## Hairy Math Stuff (my number study, April 2022)

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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.

### 3 responses to “Hairy Math Stuff (my number study, April 2022)”

1. Sent this to my PhD Math prof Linda!

2. Sent this to my PhD math prof wife Linda. She can torture her students with it😂!

3. Thank you! As I indicated, there’s a large probability in my mind that I have not yet discovered anything new. Your PhD math prof would definitely know that better than I do. In any case, I’m continuing to work on it with my little gray cells.