Number Theory Proof Portfolio
Jordan Rock
May 12, 2015
This portfolio is a collection of Number Theory proofs and problems done by Jordan
Rock in the Spring of 2014. The problems are organized first by general problems, individually assigned problems, and then problems I choose myself. Each section is also in
numerical order by the text that they come from.
The first four problems are the generally assigned problems to the class. They are in
numerical order.
1
General
Exercise 1 (5.4). A number L is called a common multiple of m and n if both m and
n divide L. The smallest such L is called the least common multiple of m and n and is
denoted by LCM(m, n). For example, LCM(3,7)=21 and LCM(12,66)=132.
(c) Give an argument proving that the relationship you found is correct for all m and n.
Proof. First, let’s start with the gcd. Let the gcd(a, b) = k for a, b, k ∈ N. By definition,
k|a and k|b therefore k|ab. Let’s assume then that ab
= l for l ∈ N. This means that l is
k
a multiple of a and l is a multiple of b. Therefore l is a common multiple of a and b. Now
suppose a and b have another common multiple, let’s call this m for m ∈ N. This means
that m = ab
where j is another divisor of a and b and j ∈ N. Notice that ab
= l where k
j
k
is the greatest common divisor, therefore k>l. This implies that m>l, making l the least
ab
common multiple. Thus lcm(a, b) = gcd(a,b)
.
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Jordan Rock
For the following exercise, we will be looking at modulo arithmetic, and what it means
for a number to be congruent to a statement. We will be looking at how to solve these
congruences for a variable.
Exercise 2 (8.3). Find all congruent solutions to each of the following congruences.
(a) 7x ≡ 3 mod 15
Notice that the inverse of 7 in mod 15 is 13. Therefore If we multiply the entire congruence,
we will see that
13(7x ≡ 3 mod 15)
91x ≡ 39 mod 15
x ≡ 39 mod 15.
Since the value of x for this congruence is 39, we can reduce this for modulo 15, therefore
39 ≡ 9 mod 15.
(c) x2 ≡ 1 mod 8
The values 1,3,5,and 7 are all solutions for this congruence because when these values are
squared and divided by 8, they will have a remainder of 1.
(e) x2 ≡ 3 mod 7
There are no solutions for this x because there is no integer that when squared has a remainder of 3 when divided by 7.
For the following exercise, we will be introduced to the phi function. We will prove how
to solve the phi function, and then do an example.
Exercise 3 (11.3). Suppose that p1 , p2 , . . . , pr are the distinct primes that divide m. Show
that the following formula for φ(m) is correct.
φ(m) = m(1 −
1
)(1
p1
Use this formula to compute φ(1000000).
2
−
1
) . . . (1
p2
−
1
).
pr
Jordan Rock
Proof. Base cases: Let m = pk11 . Then φ(m) = pk11 −pk11 −1 = pk11 (1− p11 ). Now let m = p1k1 pk22 ,
then φ(m) = (pk11 − pk11 −1 )(pk22 − p2k2 −1 ) = pk11 pk22 (1 − p11 )(1 − p12 ) = m(1 − p11 )(1 − p12 ).
Induction Hypothesis: If m1 = pk11 pk22 . . . pkr r , then φ(m1 ) = m1 (1 − p11 )(1 − p12 ) . . . (1 − p1r ).
k
k
r+1
r+1
). We
. Therefore, φ(m2 ) = φ(pk11 )φ(pk22 ) . . . φ(pkr r )φ(pr+1
Now let m2 = pk11 pk22 . . . pkr r pr+1
k2 −1
k2
k1 −1
k1
can then solve the phi functions, therefore φ(m2 ) = (p1 − p1 )(p2 − p2 ) . . . (pkr r −
kr+1
kr+1 −1
pkr r −1 )(pr+1
− pr+1
). Here we will factor out all of the pi ’s up until pr such that φ(m2 ) =
kr+1
kr+1 −1
kr+1
k1 k2
1
kr
(p1 p2 . . . pr )(1 − p1 )(1 − p12 ) . . . (1 − p1r )(pr+1
− pr+1
). Finally, we can factor out a pr+1
k
r+1
)(1 −
such that φ(m2 ) = (pk11 pk22 . . . pkr r pr+1
k
r+1
, so by substitution,
m2 = pk11 pk22 . . . pkr r pr+1
Thus we can conclude that φ(m) = m(1 −
1
)(1
p1
1
1
)(1 − pr+1
). Notice that
pr
1
φ(m2 ) = (m2 )(1− p11 )(1− p12 ) . . . (1− p1r )(1− pr+1
).
