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Math 412: Number TheoryLecture 2: GCD and linear diophantine equations
Gexin [email protected]
College of William and Mary
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
GCD and Euclidean Algorithm
Let (a, b) be the greatest common divisor (gcd) of a and b.
Lemma: if a = bq + r , then (a, b) = (b, r).
Euclidean Algorithm to find (a, b)
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
GCD and Euclidean Algorithm
Let (a, b) be the greatest common divisor (gcd) of a and b.
Lemma: if a = bq + r , then (a, b) = (b, r).
Euclidean Algorithm to find (a, b)
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
GCD and Euclidean Algorithm
Let (a, b) be the greatest common divisor (gcd) of a and b.
Lemma: if a = bq + r , then (a, b) = (b, r).
Euclidean Algorithm to find (a, b)
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Ex: Find (2134, 3246).
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Theorem: there exist integers m, n so that
(a, b) = ma + bn.
Thm: (a, b), when ab 6= 0, is the least positive integer of the formma + nb. In fact, {ma + nb : m, n ∈ Z} = {kb : k ∈ Z}.
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Theorem: there exist integers m, n so that
(a, b) = ma + bn.
Thm: (a, b), when ab 6= 0, is the least positive integer of the formma + nb. In fact, {ma + nb : m, n ∈ Z} = {kb : k ∈ Z}.
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Coprime Integers
The integers a and b are coprime (relatively prime) if (a, b) = 1.
Then a and b are coprime if and only if there exist m, n ∈ Z so that
ma + nb = 1.
Ex: If (a, b) = 1, and a|n and b|n, then ab|n.
Thm: if (a, b) = d , then (a/d , b/d) = 1. From this, we can reduceevery fraction into its lowest term.
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Coprime Integers
The integers a and b are coprime (relatively prime) if (a, b) = 1.
Then a and b are coprime if and only if there exist m, n ∈ Z so that
ma + nb = 1.
Ex: If (a, b) = 1, and a|n and b|n, then ab|n.
Thm: if (a, b) = d , then (a/d , b/d) = 1. From this, we can reduceevery fraction into its lowest term.
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Coprime Integers
The integers a and b are coprime (relatively prime) if (a, b) = 1.
Then a and b are coprime if and only if there exist m, n ∈ Z so that
ma + nb = 1.
Ex: If (a, b) = 1, and a|n and b|n, then ab|n.
Thm: if (a, b) = d , then (a/d , b/d) = 1. From this, we can reduceevery fraction into its lowest term.
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Coprime Integers
The integers a and b are coprime (relatively prime) if (a, b) = 1.
Then a and b are coprime if and only if there exist m, n ∈ Z so that
ma + nb = 1.
Ex: If (a, b) = 1, and a|n and b|n, then ab|n.
Thm: if (a, b) = d , then (a/d , b/d) = 1. From this, we can reduceevery fraction into its lowest term.
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Euclid Lemma: If a|bc and (a, b) = 1, then a|c .
Ex: (Euclid Lemma, special case) if p|ab then p|a or p|b.
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Euclid Lemma: If a|bc and (a, b) = 1, then a|c .
Ex: (Euclid Lemma, special case) if p|ab then p|a or p|b.
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Linear Diophantine Equation
Consider equation ax + by = c , where a, b, c are integers and d = (a, b).
The equation has no integral solutions if d 6 |c .
If d |c, then there are infinitely many integral solutions.
Moreover, if x = x0, y = y0 is a particular solution, then all solutionsare given by
x = x0 + (b/d)n, y = y0 − (a/d)n,
where n is an integer.
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Lemma:a1x1 + a2x2 + . . .+ anxn + an+1xn+1 = c
if and only if
a1x1 + a2x2 + . . .+ an−1xn−1 + (an, an+1)y = c
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Prime numbers
Definition: a (positive) integer p is prime if p has no divisor otherthan 1 and p. Otherwise it is a composite number.
Ex: Every composite number has a prime divisor.
Ex: every composite integer n has a prime divisor at most√n.
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Prime numbers
Definition: a (positive) integer p is prime if p has no divisor otherthan 1 and p. Otherwise it is a composite number.
Ex: Every composite number has a prime divisor.
Ex: every composite integer n has a prime divisor at most√n.
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Prime numbers
Definition: a (positive) integer p is prime if p has no divisor otherthan 1 and p. Otherwise it is a composite number.
Ex: Every composite number has a prime divisor.
Ex: every composite integer n has a prime divisor at most√n.
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Thm: there are infinite many prime numbers.
Sieve of Eratosthenes: a method to find all prime less than an integer.
Prime Number Theorem: Let π(x) be the number of primes at mostx , then π(x)→ x
ln x as x →∞.
Primitive Test: a polynomial time algorithm was found in 2002 byAgrawal, Kayal, and Saxena (an India CS professors with two understudents).
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Thm: there are infinite many prime numbers.
Sieve of Eratosthenes: a method to find all prime less than an integer.
Prime Number Theorem: Let π(x) be the number of primes at mostx , then π(x)→ x
ln x as x →∞.
Primitive Test: a polynomial time algorithm was found in 2002 byAgrawal, Kayal, and Saxena (an India CS professors with two understudents).
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Thm: there are infinite many prime numbers.
Sieve of Eratosthenes: a method to find all prime less than an integer.
Prime Number Theorem: Let π(x) be the number of primes at mostx , then π(x)→ x
ln x as x →∞.
Primitive Test: a polynomial time algorithm was found in 2002 byAgrawal, Kayal, and Saxena (an India CS professors with two understudents).
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
Thm: there are infinite many prime numbers.
Sieve of Eratosthenes: a method to find all prime less than an integer.
Prime Number Theorem: Let π(x) be the number of primes at mostx , then π(x)→ x
ln x as x →∞.
Primitive Test: a polynomial time algorithm was found in 2002 byAgrawal, Kayal, and Saxena (an India CS professors with two understudents).
Gexin Yu [email protected] Math 412: Number Theory Lecture 2: GCD and linear diophantine equations
