Sines and Cosines of Angles in Arithmetic Progression
... Proof of the Theorem. We wish to reiterate that this proof is not original with the author. In [1], Greitzer gives a proof of the first formula and leaves the second as an exercise for the reader. Therefore, in this note we will prove the second formula and refer the reader to [1] for the proof of t ...
... Proof of the Theorem. We wish to reiterate that this proof is not original with the author. In [1], Greitzer gives a proof of the first formula and leaves the second as an exercise for the reader. Therefore, in this note we will prove the second formula and refer the reader to [1] for the proof of t ...
Prime Numbers - Math Talent, Math Talent Quest Home
... Number Theory-deals with using the traits of numbers to solve problems. Common topics related to in number theory include primes, divisibility, mods, and powers. Primes-Numbers with only divisors of 1 and the number itself. All numbers can be expressed as a number of primes multiplied together. Bein ...
... Number Theory-deals with using the traits of numbers to solve problems. Common topics related to in number theory include primes, divisibility, mods, and powers. Primes-Numbers with only divisors of 1 and the number itself. All numbers can be expressed as a number of primes multiplied together. Bein ...
Marshall Hall`s Conjecture and Gaps Between Integer Points on
... are known to date (see [HTTP1] for a listing of some of these numbers). The first three such numbers are 2, 5234, and 8158. All members d of the Danilov-Elkies infinite sequence ([Danl], [HTTP6],[Elks1]) have rd slightly greater than 1, and there is no other such infinite sequence known. The highest ...
... are known to date (see [HTTP1] for a listing of some of these numbers). The first three such numbers are 2, 5234, and 8158. All members d of the Danilov-Elkies infinite sequence ([Danl], [HTTP6],[Elks1]) have rd slightly greater than 1, and there is no other such infinite sequence known. The highest ...
Congruent Numbers and Heegner Points
... L(C, s) = c(s − 1)r + higher order terms with c 0 and r = rank C(Q). In particular L(C, 1) = 0 if and only if C(Q) is infinite. ...
... L(C, s) = c(s − 1)r + higher order terms with c 0 and r = rank C(Q). In particular L(C, 1) = 0 if and only if C(Q) is infinite. ...
Introduction to Discrete Mathematics
... Proof reduced the infinitude of possible maps to 1,936 reducible configurations (later reduced to 1,476) which had to be checked one by one by computer. The computer program ran for hundreds of hours. The first significant computer-assisted mathematical proof. Write-up was hundreds of pages includin ...
... Proof reduced the infinitude of possible maps to 1,936 reducible configurations (later reduced to 1,476) which had to be checked one by one by computer. The computer program ran for hundreds of hours. The first significant computer-assisted mathematical proof. Write-up was hundreds of pages includin ...
Math 2800 Math Majors Seminar
... product of primes ( the Fundamental practically computable for large integers a ...
... product of primes ( the Fundamental practically computable for large integers a ...
On Generalized Fermat Numbers 32n + 1 1 Background
... Eq. (3.2) poses the biggest problem is applying Lucas’ test for the number En , since the exponent En − 1 factors into many distinct primes, increasing with n. In fact to employ Lucas’ test, as stated in Theorem 2.2, it would require factoring all the numbers from F3 (0) to F3 (n − 1). This concern ...
... Eq. (3.2) poses the biggest problem is applying Lucas’ test for the number En , since the exponent En − 1 factors into many distinct primes, increasing with n. In fact to employ Lucas’ test, as stated in Theorem 2.2, it would require factoring all the numbers from F3 (0) to F3 (n − 1). This concern ...
The prime divisors of the number of points on abelian
... Note, this map distinguishes for each p ∈ S whether ` divides or not the positive integer #A(kp ). We also write E := EndK̄ (A) ⊗ Q. We repeatedly make use of the following: If A is an elliptic curve without CM then for all but finitely many ` we have G` = GL2 (F` ), see [9, Thm. 2]. If A is an elli ...
... Note, this map distinguishes for each p ∈ S whether ` divides or not the positive integer #A(kp ). We also write E := EndK̄ (A) ⊗ Q. We repeatedly make use of the following: If A is an elliptic curve without CM then for all but finitely many ` we have G` = GL2 (F` ), see [9, Thm. 2]. If A is an elli ...
Square Roots practice and Pythagorean Theorem
... Exploring Square Roots and the Pythagorean Theorem ...
... Exploring Square Roots and the Pythagorean Theorem ...
Primality Testing
... – If the testing number satisfies either cases, it will be said as “inconclusive”. That means it could be a prime number. – From Fermat’s Theorem, it concludes 341 is a prime but it is 11 * 31! – Now try to use Rabin-Miller’s Algorithm. ...
