Math 25 Number Theory Midterm Review
Lagrange polynomial root theory
- if f(x)∈Z and degree of f(x) is d, and p is a prime not dividing the leading coefficient of f(x), then f(x)=0mod(p) has at most d solutions
Wilson's Theorem
(p-1)!=-1mod(p)
Running time of the Euclidean Algorithm
- Euclidean algorithm is used to find the GCD- it is the division and remainder approach - running time of the euclidean algorithm is at most 2log₂(a) steps if a≥b
naive approach to GCD
- If a is the smaller integer, the it will take at the very least a/2 steps to find the GCD, and this process is through guess and check for the first a/2 numbers
How to prove Bezouts identity
- You basically have to do the Euclidean algorithm backwards - make sure you can prove with an actual example and a general case with variables
complete set of residues mod(n)
- a set of n integers containing one representative from each of the n congruence classes is called the complete set of residues mod(n)
modular arithmetic tables
- addition and multiplication tables of mod classes use the addition and multiplication properties discussed and simplify to identify/represent the class by its least positive member
properties of equivalence classes
- addition: [a]₇+[b]₇= [a+b]₇ - multiplication: [a]₆*[b]₆= [ab]₆
composite numbers
- composite number, n can be written as the product of two or more numbers which are <n - prime numbers are not composite
equivalence class
- denoted as [7]₄, which in this case would mean the set of all integers, b∈Z, such that b=7mod(4). - Any member of this set can serve as the representative of the set, therefore [7]₄=[-1]₄
equivalence classes mod(n)
- either identical or disjoint, therefore if two equivalence classes mod(n) have one number in common, then they are the same class. - there are exactly n equivalence classes mod(n)
Fundamental Theorem of Arithmetic
- every positive integer is uniquely represented as a product of primes - prove by contradiction, therefore first assume there is a least integer n such that n has two prime power factorizations - p1 is a divisor of both factorizations, therefore must divide some factor of other factorization, qi. But qi is also prime, therefore p1=qi for some i. - repeat this method
factoring approach to the GCD
- factor a and b down to their prime number factor tree - circle all prime numbers they have in common - multiply these prime numbers together to get the GCD
If given two modular equations, x≡amod(m) and x≡bmoc(n), then
- for there to be a solution, it is necessary that a≡bmod(G) where G=gcd(m,n)
Rational Root Criterion
- given a polynomial f(x), a/b is a rational root if f(a/b)=0 and a divides the constant term, and b divides the first term of the polynomial
rational number
- has a numerator and a denominator which are integers, a/b where a,b∈Z and b≠0
if gcd(a,n)=1, then the linear congruence ax≡bmod(n) has a solution ...
- has a solution because we know 1|b, and therefore general case holds - however, there will only be d congruence classes, and in this case, d=1, so solution will be a single congruence class mod(n)
If a,b,c∈Z with a≠0 and b&c coprime, then if a|bc
- if a|bc, then we know a|c - this is because if a and b are coprime, they have no factors in common, therefore a must divide c - we can prove this using Bezout's Identity and multiplying both sides by c and then using what we know to manipulate equation
finding prime factorization of n: difference of squares method
- if given a large number, n, find the closest perfect square to n. - see if the difference between the closest perfect square and n is also a perfect square - if it is, then factor out the equation for n in terms of the squares to find the prime factorization
Fermat's Little Theorem Form 2
- if n is a prime and a and n are coprime, then aⁿ⁻¹=1mod(n) - a and n are coprime in this example - use this theorem when asked to compute a large number mod(p) or mod(n), in which case factor the n into its prime factorization
Fermat's Little Theorem
- if n is prime, then aⁿ=amod(n) for all integers, a - therefore, this means aⁿ-a has every residue class mod(n) as a root - we prove this using induction, because it definitely works for the residue class where a=0 - so then assume this works for some a≥0. Then consider (a+1)ⁿ. Factor this out and then take the equation mod(n) to prove
Rational roots rule with coprime mod(n)
- if there is some n>0 coprime to the leading coefficient of f(x), and if f(x) has no roots mod(n), then it has no rational roots - try to look up proof of this
if d=gcd(a,n), then the linear congruence ax≡bmod(n) has a solution iff
- iff d divides b - in this case, there are d congruence classes mod(n)
Fundamental problem of arithmetic
- look up to make sure - this is the problem of actually finding the prime factorization for a number
can two different products of primes represent the same integer?