1
1
1
)(1 − p2 ) . . . (1 − pr ).
p1
−
1
) . . . (1
p2
−
To find φ(1000000), we need to know the prime factors of 1000000, which are 5 and 2.
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Therefore, φ(1000000) = (1000000)(1− 15 )(1− 21 ) = 1000000( 45 )( 12 ) = 1000000( 10
) = 400000.
In this next exercise, we are still working with modular arithmetic, and solving congruences.
Exercise 4 (21.1). Determine whether each of the following congruences has a solution
(all moduli are prime).
(a) x2 ≡ −1 mod 5987
Using Euler’s Criterion, we see that, −1(5987−1)/2 ≡ −12993 = −1. Therefore, the legendre
−1
symbol ( 5987
) = −1. We know that -1 means this is a non residue mod 5987, so there are
no solutions.
(b) x2 ≡ 6780 mod 6781
First, we notice that 6780 ≡ −1 mod 6781 because if we were to add 1 to both sides of
the congruence, we would see that 6781 ≡ 0 mod 6781. Here we can use Euler’s Criterion
to see that −1(6781−1)/2 ≡ −13390 = 1. Seeing that this equals 1, we know that 6780 is a
quadratic residue in mod 6781, and therefore there is a solution.
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Jordan Rock
The following section contains three problems that were given to be individually. They
are also in numerical order.
2
Individual Assignments
This first exercise works with M-world, which is a different number system than the one
we are used. We will explore what it means to be prime in this different world, and how
to find primes.
Exercise 5 (7.6). Welcome to M-world, where the only numbers that exist are positive
integers that leave a remainder of 1 when divided by 4. In other words, the only Mnumbers are {1,5,9,13,17,21 . . . }. In the M-world, we cannot add numbers, but we can
multiply them, since if a and b both leave a remainder of 1 when divided by 4, so do their
product. We day that m M-divides n if n = mk for some M-number k. And we say that n
is an M-prime if its only M-divisors are 1 and itself.
(a) Find the first six M-primes.
The first six M-primes are 5,9,13,17,21, and 29.
(b) Find a M-number n that has two different factorizations as a product of M-primes.
The M-number 405 can be factored in these two ways, 405 = 5 × 81 = 9 × 45.
Similar to the first section, in this exercise we will be working with more modular arithmetic, but this time we are finding how many (if any) solutions exist for each congruence.
Exercise 6 (8.6). Determine the number of incongruent solutions for each of the following
congruences.
(a) 72x ≡ 47 mod 200
First we are going to find the gcd(72, 200) and if that number divides 47, then this congruence has that many solutions. If it does not divide 47, then there are no solutions. So
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Jordan Rock
first, we use the Euclidean Algorithm to find the greatest common divisor like so,
200 = 72 × 2 + 54
72 = 54 × 1 + 18
54 = 18 × 3 + 0.
Notice that since 18 is the last remainder before 0, that it is the greatest common divisor
of 200 and 72. Unfortunately, 18 - 47, so this congruence has no solutions.
(b) 4183x ≡ 5781 mod 15087
First, we need the gcd(15087, 4183) and we will find this using the Euclidean Algorithm.
We can see that
15087 = 8143 × 3 + 2538
4183 = 2538 × 1 + 1695
2538 = 1695 × 1 + 893
1695 = 893 × 1 + 752
893 = 752 × 1 + 191
752 = 191 × 5 + 47
191 = 47 × 3 + 0.
Notice that 47 is the last nonzero remainder, and therefore gcd(15087, 4183) = 47. Since
47 | 5781, we know that this congruence has 47 solutions.
(c) 1537x ≡ 2863 mod 6731
First, we need to find the gcd(6731, 1537) using the Euclidean Algorithm. We can see that
6731 = 1537 × 4 + 583
1537 = 583 × 2 + 371
583 = 371 × 1 + 212
371 = 212 × 1 + 159
212 = 159 × 1 + 53
159 = 53 × 3 + 0.