... – If the testing number satisfies either cases, it will be said as “inconclusive”. That means it could be a prime number. – From Fermat’s Theorem, it concludes 341 is a prime but it is 11 * 31! – Now try to use Rabin-Miller’s Algorithm. ...
Math 75 notes, Lecture 25 P. Pollack and C. Pomerance What about
... As an example, take n = 91. We can easily compute (by the repeated squaring algorithm) that 290 ≡ 64 (mod 91). So 91 cannot be a prime! (In fact, as you may have already noticed, 91 = 7 · 13.) What is notable about the proof via Fermat’s little theorem is that it shows 91 composite without factoring ...
... As an example, take n = 91. We can easily compute (by the repeated squaring algorithm) that 290 ≡ 64 (mod 91). So 91 cannot be a prime! (In fact, as you may have already noticed, 91 = 7 · 13.) What is notable about the proof via Fermat’s little theorem is that it shows 91 composite without factoring ...
Inclusion-Exclusion Principle and Applications
... Answer: Let S = {1, 2, 3, . . . , 2009}. A1 = {4, 8, 12, . . . , 2008}, |A1 | = b 2009 4 c = 502. A2 = {5, 10, 15, . . . , 2005}, |A2 | = b 2009 5 c = 401. A3 = {6, 12, 18, . . . , 2004}, |A3 | = b 2009 6 c = 334. The least common multiple of some numbers a1 , a2 , . . . , an is the smallest positiv ...
... Answer: Let S = {1, 2, 3, . . . , 2009}. A1 = {4, 8, 12, . . . , 2008}, |A1 | = b 2009 4 c = 502. A2 = {5, 10, 15, . . . , 2005}, |A2 | = b 2009 5 c = 401. A3 = {6, 12, 18, . . . , 2004}, |A3 | = b 2009 6 c = 334. The least common multiple of some numbers a1 , a2 , . . . , an is the smallest positiv ...
Lecture 5: Ramsey Theory 1 Ramsey`s theorem for graphs
... clique) of a certain size such that all its edges have the same color. We shall see that this is always true for a sufficiently large n. Note that the question about frienships corresponds to a coloring of K6 with 2 colors, “friendly” and “unfriendly”. Equivalently, we start with an arbitrary graph ...
... clique) of a certain size such that all its edges have the same color. We shall see that this is always true for a sufficiently large n. Note that the question about frienships corresponds to a coloring of K6 with 2 colors, “friendly” and “unfriendly”. Equivalently, we start with an arbitrary graph ...
Totient Theorem
... Take two numbers a and p which are relatively prime, and where p itself is prime. Consider the set of the multiples of a { a, 2a, 3a, 4a, 5a ..... (p-1)a } Consider the set of numbers { 1, 2, 3, 4, 5 ..... (p-1) } If taken to the modulus p each element of the first set will be congruent to an elemen ...
... Take two numbers a and p which are relatively prime, and where p itself is prime. Consider the set of the multiples of a { a, 2a, 3a, 4a, 5a ..... (p-1)a } Consider the set of numbers { 1, 2, 3, 4, 5 ..... (p-1) } If taken to the modulus p each element of the first set will be congruent to an elemen ...
“No professor has been asked questions by all of his students
... Difference between contradiction and contrapositive proofs Prove that if n is an integer and n3 + 5 is odd, then n is even. Contrapositive Proof: Suppose n is odd. ...
... Difference between contradiction and contrapositive proofs Prove that if n is an integer and n3 + 5 is odd, then n is even. Contrapositive Proof: Suppose n is odd. ...
Full text
... integer ap with |ap | < 2 p. Gordon [3], used the simple formula for #E(Fp ) in the case (−d|p) = −1 to define the following test of compositeness: Let Q be a point in E(Q) of infinite order. Let N > 163 be a number coprime to 6 to be tested. We compute (−d|N ). If it is 1 we do not test and if it i ...
... integer ap with |ap | < 2 p. Gordon [3], used the simple formula for #E(Fp ) in the case (−d|p) = −1 to define the following test of compositeness: Let Q be a point in E(Q) of infinite order. Let N > 163 be a number coprime to 6 to be tested. We compute (−d|N ). If it is 1 we do not test and if it i ...
Modular Arithmetic - svmoore
... • First published proof by Leonhard Euler in 1736 • Highly useful for simplifying the computation of exponents in modular arithmetic • Corollary by Euler serves as the basis for RSA encryption • Theorem: If p is a prime number and p does not divide a, then ap-1 = 1 (mod p) ...
... • First published proof by Leonhard Euler in 1736 • Highly useful for simplifying the computation of exponents in modular arithmetic • Corollary by Euler serves as the basis for RSA encryption • Theorem: If p is a prime number and p does not divide a, then ap-1 = 1 (mod p) ...