- no, each integer has a unique prime power factorization
Are there infinitely many primes with a remainder of 3 when divided by 4?
- once again prove by contradiction, therefore assume there is a largest prime p with a remainder of 3 when divided by 4: p=4q+3 - let n be the product of all primes <p. Consider 4n-1 where n≥2 since 2 is the smallest prime - all the primes that divide n are odd since we can see it is not divisible by 2 - therefore, all the primes are in the 1 or 3 category when divided by 4 since they are odd - the product of numbers in the "1" category also have a remainder of 1 when divided by 4- show with variables - therefore, we must have one number in the "3" category that divides 4n-1 - however, this prime in the "3" category cannot be on our list since n is divisible by all primes in the 3 category and (4n-1) is relatively prime to n
Corollary of Bezouts identity
- the least positive integer of the form au+bv where u,v∈Z is the GCD(a,b)
prime numbers
- the only divisors of prime numbers are the number itself and 1 - the only even prime number is 2 - 2 is also the smallest prime number
if a prime number divides the product of two numbers
- then it must divide one of the two numbers
If GCD(a,b)=d, then we know a/d and b/d
- then we know a/d and b/d are relatively prime with each other/ coprime - to prove this, use the multiplication theorem for the GCD(ma,mb) and have m serve as d
If a and b are relatively prime and a|c and b|c, then
- then we know ab|c because a and b have no factors in common - to prove, manipulate division algorithm equations
Twin prime conjecture
- there are infinitely many pairs of primes different by 2 - this is not proven. A conjecture is a theory where no one has found any counterexamples but it has also never been solved
If two numbers a and n are coprime, then we know there exists
- there exists an integer u such that au=1mod(n) - if gcd(a,n)=1, then there exists integers u and v such that au+nv=1. Modify this equation by n and we have proved theorem above
If two integers a and b are relatively prime, then we know
- they have gcd(a,b)=1 - we can write any integer as au+bv where a and b are relatively prime and u,v∈Z
if ac≡bcmod(n), then a≡bmod(n) iff
- this only holds true if a,b,c are integers and c≠0mod(n) - therefore, in order for this to be true, c and n must be coprime - this is only true when gcd(c,n)=1
find u with au≡1mod(n)
- to solve, we must solve the equation au+nv=1. We will be given a and n, therefore must find u and v to satisfy equation - once you find u, solve the congruence and check to make sure mod(n) the equation holds
Find x with ax≡bmod(n)
- to solve, we want to try and manipulate modulus equation to be 1mod(n) - if you multiply, multiply both sides of the modulus - if you can get either side to be 1mod(n) or 1xmod(n), then you will be able to solve
Every positive integer is the product of primes: true/false and prove
- true, 1 is the empty power: x⁰ and every prime is itself to the first power, p¹ - otherwise, the theorem is true if n is composite, because even if we call n the least counterexample, then since we said n is composite, n=a*b and a,b<n, therefore a and b are not counterexamples, but instead are the product of primes, and therefore so is n
When trying to prove something, different things to try include
- try proving by induction if you can prove for n=1, n=2 - try proving by contradiction if it is easier to start with "let____hypothesis_____ be true" Whenever it fails, you have proven by contradiction - try proving with the contrapositive statement if we know more about proving that. Contrapositive is true iff regular statement is true - try proving by strong induction if we want to assume it true for a whole set of integers, {1,2,...,k} and therefore, using that it is true for this set when k≥2, we show that it holds true for k+1
Relatively Prime
- two integers a and b are relatively prime or coprime if GCD(a,b)=1, therefore they have no other divisors in common
What is the remainder when 29 is divided into 34*17?
