To factor an integer N, given that it is equal to the product of two primes, it suffices to find an integer d satisfying a certain simple numerical test. In this approach, the factorization problem equates to the problem of designing an optimal data base of values d to be tested.
If x is a positive real number, let C(x) denote the smallest integer greater than or equal to x (the integer ceiling function). Let the function f (x) be defined by the formula f (x) = C(2 √ x) 2 -4x and let f 2 (x) denote the composition f (f (x)) .
PROPOSITION 1.1. Let n be a positive integer. Then f (n) = 0 if and only if n is a perfect square ( √ n is a whole number).
n is an even integer. Therefore, √ n is an integer. Conversely, if √ n is an integer, then so is 2 √ n . Thus, C(2 PROOF. Suppose that f 2 (n) = 0 . By Proposition 1.1, f (n) = t 2 for some positive integer t . Thus,
) and t must be both even or both odd. Therefore, if we let
, then both u and v are integers and n = uv . To show that
then n is a perfect square and there is nothing to prove. So we may assume that u = v . Let t = |u -v| .
Then
Thus, f 2 (n) = 0 .
COROLLARY 1.3. Suppose N is the product of two primes p and q where | √ p -√ q| ≤ 1 . Then the prime factors can be recovered explicitly in terms of N by way of the formulas: PROPOSITION 1.5. To factor an integer N , given that it is equal to the product of two primes p and q , it suffices to find an integer d < 1 2 N satisfying the test
The prime factors p and q can be recovered separately as factors of u and v through the formulas p = gcd(N, u) and q = gcd(N, v) .
It remains to show that p and q are factors of u and v separately. A rough estimate is enough to prove this. Since, by
u, so that, at least whenever u ≥ 16, we have
If p and q both were factors of u , we would have u = apq = aN for some integer a which would imply that N d ≥ 1 2 a 2 N 2 or d ≥ 1 2 a 2 N , contradicting the assumption that d < 1 2 N . A similar contradiction occurs if we suppose that p and q both divide v . At this stage, we will attempt to supply at least a rough estimate of the effectiveness of a data base. Suppose N = pq , where p < q . In D we want to find d = xy such that
This inequality is satisfied if the set of fractions x y in the interval [0, 1] is so numerous that, for at least one of them, x y comes within a distance of 1 √ qy of the fixed quantity p q . This will have a high probability of happening if
At this point, we need to formulate a reasonable estimate for √ qy . However, without any specific knowledge of the structure of D this is difficult to do. We turn then to a discussion of some specific choices of data base D .
The simplest data base is a list of consecutive integers starting at 1. Let D 0 (m) = [1, 2, 3, . . . , m] .
To refine the inequality (2), note that y ≤ m . However, the median divisor of a typical d ∈ D is
The explicit dependence on q can be removed by setting R = q p > 1 , so that
A sharp lower bound estimate of Y (D 0 (m)) is given by m itself. To see this, note that d∈D 0 (m) Y (d) is bounded above by 1≤k≤m τ (k) = O(n ln(n)) [1]. Thus, the data base D 0 (m) has a good chance of
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