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Then, by strong induction, assume this is true for all numbers greater than 1 and less than n. If n is prime, there is nothing more to prove. Otherwise, there are integers a and b, where n = a b, and 1 < a ≤ b < n. By the induction hypothesis, a = p 1 p 2 ⋅⋅⋅ p j and b = q 1 q 2 ⋅⋅⋅ q k are products of primes.
To factorize a small integer n using mental or pen-and-paper arithmetic, the simplest method is trial division: checking if the number is divisible by prime numbers 2, 3, 5, and so on, up to the square root of n. For larger numbers, especially when using a computer, various more sophisticated factorization algorithms are more efficient.
Hensel's original lemma concerns the relation between polynomial factorization over the integers and over the integers modulo a prime number p and its powers. It can be straightforwardly extended to the case where the integers are replaced by any commutative ring, and p is replaced by any maximal ideal (indeed, the maximal ideals of have the form , where p is a prime number).
An n th root of a number x, where n is a positive integer, is any of the n real or complex numbers r whose nth power is x: r n = x . {\displaystyle r^{n}=x.} Every positive real number x has a single positive n th root, called the principal n th root , which is written x n {\displaystyle {\sqrt[{n}]{x}}} .
The other roots of the equation are obtained either by changing of cube root or, equivalently, by multiplying the cube root by a primitive cube root of unity, that is . This formula for the roots is always correct except when p = q = 0 , with the proviso that if p = 0 , the square root is chosen so that C ≠ 0 .
The principal cube root is the cube root with the largest real part. In the case of negative real numbers, the largest real part is shared by the two nonreal cube roots, and the principal cube root is the one with positive imaginary part. So, for negative real numbers, the real cube root is not the principal cube root. For positive real numbers ...
The conductor of N over k is f, and f 2 is the relative discriminant of N over K. The discriminant of N is d 3 f 4. [6] [7] The field K is a pure cubic field if and only if d = −3. This is the case for which the quadratic field contained in the Galois closure of K is the cyclotomic field of cube roots of unity. [7]
As far as is known, this is not possible using classical (non-quantum) computers; no classical algorithm is known that can factor integers in polynomial time. However, Shor's algorithm shows that factoring integers is efficient on an ideal quantum computer, so it may be feasible to defeat RSA by constructing a large quantum computer.