3.3. Marco referencial
3.3.6. Análisis del comportamiento del consumidor de
Integers with certain properties were studied extensively over the centuries. We present some examples of such integers and prove theorems related to these inte- gers and their properties.
We start by defining perfect numbers.
4.4. PERFECT, MERSENNE, AND FERMAT NUMBERS 83 In other words, a perfect number is a positive integer which is the sum of its proper divisors.
Example 43. The first perfect number is 6, sinceσ(6) = 12. You can also view this as 6 = 1 + 2 + 3. The second perfect number is 28, since σ(28) = 56or
28 = 1 + 2 + 4 + 7 + 14.
The following theorem tells us which even positive integers are perfect. Theorem 49. The positive integernis an even perfect number if and only if
n= 2l−1(2l−1),
wherelis an integer such thatl≥2and2l−1is prime.
Proof. We show first that if n = 2l−1(2l −1) where l is an integer such that
l ≥ 2and 2l −1 is prime then n is perfect. Notice that 2l −1is odd and thus
(2l−1,2l−1) = 1. Also, notice thatσis a multiplicative function and thus
σ(n) =σ(2l−1)σ(2l−1).
Notice thatσ(2l−1) = 2l−1and since2l−1is prime we getσ(2l−1) = 2l. Thus
σ(n) = 2n.
We now prove the converse. Suppose that n is a perfect number. Letn = 2rs,
whererandsare positive integers andsis odd. Since(2r, s) = 1, we get
σ(n) =σ(2r)σ(s) = (2r+1−1)σ(s).
Sincenis perfect, we get
(2r+1−1)σ(s) = 2r+1s.
Notice now that(2r+1−1,2r+1) = 1and thus2r+1 |σ(s). Therefore there exists an integerqsuch thatσ(s) = 2r+1q. As a result, we have
and thus we get
(2r+1−1)q=s
So we get thatq |s. We addqto both sides of the above equation and we get
s+q = (2r+1−1)q+q = 2r+1q =σ(s).
We have to show now that q = 1. Notice that if q 6= 1, thens will have three divisors and thusσ(s) ≥ 1 +s+q. Henceq = 1and as a result s = 2r+1 −1. Also notice thatσ(s) =s+ 1. This shows thatsis prime since the only divisors ofsare1ands. As a result,
n= 2r(2r+1−1),
where(2r+1−1)is prime.
In theorem 50, we see that to determine even perfect numbers, we need to find primes of the form2l−1. It is still unknown whether there are odd perfect
numbers or not.
Theorem 50. If2l−1is prime wherelis a positive integer, thenlmust be prime. Proof. Suppose thatlis composite, that isl =rswhere1< r < mand1< s < m. Thus after factoring, we get that
2m−1 = (2r−1)(2r(s−1)+ 2r(s−2)+...+ 2r+ 1)
Notice that the two factors above are both greater than 1. Thus2m−1is not prime. This is a contradiction.
The above theorem motivates the definition of interesting numbers called Mersenne numbers.
Definition 25. Letl be a positive integer. An integer of the formMl = 2l−1is
called thelth Mersenne number; ifl is prime thenMl = 2l −1is called the lth
4.4. PERFECT, MERSENNE, AND FERMAT NUMBERS 85 Example 44. M3 = 23−1 = 7is the third Mersenne prime.
We prove a theorem that help decide whether Mersenne numbers are prime. Theorem 51. Divisors ofMp = 2p−1for primepis of the form2mp+ 1, where
mis a positive integer.
Proof. Letp1 be a prime dividingMp = 2p−1. By Fermat’s theorem, we know
thatp1 |(2p1−1−1). Also, it is easy to see that
(2p−1,2p1−1−1) = 2(p,p1−1) −1.
Sincep1is a common divisor of2p−1and2p1−1−1and thus not relatively prime.
Hence(p, p1−1) = p. Hencep|(p1−1)and thus there exists a positive integer
k such thatp1−1 = kp. Sincep1 is odd, thenkis even and thusk= 2m. Hence
p1 =kp+ 1 = 2mp+ 1.
Because any divisor ofMpis a product of prime divisors ofMp, each prime divisor
ofMpis of the form2mp+ 1and the result follows.
Example 45. M23 = 223−1is divisible by47 = 46k+ 1. We know this by trial
and error and thus looking at all primes of the form 46k + 1 that are less than √
M23.
