CAVEAT: Serious Math Stuff Ahead (Re: My Current Study)

Fri, 15 Apr 2022 21:20:47 GMT

Let positive integer m = p*q where p, q are prime numbers and p is not equal to q.

Let s be the smallest positive integer such that s^2 > m, and let r be the difference between them so that:

m = s^2 - r.

Consider x^2 - r to be a quadratic polynomial in x that evaluates to a value of m at x=s.

Construct a series of quadratic polynomials in x that also evaluate to a value of m at x=s by setting n = 0, 1, 2, ... and forming each polynomial as:

x^2 + n*x - (r + n*s).

Verification that each of these will evaluate to a value of m is an exercise for the reader.

Using the Quadratic Formula to determine the roots of y = x^2 + n*x - (r + n*s), construct the Discriminant (the part under the square root sign) as follows, for each such polynomial corresponding to a value of n:

Discriminant = n^2 + 4*(r+n*s) = n^2 + 4*s*n + 4*r.

When the Discriminant is a perfect square, then the polynomial will factor, and since the value of each such polynomial at x=s is equal to m, the integer roots of the polynomial thus discovered via the Quadratic Formula, give the factors of m.

So the problem is: Given positive integers r and s, series n = 0, 1, 2... and the quadratic sequence of Discriminant values n^2 + 4*s*n + 4*r, can we quickly and/or formulaically determine a value of n that makes that Discriminant a perfect square?

The difficulty of the integer factorization problem tells me the answer is probably no, but I am not certain anyone has looked at the problem this way, and a mathematics instructor once told me WE DO NOT KNOW how simple solutions to unknown problems like this could be.

ADDENDUM: Expressing the discriminant as a quadratic polynomial in n with integer coefficients, n^2 + 4*s*n + 4*r, the Quadratic formula gives the greater root (zero) of that polynomial at the real number 2*(SQRT(m) - s). The minimum of that polynomial occurs when n = -2*s.

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jcsbimp01

CAVEAT: Serious Math Stuff Ahead (Re: My Current Study)