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One way to classify composite numbers is by counting the number of prime factors. A composite number with two prime factors is a semiprime or 2-almost prime (the factors need not be distinct, hence squares of primes are included). A composite number with three distinct prime factors is a sphenic number. In some applications, it is necessary to ...
All of these cases take the form "if a prime is a residue (mod a composite), then the composite is a residue or nonresidue (mod the prime), depending on the congruences (mod 4)". He proves that these follow from cases 1) - 8). Gauss needed, and was able to prove, [18] a lemma similar to the one Legendre needed: Gauss's Lemma.
The quadratic residuosity problem (QRP [1]) in computational number theory is to decide, given integers and , whether is a quadratic residue modulo or not. Here = for two unknown primes and , and is among the numbers which are not obviously quadratic non-residues (see below).
Modulo 2, every integer is a quadratic residue. Modulo an odd prime number p there are (p + 1)/2 residues (including 0) and (p − 1)/2 nonresidues, by Euler's criterion.In this case, it is customary to consider 0 as a special case and work within the multiplicative group of nonzero elements of the field (/).
(In modern terminology: if a prime p divides the product ab, then p divides either a or b or both.) Proposition 30 is referred to as Euclid's lemma, and it is the key in the proof of the fundamental theorem of arithmetic. Any composite number is measured by some prime number. —
For example, the AP-3 {3, 7, 11} does not qualify, because 5 is also a prime. For an integer k ≥ 3, a CPAP-k is k consecutive primes in arithmetic progression. It is conjectured there are arbitrarily long CPAP's. This would imply infinitely many CPAP-k for all k. The middle prime in a CPAP-3 is called a balanced prime.
A cluster prime is a prime p such that every even natural number k ≤ p − 3 is the difference of two primes not exceeding p. 3, 5, 7, 11, 13, 17, 19, 23, ... (OEIS: A038134) All odd primes between 3 and 89, inclusive, are cluster primes. The first 10 primes that are not cluster primes are: 2, 97, 127, 149, 191, 211, 223, 227, 229, 251.
Hence, for a highly composite number n, the k given prime numbers p i must be precisely the first k prime numbers (2, 3, 5, ...); if not, we could replace one of the given primes by a smaller prime, and thus obtain a smaller number than n with the same number of divisors (for instance 10 = 2 × 5 may be replaced with 6 = 2 × 3; both have four ...