to both ni and mi. Therefore the greatest common factor of a and b is
gcf( a , b ) = pe 11 pe 22 pe 33... pe kk
where ei is the lesser of ni and mi.
PROPOSITION 2.21. Multiples.
(1) The positive multiples of an integer a are the integers a · b where b is any positive integer. (2) If a = pn 11 pn 22 pn 33... pn kk is the factorization of a into a product of primes, then making the exponents ni larger produces multiples, also bringing in additional primes produces multiples. (3) The multiples of a are all integers of the form m = pm 1 1 pm 2 2 pm 3 3... pm k kqs 11 qs 22 qs 33 · · · where mi ≥ ni and si ≥ 0_._
PROPOSITION 2.22. Common Multiples. Suppose
a = pn 11 pn 22 pn 33... pn kk
is the factorization of the integer a into a product of primes, and
b = pm 1 1 pm 2 2 pm 3 3... pm kk
is the factorization of the integer b into a product of primes, then the common multiples of a and b are all integers
m = pM 1 1 pM 2 2 pM 3 3... pM k kqs 11 qs 22 qs 33 · · ·
- EXERCISES 13
EXERCISE 3.4. Find the greatest common factor of the following integers a and b using the method of Euclid.
(1) a = 543 , b = 113_._ (2) a = 4563 , b = 981_._ (3) a = 451 , b = 85_._ (4) a = 1111 , b = 11111_._ (5) a = 12345 , b = 1234_._
EXERCISE 3.5. Find the greatest common factor of the following integers a and b and find integers u, v such that gcf( a , b ) = ua + vb.
(1) a = 543 , b = 113_._ (2) a = 4563 , b = 981_._ (3) a = 451 , b = 85_._ (4) a = 1111 , b = 11111_._ (5) a = 12345 , b = 1234_._
EXERCISE 3.6. Find the least common multiple of the following integers a and b by listing common multiples and looking for the least one among these.
(1) a = 543 , b = 113_._ (2) a = 56 , b = 98_._ (3) a = 45 , b = 85_._ (4) a = 111 , b = 111_._ (5) a = 123 , b = 234_._
EXERCISE 3.7. Find the least common multiple of the following integers a and b using Theorem ??.
(1) a = 543 , b = 112_._ (2) a = 4563 , b = 981_._ (3) a = 451 , b = 85_._ (4) a = 1111 , b = 11111_._ (5) a = 12345 , b = 1234_._
EXERCISE 3.8. Compute the following sums and differences using least common denomi- nators.
(1)
14 3. NUMBER THEORY
EXERCISE 3.9. Which ones of the following numbers are prime and which ones are com- posite?
211 , 373 , 453 , 565 , 463 , 371 , 637 , 343 , 111111 , 3487 + 539.
EXERCISE 3.10. How many positive factors do the following integers possess?
24 , 67 , 69 , 2445 , 1111 , 999 ,
56 , p^7 where p is a prime , 7 n , pn^ where p is a prime.
EXERCISE 3.11. Using your calculator find integers q and r such that
1238459 = q 4593 + r , 0 ≤ r < 4593.
EXERCISE 3.12. Let x and y be (unknown) integers that are related by the equality
x − 130 = 75 y.
Why are the following statements true?
(1) 5 is a factor of x. (2) 3 is not a factor of x. (3) If y has a factor of 13 , then x has a factor of 13_._ (4) If y is even, then x is also even. (5) If y is odd, then x is also odd. (6) If gcf( 13 , y ) = 1 , then 13 cannot be a factor of x.
EXERCISE 3.13. Verify the following statements.
(1) odd × odd = odd. (2) even × even = even. (3) even × odd = even. (4) odd + odd = even. (5) even + even = even.
EXERCISE 3.14. Let a = 25 × 42 × 19 × b where b is some unknown positive integer such that gcf( 2 × 5 × 19 , b ) = 1_. Which of the following numbers are factors of a and which ones are not?_
16 , 80 , 95 , 361 , 100 , 76.
