Partitions into three triangular numbers
... numbers rather than the representations of n into three triangular numbers. For instance, the three representations 28 + 1 + 1, 1 + 28 + 1, and 1 + 1 + 28 stem from one partition, 28 + 1 + 1. Thus the integer 30 can be partitioned into three triangular numbers in only four ways. In this note, we wil ...
... numbers rather than the representations of n into three triangular numbers. For instance, the three representations 28 + 1 + 1, 1 + 28 + 1, and 1 + 1 + 28 stem from one partition, 28 + 1 + 1. Thus the integer 30 can be partitioned into three triangular numbers in only four ways. In this note, we wil ...
10 - Harish-Chandra Research Institute
... modulo p) and the complement set contains all the non-residues which are not primitive roots modulo p. In 1927, E. Artin [1] conjectured the following; Artin’s primitive root conjecture. Let g 6= ±1 be a square-free integer. Then there are infinitely many primes p such that g is a primitive root mod ...
... modulo p) and the complement set contains all the non-residues which are not primitive roots modulo p. In 1927, E. Artin [1] conjectured the following; Artin’s primitive root conjecture. Let g 6= ±1 be a square-free integer. Then there are infinitely many primes p such that g is a primitive root mod ...
19 4|( + 1)
... 2008). All primes, prime powers and all positive integers , , ( ) = 1 are solitary (Dris, 2008). There are also numbers such as = 18, 45, 48, and 52 which are solitary but for which , ( ) ≠ 1 (Dris, 2008). However, there exist numbers such as 10, 14, 22, 26, 34 and 38, whose categorization as eithe ...
... 2008). All primes, prime powers and all positive integers , , ( ) = 1 are solitary (Dris, 2008). There are also numbers such as = 18, 45, 48, and 52 which are solitary but for which , ( ) ≠ 1 (Dris, 2008). However, there exist numbers such as 10, 14, 22, 26, 34 and 38, whose categorization as eithe ...
Mathematical Reasoning (Part III)
... then the proof begins by assuming the existence of a counterexample of this statement. Therefore, the proof might begin with: – Assume, to the contrary, that there exists some element x ∈ D for which P (x) is true and Q(x) is false. or – By contradiction, assume, that there exists an element x ∈ D s ...
... then the proof begins by assuming the existence of a counterexample of this statement. Therefore, the proof might begin with: – Assume, to the contrary, that there exists some element x ∈ D for which P (x) is true and Q(x) is false. or – By contradiction, assume, that there exists an element x ∈ D s ...
ON DICKSON`S THEOREM CONCERNING ODD PERFECT
... If N is a natural number, we write σ(N ) := d|N d for the sum of the divisors of N . We call N perfect if σ(N ) = 2N , i.e., if N is equal to the sum of its proper divisors. The even perfect numbers were completely classified by Euclid and Euler, but the odd perfect numbers remain utterly mysterious ...
... If N is a natural number, we write σ(N ) := d|N d for the sum of the divisors of N . We call N perfect if σ(N ) = 2N , i.e., if N is equal to the sum of its proper divisors. The even perfect numbers were completely classified by Euclid and Euler, but the odd perfect numbers remain utterly mysterious ...
Solutions - math.miami.edu
... There are various geometric ways to do this. The easiest way is to consider point M as the origin (0, 0) of a Cartesian plane. Recall that the equation of a circle with radius ρ and center (α, β) is (x − α)2 + (y − β)2 = ρ2 . Our circle has center (−b, a/2) and radius a/2, so it has equation (x + b) ...
... There are various geometric ways to do this. The easiest way is to consider point M as the origin (0, 0) of a Cartesian plane. Recall that the equation of a circle with radius ρ and center (α, β) is (x − α)2 + (y − β)2 = ρ2 . Our circle has center (−b, a/2) and radius a/2, so it has equation (x + b) ...
CHECKING THE ODD GOLDBACH CONJECTURE UP TO 10 1
... values greater than 33 . This bound was then reduced to 1043000 . In this paper we investigate this conjecture numerically and prove it to be true for all integers less than 1020 . 2. Principle of the algorithm Because of the huge size of the set of odd integers considered, systematic verification f ...
... values greater than 33 . This bound was then reduced to 1043000 . In this paper we investigate this conjecture numerically and prove it to be true for all integers less than 1020 . 2. Principle of the algorithm Because of the huge size of the set of odd integers considered, systematic verification f ...
The Riddle of the Primes - Singapore Mathematical Society
... Father Marin Mersenne (1588-1648). We write 2P -1 as·M (p). The French mathematician Edouard Lucas showed in 1914 that M(127) is prime. This to my knowledge is the largest prime ever found by noncomputer methods. With computer searches, a number of Mersenne primes have been found: M(p) is prime for ...
... Father Marin Mersenne (1588-1648). We write 2P -1 as·M (p). The French mathematician Edouard Lucas showed in 1914 that M(127) is prime. This to my knowledge is the largest prime ever found by noncomputer methods. With computer searches, a number of Mersenne primes have been found: M(p) is prime for ...
A CHARACTERIZATION OF ALL EQUILATERAL TRIANGLES IN Z3
... The connection with Carmichael numbers goes a little further. Carmichael numbers have at least three prime factors and numerical evidence suggests that the following conjecture is true: Conjecture: The Diophantine equation (4) has degenerate solutions if and only if d has at least three distinct pri ...
... The connection with Carmichael numbers goes a little further. Carmichael numbers have at least three prime factors and numerical evidence suggests that the following conjecture is true: Conjecture: The Diophantine equation (4) has degenerate solutions if and only if d has at least three distinct pri ...
Congruent Numbers and Heegner Points
... triangle with rational sides and area n. This was considered as a principle object of the theory of rational triangles in 10th century. The equivalence of the two forms is not difficult to prove: Suppose we are given an arithmetic progression α2 , β2 , γ2 with common difference n then we have the fo ...
... triangle with rational sides and area n. This was considered as a principle object of the theory of rational triangles in 10th century. The equivalence of the two forms is not difficult to prove: Suppose we are given an arithmetic progression α2 , β2 , γ2 with common difference n then we have the fo ...
Lecture 3: Principle of inclusion and exclusion 1 Motivation 2
... How many positive integers less than 100 is not a factor of 2,3 and 5? For solving this problem at first we have to find the number of positive integers less than 100 which are divisible by 2 or 3 or 5. Let A = The set of elements which are divisible by 2 Let B = The set of elements which are divisi ...
... How many positive integers less than 100 is not a factor of 2,3 and 5? For solving this problem at first we have to find the number of positive integers less than 100 which are divisible by 2 or 3 or 5. Let A = The set of elements which are divisible by 2 Let B = The set of elements which are divisi ...
Fermat's Last Theorem
In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two. The cases n = 1 and n = 2 were known to have infinitely many solutions. This theorem was first conjectured by Pierre de Fermat in 1637 in the margin of a copy of Arithmetica where he claimed he had a proof that was too large to fit in the margin. The first successful proof was released in 1994 by Andrew Wiles, and formally published in 1995, after 358 years of effort by mathematicians. The theretofore unsolved problem stimulated the development of algebraic number theory in the 19th century and the proof of the modularity theorem in the 20th century. It is among the most notable theorems in the history of mathematics and prior to its proof it was in the Guinness Book of World Records for ""most difficult mathematical problems"".