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Re:Factoring composite numbers into nearly equal factors by dmdox on Friday February 11, 2022 @03:12PM (#30306442)
Right.

Define L = (a+b)/2 and R = (a-b)/2, so that a * b = L² - R² for all a, b.

If there exist positive integers a, b such that a * b = n with |a-b| <= k n^(1/4) [constant k]

Then there exist L, R (either both integers or both half-integers) with L² - R² = n and |R| <= (1/2) k n^(1/4)

From which one can deduce:

R² <= (k²/4) √n

L² - R² >= L² - (k²/4) √n

n >= L² - (k²/4) √n

L² <= n + (k²/4) √n

L <= √(n + (k²/4) √n)

L <= √n + k²/8

If k <= 1, then L must be ceil(√n) or ceil(√n) - 1/2

If k <= 2√2, then L must be ceil(√n), ceil(√n) - 1/2, or ceil(√n) + 1/2

(my original estimate of 2 steps was due a slight miscalculation)

Adding epsilon would re-introduce an exponential component to the search. In fact, epsilon = 3/4 gives you total power. But this is only relevant if the reduction (https://cstheory.stackexchange.com/a/4785) can be made precise enough to bound L, U to within a fixed small epsilon. It could even be the case that you can bound to an arbitrarily small epsilon, but only at the expense of exploding other parameters. (B, T, k).

Re:Factoring composite numbers into nearly equal factors by svat on Friday February 11, 2022 @05:59PM (#30308039)
In the meantime, I wrote this up in (too much elementary) detail: https://colab.research.google.com/drive/10t4WRhyAPK1pCzBkaau...

The funny thing is, it's often the case that already √(m^2 - 4N) > N^(1/4), so we can answer "No" to whether such a factorization exists, without even trying a single attempt.

Re:Factoring composite numbers into nearly equal factors by dmdox on Sunday February 13, 2022 @06:30PM (#30326681)
If you're interested in investigating more, take a look at "Continued Fraction Factorization". This is essentially the first step of that more advanced algorithm.

From the continued fraction of √n, you get a series of best rational approximations a/b to √n. Each such approximation gives you:

a/b ~ √n

a² ~ b² n

The first approximation is a = floor(√n), b = 1

These approximations satisfy |a² - b² n| < 2 √n.

(this can be used for generating relations as input to Dixon's method)