EXERCISE 3.15. The integer 378 has 16 different positive factors and a partial list of factors is
1 , 2 , 3 , 6 , 7 , 9 , 14 , 18 , 21 , 63 , 378.
Complete the list.
16 3. NUMBER THEORY
?? These problems are very time consuming and unpleasant. The lesson is that with more mathematics life gets much easier. The answers given are obtained in the “advanced fash- ion”. ?? gcf( 543 , 112 ) = 1, lcm( 543 , 112 ) = 543 × 112 / gcf( 543 , 112 ) = 60 816, gcf( 4563 , 981 ) = 9; lcm( 4563 , 981 ) = 4563 × 981 / gcf( 4563 , 981 ) = 497 367 gcf( 451 , 85 ) = 1; 451 × 85 = 38 335 gcf( 1111 , 11111 ) = 1; lcm( 1111 , 11111 ) = 1111 × 11111 = 12 344 321 gcf( 12345 , 1234 ) = 1; lcm( 12345 , 1234 ) = 12345 × 1234 = 15 233 730 ?? 28 543 +^
25 213 =^
2171 12 851 23 323 +^
12 437 =^
733 7429 61 245 −^
51 300 =^
387 4900 14 289 +^
23 221 =^
573 3757 1 2599 −^
1 12769 =^
90 293 687 ?? 211 = 211 is prime; 373 = 373 is prime; 453 = 3 × 151 is composite; 565 = 5 × 113 is composite; 463 = 463 is prime; 371 = 7 × 53 is composite 637 = 72 13 is composite; 343 = 73 is composite; 111111 = 3 × 7 × 11 × 13 × 37 is comp[osite; 3487 + 539 = 4026 = 2 × 3 × 11 × 61 is composite. ?? 1238459 ÷ 4593 = 269. 6.. .. So q = 269 and r = 1238459 − 269 × 4593 = 2942. ??
(1) 5 is a factor of x because 5 is a factor of 75 y and 130. (2) 3 is not a factor of x because if it were, then 3 would be a factor of x and 75, hence of 75 y and x , hence of 130 which is not true. (3) If y has a factor of 13, then x has a factor of 13. True. (4) If y is even, then x is also even. True. (5) If y is odd, then x is also odd. True. (6) If gcf( 13 , y ) = 1, then 13 cannot be a factor of x. True. Assume to the contrary that 13 is a factor of x. Then 13 is a factor of x and 130 hence of 75 y. Since gcf( 13 , y ) = 1, we know that 13 is not a factor of y so it would have to be a factor of 75, which is false.
??
(1) odd × odd = odd. This was done in class. (2) even × even = even. Take two even numbers x and y. Being even they are of the form x = 2 x ′, y = 2 y ′. Hence xy = 2 x ′^ · 2 y ′^ = 4 x ′ y ′, so xy even has a factor 4 which is more than having a factor 2. (3) even × odd = even. Let x be even and y be odd. Then x = 2 x ′^ and y = 2 y ′^ + 1. Hence xy = 2 x ′ y which is even. (4) odd + odd = even. True. (5) even + even = even. True.
- SOLUTION TO EXERCISES 17
?? Let a = 25 × 42 × 19 × b where b is some unknown positive integer such that Since gcf( 2 × 5 × 19 , b ) = 1 any integer having only factors 2, 5, and 19 must be a factor of 25 × 42 × 19 = 52 × 24 × 19. So
16 : Yes, 80 Yes, 95 Yes, 361 = 192 No, 100 Yes, 76 = 4 × 19 Yes.
?? Look for the complementary factors 378/ 1 = 378, 378/ 2 = 189, 378/ 3 = 126, 378/ 6 = 63, 378/ 7 = 54, 378/ 9 = 42, 378/ 14 = 27, 378/ 18 = 21. Hence the complete list is
1 , 2 , 3 , 6 , 7 , 9 , 14 , 18 , 21 , 27 , 42 , 54 , 63 , 126 , 189 , 378.