Notice that the gcd(6731, 1537) = 53. Since 53 - 2863, we know that this congruence has
no solutions.
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Jordan Rock
When working with congruences, to solve for quadratics, we cannot simply take the
square root of both sides and call it a day. Here we will be working with roots of a
congruence.
Exercise 7 (17.3). In this chapter we described how to compute a k t h root of b mod m,
but you may well have asked yourself if b can have more than one k t h root. It Can! If a is
a square root of b mod m, then −a is also a square root of b mod m.
(a) Let b, k and m be integers that satisfy gcd(b, m) = 1 and gcd(k, φ(m)) = 1. Show that
b has exactly one k t h root mod m.
Proof. Let b, k, m ∈ Z such that gcd(b, m) = 1 and gcd(k, φ(m)) = 1. By way of contradiction, suppose x, y ∈ Z are unique solutions for the congruence xk ≡ b mod m
and y k ≡ b mod m. We know that gcd(k, φ(m)) = 1 so by Bezout’s lemma, there exist u, v ∈ Z such that ku + φ(m)v = 1. By Euler’s Formula, we know that for integer a, aφ(m) ≡ 1 mod m. We can rewrite the equation as ku = 1 − φ(m)v. thus,
xku ≡ xφ(m)v x mod m ≡ x mod m. Therefore, xku ≡ bu mod m. By Euler’s formula,
we see that x−φ(m)v+1 ≡ bu mod m, so x ≡ bu mod m.
Without loss of generality, y ≡ bu mod m. Since gcd(b, m) = 1 and by the linear
congruence theorem, there exists only one solution for x ≡ bu mod m and y ≡ bu mod m.
Thus we have our contradiction and x ≡ y mod m.
This last section contains three problems that I choose would be beneficial to this portfolio. I choose them because I believed they were a good representation of my learning for
this course, and I felt that I had a good understanding of each of these exercises.
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Free Choice
The first exercise uses the greatest common divisor and Bezout’s lemma to show that there
exists solutions x, y ∈ Z for the equation ax + by = c when a and b are relatively prime to
one another.
Exercise 8 (6.5). Suppose that gcd(a, b) = 1. Prove that for every integer c, the equation
ax + by = c has a solution in integers x and y. [Hint. Find a solution to au + bv = 1
and multiply by c.] Find a solution to 37x + 47y = 103. Try to make x and y as small as
possible.
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Jordan Rock
Proof. By the linear equation theorem, know there exist u, v ∈ Z such that au + bv = 1 for
gcd(a, b) = 1. If we multiply both sides of this equation by a number c, we see that
c(au + bv =1)
auc + bvc =c
We can now say that x = uc and y = vc such that ax + by = c.
For example, let’s consider 37x + 47y = 103. First we will find the gcd(37, 47) with the
euclidean algorithm. Notice that
47 = 1 × 37 + 10
37 = 3 × 10 + 7
10 = 1 × 7 + 3
7=2×3+1
3=3×1+0
Notice that the gcd(37, 47) = 1. If we take each step of previous process and solve for the
remainder, we can determine values for x and y in the equation 37x + 47y = 1. Let a = 47
and b = 37 such that
10 = a − b
7 = b − 3(a − b)
7 = 4b − 3a
3 = (a − b) − (4b − 3a)
3 = 4a − 5b
1 = (−3a + 4b) − 2(4a − 5b)
1 = −11a + 14b.
Notice here that the coefficients of a and b will be v and u respectively. Therefore
vc = (−11)(103) = −1, 133 and uc = (14)(103) = 1, 442 and hence for this solution,
x = 1442 and y = −1133.
The second exercise is a lengthy proof of the Chinese Remainder Theorem for 3 congruences.
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Jordan Rock
Exercise 9 (11.9). In this exercise you will prove a version of the Chinese Remainder Theorem for 3 congruences. Let m1 , m2 , m3 be positive integers such that each pair is relatively
prime;
gcd(m1 , m2 ) = 1
gcd(m2 , m3 ) = 1
gcd(m1 , m3 ) = 1.
Let a1 , a2 , a3 be any 3 integers. Show that there is exactly one integer k in the interval
0 ≤ x < m1 , m2 , m3 that solves these:
x ≡ a1 mod m1
x ≡ a2 mod m2
x ≡ a3 mod m3 .
