>>7681

>>7682

This is amazing work. I'm going to integrate this into my current research, may solve something.

>>7428

Can you elaborate more on this?

>>9081

getElement(-f, 1, getT(x[t]+1))

Where c is the product of two primes:

c % (N-1) = 2d+1

If n is too small, the large square appears as an exact value of i[t] in e or -f

That is what the algorithm from earlier is showing, it can be extended.

>>9125

public VQCElement getElementBelowRoot(BigInteger e, BigInteger n, BigInteger d) { BigInteger c = d.multiply(d).add(e); BigInteger f = (d.add(n)).pow(2).subtract(c); BigInteger xpn = sqrt(f); BigInteger x = xpn.subtract(n); boolean eIsEven = isEven(e); boolean xIsEven = isEven(x); if (!(eIsEven == xIsEven)) { x = x.subtract(one); } return getElement(e, n, getT(e, x)); }

>>9126

If we input -f, 1, d into this method, calculates

c = d^2 + f

(d+1)^2 - c = e

xpn = sqrt(e)

x = xpn - 1

So in this way, taking sqrt(e) and sqrt(f) give us the elements that d is between.

>>9127

I was wrong about half of my observation. sqrt(f) gives you where d is between in (e,1), but sqrt(2f + e) is what gives you where d is in (-f, 1)

I misinterpreted an addition operation for a subtraction operation in my program. What it for -f is this:

c' = d^2 - f

(d+1)^2 - c' = f'

xpn = sqrt('f)

x = xpn - 1

I noticed my mistake because sqrt(e) was giving the wrong x value, but sqrt(2f+e) wasn't.

The importance of column zero is all in the fact that it encodes c value whose remainder is 0, squares

It's filled with alternate factorisations of squares

Shouldn't such an alternate factorisation exist for i or squared?