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m is a divisor of n (also called m divides n, or n is divisible by m) if all prime factors of m have at least the same multiplicity in n. The divisors of n are all products of some or all prime factors of n (including the empty product 1 of no prime factors). The number of divisors can be computed by increasing all multiplicities by 1 and then ...
More generally, a positive integer c is the hypotenuse of a primitive Pythagorean triple if and only if each prime factor of c is congruent to 1 modulo 4; that is, each prime factor has the form 4n + 1. In this case, the number of primitive Pythagorean triples (a, b, c) with a < b is 2 k−1, where k is the number of distinct prime factors of c ...
If none of its prime factors are repeated, it is called squarefree. (All prime numbers and 1 are squarefree.) For example, 72 = 2 3 × 3 2, all the prime factors are repeated, so 72 is a powerful number. 42 = 2 × 3 × 7, none of the prime factors are repeated, so 42 is squarefree. Euler diagram of numbers under 100:
where the denominator 42 was used, because it is the least common multiple of 21 and 6. ... Factor each number and express it as a product of ... The following pairs ...
The greatest common divisor (GCD) of integers a and b, at least one of which is nonzero, is the greatest positive integer d such that d is a divisor of both a and b; that is, there are integers e and f such that a = de and b = df, and d is the largest such integer.
From this we see that r is any even integer and that s and t are factors of r 2 /2. All Pythagorean triples may be found by this method. When s and t are coprime, the triple will be primitive. A simple proof of Dickson's method has been presented by Josef Rukavicka, J. (2013). [7] Example: Choose r = 6. Then r 2 /2 = 18. The three factor-pairs ...
For example, the "primitive" friendly pair 6 and 28 gives rise to friendly pairs 6n and 28n for all n that are congruent to 1, 5, 11, 13, 17, 19, 23, 25, 29, 31, 37, or 41 modulo 42. [4] This shows that the natural density of the friendly numbers (if it exists) is positive.
If the Euclidean algorithm requires N steps for a pair of natural numbers a > b > 0, the smallest values of a and b for which this is true are the Fibonacci numbers F N+2 and F N+1, respectively. [98] More precisely, if the Euclidean algorithm requires N steps for the pair a > b, then one has a ≥ F N+2 and b ≥ F N+1.