- we can phrase this question with modular arithmetic: 34*17mod(29)= 5*(-12)mod(29)=-60mod(29)= -2mod(29)=27mod(29) Therefore, the answer is 27
(x²-1)≡0mod(p)
- we know that p must divide (x+1)(x-1) - therefore, there are two solutions: x≡1mod(p) and x≡-1mod(p) - we can do this for any factoring as long as moduli is prime
Euclid's proof for infinitely many primes
- we prove by contradiction: so let there be a prime, p such that p is the largest prime - let "n" be the product of all primes which are <p - consider n+1. If you divide n+1 by any prime, you will have a remainder of one. n+1≥2 because the smallest prime is 2 - therefore, this contradicts that every number is the nonempty product of a number of primes because (n+1) is not. - therefore, we prove false by contradiction, there is no largest prime number
Testing if a number is a perfect square
- when you divide a perfect square by 3, you should get a remainder of 0 or 1 - a perfect square can only end in a 0,1,4,5,6 or 9 - digital root must be 1,4,7, or 9. We compute a digital root by adding all the digits of the number, and keep doing this until you get a single digit. - last two digits of a perfect square cannot both be odd -when you divide a perfect square by 4, you must get a remainder of 0 or 1 - for number to be perfect square, all prime factors must be of even multiplicities - total number of prime factors of a perfect square is always odd
Are there infinitely many primes?
- yes, and we can prove this by contradiction - we can let n be the least integer which is neither prime or the product of primes - if we call d the least divisor of n, then d<n because otherwise it could not divide n. we know d must either be the product of primes or a prime since n is the smallest number that is neither. However, if n= d(n/d), then we know d is product of primes or prime and so must n/d be. Therefore, so is n Proof by contradiction
are there infinitely many primes of the form 2ⁿ-1?
- yes, if 2ⁿ-1 is prime, then 2ⁿ⁻¹p is perfect - every even perfect number is of Euclid's form 2ⁿ⁻¹p - we do not know if there are any odd perfect numbers
if n ends in....
-if n ends in 1 or 9, then n² ends in 1 - if n ends in 2 or 8, then n² ends in 4 - if n ends in 3 or 7, then n² ends in 9 - if n ends in 5, then n² ends in 25 - if n ends in 4 or 6 then n² ends in 6 - if n ends in 0, then n² must have an even number of 0's
what is the difference between a rational root and integral root?
-rational root is of the form a/b, so it is a number such that f(a/b)=0 - integral root is an integer root, f(x)=0 - if we can prove that a function has no rational roots, then it also has no integer roots. If a function has an integer root, then it has a rational root
the remainder of a square when divided by 4
-the remainder of a square when divided by 4 is always 0 or 1 - to prove, we know every integer can be written as b= 2q+r where r=0 if even and r=1 if odd. Therefore, square this equation and show that there will be a remainder of 0 or 1 when divide by 4
Let n have the prime power factorization n=p₁ⁿ¹p₂ⁿ²...., then we know that if a=bmod(n)
-then a=bmod(p₁ⁿ¹) for all primes that are factors of n. -This also works the other way around, a=bmod(n) iff for all primes that are factors of n, a=bmod(p₁ⁿ¹)
0 can only be a divisor of
0 is only a divisor of itself
Using Chinese Remainder Theorem to solve polynomial congruences-
1.) Factor polynomial as well as factor the modulus into primes to work with 2.) we should be able to solve the polynomial down to the form x=amod(m) and x=bmod(n) for m,n being prime factors of the modulus 3.) Therefore we find mu+nv=1 and find values of u and v that work for this 4.) Then, we calculate Ab+Ba to find solution, where A=mu, B=nv and a=a and b=b as given