We now define Fermat numbers and prove some theorems about the properties of these numbers.
Definition 26. Integers of the formFn = 22
n
+ 1are called Fermat numbers. Fermat conjectured that these integers are primes but it turned out that this is not true. Notice thatF0 = 3,F1 = 5,F2 = 17,F3 = 257andF4 = 65,537while
F5is composite. It turned out theF5is divisible by641. We now present a couple
Theorem 52. For all positive integersn, we have
F0F1F2...Fn−1 =Fn−2
Proof. We will prove this theorem by induction. Forn = 1, the above identity is true. Suppose now that
F0F1F2...Fn−1 =Fn−2
holds. We claim that
F0F1F2...Fn=Fn+1−2.
Notice that
F0F1F2...Fn= (Fn−2)Fn = (22
n
−1)(22n + 1) = 22n+1−1 =Fn+1−2.
Using Theorem 53, we prove that Fermat numbers are relatively prime. Theorem 53. Lets6=tbe nonnegative integers. Then(Fs, Ft) = 1.
Proof. Assume without loss of generality thats < t. Thus by Theorem 52, we have
F0F1F2...Fs...Ft−1 =Ft−2
Assume now that there is a common divisord of Fs andFt. thus we see thatd
divides
Ft−F0F1F2...Fs...Ft−1 = 2.
Thusd = 1ord = 2. But sinceFtis odd for allt. We haved = 1. ThusFs and
Ftare relatively prime.
Exercises
1. Find the six smallest even perfect numbers. 2. Find the eighth perfect number.
4.4. PERFECT, MERSENNE, AND FERMAT NUMBERS 87 3. Find a factor of21001 −1.
4. We saynis abundant ifσ(n)>2n. Prove that ifn = 2m−1(2m−1)where
mis a positive integer such that2m−1is composite, thennis abundant.
5. Show that there are infinitely many even abundant numbers. 6. Show that there are infinitely many odd abundant numbers. 7. Determine whetherM11is prime.
8. Determine whetherM29is prime.
Chapter 5
Primitive Roots and Quadratic
Residues
In this chapter, we discuss the multiplicative structure of the integers modulo n. We introduce the concept of the order of integer modulo nand then we study its properties. We then define primitive roots modulon and show how to determine whether an integer is primitive modulonor not. We later find all positive integers having primitive roots and prove related results.
We define the concept of a quadratic residue and establish its basic properties. We then introduce Legendre symbol and also develop its basic properties. We also introduce the law of quadratic reciprocity. Afterwards, we generalize the notion of Legendre symbol to the Jacobi symbol and discuss the law of reciprocity related to Jacobi symbol.
5.1
The order of Integers and Primitive Roots
In this section, we study the order of an integer modulon, wherenis positive. We also define primitive roots and related results. Euler’s theorem in Chapter 4 states that if a positive integera is relatively prime ton, thenaφ(n) ≡ 1(mod n). Thus
by the well ordering principle, there is a least positive integerxthat satisfies this congruenceax ≡1(mod n).
Definition 1. Let (a, b) = 1. The smallest positive integer x such that ax ≡
1(mod b)is called the order ofamodulob. We denote the order ofamodulobby
ordba.
Example 46. ord72 = 3 since23 ≡ 1(mod7)while 21 ≡ 2(mod 7)and 22 ≡
4(mod7).
To find all integersxsuch thatax ≡1(mod b), we need the following theorem.
Theorem 54. If(a, b) = 1withb > 0, then the positive integer xis a solution of the congruenceax ≡1(mod b)if and only ifordba|x.
Proof. Havingordba |x, then we have thatx=k.ordbafor some positive integer
k. Thus
ax =akordba= (aordba)k≡1(mod b).
Now ifax ≡1(mod b), we use the division algorithm to write
x=qordba+r, 0≤r < ordba.
Thus we see that
ax ≡aqordba+r ≡(aordba)qar ≡ar(mod b).
Now sinceax ≡ 1(mod b),we havear ≡ 1(mod b). Sinceord
ba, we get r = 0.
Thusx=q.ordbaand henceordba|x.
Example 47. Since ord72 = 3, then 215 ≡ 1(mod 7)while 10 is not a solution
for2x ≡1(mod7).
