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A Problem on the largest divisor dd of NN with dNd\leq \sqrt{N}

Published 8 Sep 2026 in math.NT | (2609.08596v1)

Abstract: For a given number NN, we consider the problem of computing two integers $1\leq r,f < N$ such that the set X(N,r,f)=(a+b)(f+Nr+1f):ab=Nr\mathcal{X}(N,r,f) = {(a+b)-(f+\frac{Nr+1}{f}): ab=Nr} consists only of positive integers. Computing a solution to the problem is equivalent to finding a pair (r,f)(r,f) satisfying $l(Nr) < f \leq l(Nr+1)$, where l(x)l(x) is the largest divisor of xx bounded by x\sqrt{x}. This requires factoring both NrNr and Nr+1Nr+1. We present a simple randomized algorithm that - avoiding factoring - computes pairs (r,f)(r,f). We give an exact formula for the total number of possible pairs (r,f)(r,f), and with the aid of empirical data we estimate that the ratio φ(N)2F(N)\frac{φ(N)-2}{|\mathfrak{F}(N)|} to be roughly about cloglogNc*\log \log N. Here, F(N)\mathfrak{F}(N) is the set of unique rr appearing among all possible pairs (r,f)(r,f), φ(.)φ(.) is the Euler's totient function, and cc is a constant equal to 2 for prime NN and oscillates much for composite NN. As a separate and independent case, we study the same problem of computing (r,f)(r,f) with $r>N$. We present a procedure to find such an rr, which requires finding the least prime in an arithmetic progression.

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