How to test if an equation has any integral roots
1.) a is an integral root if f(a)=0, and then f(a)=0mod(n) if f(a)≠0mod(n) for all n, then f(a)≠0, and therefore a is not an integral root 2.) if asked if f(x) has a root mod(n), start testing f(0), f(1),...f(n-1) and try to find a root such that f(j)=omod(n). If it has a root, then move on. 3.) Try to pick a number n such that you cannot find a root for it when you test all possible x values. If you can find this, then you have proven f(x) has no integral root
If a=a'mod(n) and b=b'mod(n), then
1.) ab=a'b'mod(n) 2.) a+b=a'+b'mod(n)
algorithm to solve linear congruences
1.) calculate gcd(a,n)=d. If d|b, then move on, there are solutions. If d does not divide b, then no solution 2.) If d divides a,b,n, then we can divide it out! We get a'x=b'mod(n') where a and n are coprime 3.) Now that a' and n' are coprime, we take gcd(a,b)=m. We then divide by m. We get a"x≡b"xmod(n') where a"=a'/m and b"=b'/m 4.) try to reduce a" to 1 using modulus rules
how to find GCD of large numbers
1.) divide larger number by smaller 2.) take gcd of the smaller number and the remainder of the above division 3.) continue this process until you get a remainder of 0. Then, the GCD will be whatever you divided by (the smaller number in the last GCD)
find the remainder when 19 is divided into 5¹⁰
1.) first set equation 5²=25mod(19) and then reduce to 6mod(19) and square both sides again 2.) 5⁴=36mod(19)=-2mod(19) square again 3.) 5⁸=4mod(19) and then multiply by 5²mod(19) 4.) answer is 5¹⁰= 24mod(19)=5mod(19)
General Chinese Remainder Theorem
1.) if a≡bmod(G) where G=gcd(m,n), then we know there is some integer C such that x≡amod(m) and x≡bmod(n) and x≡Cmod(mn/g)
Chinese remainder theorem with more than 2 moduli
1.) if m₁,m₂,m₃...... are pairwise coprime and a₁,a₂,a₃..... are ay integers, then there is some integer C such that C≡a₁mod(m₁) for all a and m values 2.) also, all solutions to these equations are given by x≡Cmod(m₁m₂m₃...)
Properties of prime numbers (3)
1.) if p is prime and a∈Z and p does not divide a, then GCD(p,a)=1 2.) if p is prime and a,b are positive integers, with p|ab, then p|a or p|b 3.) if p and q are both primes, and p|q, then p=q
Steps to show number is irrational
1.) label number as r. This way we can manipulate equation 2.) get rid of any roots 3.) once you are rid of roots, you know number is rational if only multiplication/division/addition/subtraction, etc 4.) try to make it so irrational number is equal to rational number so proof by contradiction
properties of modulos (3)
1.) reflexive: a≡amod(n) 2.) symmetric: if a≡bmod(n) then b≡amod(n) 3.) transitive: if a≡bmoc(n) and b≡cmod(n), then a≡cmod(n)
We can only use division on modulus
1.) when we can divide across both sides AS WELL AS the modulus. Therefore, when ac≡bcmod(mc), this is equivalent to a≡bmod(m) 2.) we can also divide as long as both sides are divisible even if modulus isn't. Therefore, ac≡bcmod(n) is equivalent to a≡bmod(n). However, we can only do this if c and n are coprime, therefore if gcd(n,c)=1
Theorem 1.6 of the last non-zero remainder
GCD(a,b)= GCD(b,r)= GCD(r, r₁)= GCD(r₁, r₂)=..... and GCD=d=r↓n-1
If a,b∈Z and not both 0 and if m is nonzero ∈Z,then GCD(ma,mb)=
GCD(ma,mb) = m*GCD(a,b) - we can prove this because GCD(ma,mb) is the smallest number of the set {mau+mbv where u,v∈Z} and we can therefore factor out the m
Bezouts Identity
If a,b ∈Z and a,b not both ≠0, and GCD(a,b)=d, then there exists a u,v∈Z such that d=ua+vb
GCD division approach (lemma 1.5). This is also called the Euclidean Algorithm
If a=bq+r, then GCD(a,b)= GCD(b,r). we repeat this process until the remainder divides our "b" term