Theorem 55. If(a, b) = 1withb >0, then
5.1. THE ORDER OF INTEGERS AND PRIMITIVE ROOTS 91 whereiandj are nonnegative integers, if and only if
i≡j(mod ordba)
Proof. Suppose that
i≡j(mod ordba) and 0≤j ≤i.
Then we havei−j =k.ordba, wherekis a positive integer. Hence
ai =aj+k.ordba =aj(aordba)k ≡aj(mod b).
Assume now thatai ≡aj(mod b)withi≥j. Thus we have
ai ≡ajai−j ≡aj(mod b)
Since(a, b) = 1, we have(aj, b) = 1and thus by Theorem 22, we get
ai−j ≡1(mod b).
By theorem 54, we get thatordba|(i−j)and hencei≡j(mod b).
We introduce now primitive roots and discuss their properties. We are inter- ested in integers whose order modulo another integer isφ(b). In one of the exer- cises, one is asked to prove that ifaandb are relatively prime thenordba | φ(b).
Definition 2. If(r, m) = 1withm > 0and if ordmr = φ(m)thenris called a
primitive root modulom.
Example 48. Notice thatφ(7) = 6hence2is not a primitive root modulo7. While
ord73 = 6and thus3is a primitive root modulo7.
Theorem 56. If(r, m) = 1withm >0and ifris a primitive root modulon, then the integers{r1, r2, ...rφ(m)}form a reduced residue set modulom.
Proof. To prove that the set{r1, r2, ...rφ(m)} form a reduced residue set modulo
m we need to show that every two of them are relatively prime and that no two of them are congruent modulom. Since(r, m) = 1, it follows that(rn, m) = 1
for all positive integersn. Hence all the powers ofrare relatively prime tom. To show that no two powers in the above set are equivalent modulom, assume that
ri ≡rj(mod m).
By Theorem 55, we see that
i≡j(mod ordmφ(m)).
Notice that1≤i, j ≤φ(m)and hencei=j.
Theorem 57. Ifordma=tand ifuis a positive integer, then
ordm(au) =t/(t, u).
Proof. Let
v =ordm(au), w = (t, u), t=t1wand u=u1w.
Notice that(t1, u1) = 1.
Becauset1 = t/(t, u), we want to show thatordm(au) = t1. To do this, we
will show that(au)t1 ≡ 1(mod m) and that if (au)v ≡ 1(mod m), thent
1 | v.
First note that
(au)t1 = (au1w)(t/w) = (at)u1 ≡1(mod m). Hence by Theorem 54, we havev |t1. Now on the other hand, since
(au)v =auv ≡1(mod m),
we know thatt |uv. Hencet1w|u1wvand hencet1 | u1v. Because(t1, u1) = 1,
we see that t1 | v. Since v | t1 and t1 | v, we conclude thatv = t1 = t/w =
5.1. THE ORDER OF INTEGERS AND PRIMITIVE ROOTS 93 Example 49. We see thatord734 = 6/(6,4)sinceord73 = 6.
Corollary 2. Let rbe a primitive root modulo m, wherem is a positive integer,
m >1. Thenru is a primitive root modulomif and only if(u, φ(m)) = 1. Proof. By Theorem 57, we see that
ordmru =ordmr/(u, ordmr) =φ(m)/(u, φ(m)).
Thusordmru =φ(m)andru is a primitive root if and only if(u, φ(m)) = 1.
The above corollary leads to the following theorem
Theorem 58. If the positive integermhas a primitive root, then it has a total of
φ(φ(m))incongruent primitive roots.
Proof. Letrbe a primitive root modulom. By Theorem 56, we see that{r1, r2, ..., rφ(m)} form a reduced residue system modulon. By Corollary 1, it is known thatru is
a primitive root modulo m if and only if(u, φ(m)) = 1. Thus we have exactly
φ(φ(m)) such integers u that are relatively prime to φ(m) and hence there are exactlyφ(φ(m))primitive roots modulom.
Exercises
1. Determineord1310.
2. Determineord113.
3. Show that 5 is a primitive root of 6.
4. Show that if¯ais an inverse ofamodulon, thenordna =ordn¯a.
5. Show that ifnis a positive integer, andaandbare integers relatively prime tonsuch that(ordna, ordnb) = 1, thenordn(ab) = ordna.ordnb.
6. Show that if a is an integer relatively prime to the positive integer m and
ordma=st, thenordmat =s.