Can you figure out how to generalize this problem to deal with lots of congruences x ≡
a1 mod m1 , x ≡ a2 mod m2 . . . x ≡ ar mod mr ? In particular, what conditions do the
moduli m1 , m2 , . . . , mr need to satisfy?
Proof. Let x ≡ a1 mod m1 , x ≡ a2 mod m2 , and x ≡ a3 mod m3 for m1 , m2 , m3 ∈ +Z and
for a1 , a2 , a3 ∈ Z. Now let M1 = m2 m3 , M2 = m1 m3 , and M3 = m1 m2 . It is given that
m1 , m2 , m3 are relatively prime to each other, therefore gcd(M1 , m1 ) = 1, gcd(M2 , m2 ) = 1,
and gcd(M3 , m3 ) = 1. By Bezout’s Lemma, we can say that there exist
m1 x1 + M1 y1 = 1, m2 x2 + M2 y2 = 1, and m3 x3 + M3 y3 = 1
for y1 , y2 , y3 ∈ Z.
If we convert each of these equations back into modulo m form, we see that
M1 y1 ≡ 1 mod m1 , M2 y2 ≡ 1 mod m2 , and M3 y3 ≡ 1 mod m3 .
By the linear congruence theorem, there is exactly one solution for y1 , and for y2 , and for
y3 such that 0 ≤ y1 < m1 , 0 ≤ y2 < m2 , and 0 ≤ y3 < m3 . If we multiply both sides of
each congruence by their respective a1 , a2 , a3 , we see that
a1 M1 y1 ≡ 1 mod m1 , a2 M2 y2 ≡ 1 mod m2 , and a3 M3 y3 ≡ 1 mod m3 .
If we add a1 M1 y1 + a2 M2 y2 + a3 M3 y3 we see that
a2 M2 y2 + a3 M3 y3 ≡ 0 mod m1 , a1 M1 y1 + a3 M3 y3 ≡ 0 mod m2 , and a1 M1 y1 + a2 M2 y2 ≡
0 mod m3 .
Notice that this is because m1 |M2 M3 , m2 |M1 M3 , and m3 |M1 M3 . Therefore x = a1 M1 y1 +
a2 M2 y2 + a3 M3 y3 .
Based off of this proof, the generalized form for any number of congruences is as follows
x ≡ a1 mod m1 , x ≡ a2 mod m2 , . . . , x ≡ ar mod mr
x = a1 M1 y1 + a2 M2 y2 + · · · + ar Mr yr .
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Jordan Rock
Finally, the last exercise of this portfolio is another proof. This is on the last two cases
of Quadratic Reciprocity. I think it is very important to know all of the cases, so I knew I
wanted to include this proof in my portfolio.
Exercise 10 (21.4). Finish the proof of Quadratic Reciprocity (part II) for the other two
cases: primes congruent to 1 mod 8 and primes congruent to 5 mod 8.
Proof. First we will start with P ≡ 1 mod 8. This means p = 8k + 1 for some k ∈ Z. The
even integers between 1 and p − 1 are 2, 4, . . . , p − 1. The ”lower half” of these evens are
2, 4, . . . , 4k. The ”upper half” of these evens are 4k + 2, 4k + 4, . . . , 8k. To find the sun
in the upper half, we will take half of the highest value, 8k, and subtract the lower half
highest value, 4k. Therefore, 12 ((8k) − (4k)) = 2k. This is the value of the negative signs,
so if we use the formula 2(p−1)/2 ≡ −12k mod p we see that 2(p−1)/2 ≡ 1 mod p. Therefore
( p2 ) = 1. Thus 2 is a QR if any prime is congruent to 1 mod 8.
Now let’s work with p ≡ 5 mod 8. This means p = 8k + 5 for some k ∈ Z. All of the
evens between 1 and p − 1 are 2, 4, . . . , 8k + 4. The lower half is 2, 4, . . . , 4k + 2 and the
upper half is 4k + 4, 4k + 6, . . . , 8k + 4. If we take half of the difference of the two highest
values, 21 ((8k + 4) − (4k + 2)), we see that the number of evens in the upper region is 2k + 1.
This is also the number negative signs. Therefore 2(p−1)/2 ≡ −12k+1 mod p ≡ −1 mod p.
By Euler’s Criterion, 2 is a NR for all primes congruent to 5 mod 8.
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