b∈[a]₅ iff
[a]₅=[b]₅ because if b is in the set, b should be able to serve as the representative of the set
if a and n are coprime and m divides a and b, then ax≡bmod(n) iff
a'x≡b'mod(n) where a'=a/m and b'=b/m
if m divides a,b,n, then ax≡bmod(n) iff
a'≡b'mod(n') where a'=a/m, b'=b/m, n'=n/m
a|b and b|a iff
a= ±b
[a]=[b] iff
a=bmod(n)
If a,b∈Z and not both 0 and if d=GCD(a,b), then ax+by=
ax+by=nd such that n∈Z as long as x,y∈Z - we have to prove this in both directions, therefore name each set and show the sets are contained in each other
If b and c are coprime to a, then we know
bc is also coprime to a - we can prove this by using the identity that if two numbers are coprime, ax+by=1, and then multiply them together and manipulate to get a new equation with a(Z₁)+bc(Z₂)=1
if b₁,b₂,b₃,....bk are coprime to a, then we know
b₁b₂b₃...bk is also coprime to a - we would prove this by strong induction. Therefore, prove this is true when k=1 and 2 and then use induction using rule that if two numbers are coprime to a, then so is their product
if c|a and c|b, then we know
c|au+bv where u,v∈Z
if f(x) is a polynomial with integer coefficients, and a=bmod(n), then
f(a)=f(b)mod(n) - this is because we know multiplication and addition work over mod functions and therefore if we manipulate both sides a and b with the same polynomial, then we should get a congruent modulus
there is no non-constant polynomial f(x) with integer coefficients such that
f(x) is prime for all integers x
True/false: if a|bc, then a|b or a|c
false, think of a=6, b=3, c=4
gcd(a,b,c)=
gcd(gcd(a,b),c)
Euler's Theorem: ∑(1/p) where p is prime
∑(1/p)=∞
Fermat's theorem for primes of the form 2ⁿ+1
if 2ⁿ+1 is prime, then n= 2^k for some k, therefore n is also a power of 2 - make sure you understand how to prove this
equivalence relation
if a relationship is reflexive, symmetric and transitive, then it is called an equivalence relation
dirichlet's theory
if a,b are positive coprime integers, then there are infinitely many primes that leave a remainder of a when divided by b
division algorithm
if a,b,∈ Z, then there exists some q,r∈ Z such that a=bq+r where 0≤r<b
If n² is even/odd
if n² is even, n is even if n² is odd, n is odd
gcd(a,b)=1 iff
iff there exists x,y∈Z such that ax+by=1
If a,b ∈Z with a,b>0, then there must be
integers q,r such that a=qb+r. And q and r must be unique integers because 0<r<b
least common multiple theorem
lcm(a,b)= ab/gcd(a,b) - you can rewrite this as lcm(a,b)*gcd(a,b) = ab
Asymptotic density of a set of, S, of positive integers
lim((1/x)∑1) for n∈S and n≤x as x→∞ - so you set an x and then calculate this value as a limit
Asymptotic density of prime numbers
primes have an asymptotic density of 0
Each equivalence class mod(n) corresponds to
r=0,1,2,3,....,n-1
gcd(a,b,c,d,...n) is equal to
the least positive number of the form ax₁+bx₂+cx₃+.....+nxn which is essentially Bezout's identity expanded
least absolute residues mod(n)
the set of n integers containing representatives from each of the n congruence classes. set with the remainders -n/2<r≤n/2
least non-negative residues mod(n)
the set of n integers containing the least positive representative from each of the congruence classes. Therefore for mod(n), this is the set {0,1,2,...,n-1}
GCD is capped at
the smallest of the two integers
if a|f(x) and b|f(x) and a and b are relatively prime,
then ab|f(x) as well!
if a|b and b|c
then a|c, transitive property holds for division
if f(x)∈ Z and n is coprime to the leading coefficient of f(x), then if f(x)≡0mod(n), has no integral solution,
we know f(x) also has no rational roots
if (a-b)≠0, then we know what about f(a)-f(b)
we know, (a-b)|f(a)-f(b)