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If every pair in a set of integers is coprime, then the set is said to be pairwise coprime (or pairwise relatively prime, mutually coprime or mutually relatively prime). Pairwise coprimality is a stronger condition than setwise coprimality; every pairwise coprime finite set is also setwise coprime, but the reverse is not true. For example, the ...
Because a knot cannot be self-intersecting, the three integers ,, must be pairwise relatively prime, and none of the quantities ,, may be an integer multiple of pi.Moreover, by making a substitution of the form ′ = +, one may assume that any of the three phase shifts , , is equal to zero.
In mathematics, the Chinese remainder theorem states that if one knows the remainders of the Euclidean division of an integer n by several integers, then one can determine uniquely the remainder of the division of n by the product of these integers, under the condition that the divisors are pairwise coprime (no two divisors share a common factor other than 1).
If any prime factor were shared by any two of (and hence all three of) d, e, and f then by this last equation that prime would also divide each of a, b, and c. So if a, b, and c are in fact pairwise coprime, then d, e, and f must be pairwise coprime as well. This holds for B and C as well as for A.
A seemingly weaker yet equivalent statement to Bunyakovsky's conjecture is that for every integer polynomial () that satisfies (1)–(3), () is prime for at least one positive integer : but then, since the translated polynomial (+) still satisfies (1)–(3), in view of the weaker statement () is prime for at least one positive integer >, so ...
Let be a cyclic group of order mn, where m and n are relatively prime. Then a n {\displaystyle \langle a^{n}\rangle } and a m {\displaystyle \langle a^{m}\rangle } are cyclic subgroups of orders m and n , respectively, and a {\displaystyle \langle a\rangle } is the internal direct product of these subgroups.
But in such a large, complex market, items frequently fall through the cracks, so if you do discover a counterfeit item in your Amazon Prime box, Dimyan suggests two actions. "Utilize Amazon's ...
It is convenient at this point (per Trautman 1998) to call a triple (a,b,c) standard if c > 0 and either (a,b,c) are relatively prime or (a/2,b/2,c/2) are relatively prime with a/2 odd. If the spinor [m n] T has relatively prime entries, then the associated triple (a,b,c) determined by is a standard triple. It follows that the action of the ...