>>6160

I've been thinking along the lines of the equation for d in (e, 1).

We know these:

• Odd e: 2t^2 + (e-1)/2

• Even e: 2t(t-1) - (e/2)y

And when we look into (0, 1) and (1, 1) (by subtracting the jumping from (e, 1) >>6153) we have the following equations:

Odd e: 2t^2

Even e: 2t(t-1)

But I don't know if the d in (e, n) has any similar formula. I've looked a bit into it, but I couldn't figure any generalized solution. At least not yet (Any hints VQC?).

Anyways. I've been thinking about the following equation (for even e):

2t(t-1) - d == 0 (mod (n-1))

Now we know that: 2t(t-1) % (n-1) != 0, but we know 2t(t-1) % (n-1) = d % (n - 1).

And that's all I got for now. I've been reading up on modular arithmetics, but so far I haven't gotten anywhere useful.

One thing I've been thinking about is that VQC has been hinting at using the grid to find a shortcut, which means this D[t] - d == 0 (mod n-1) isn't an actual answer, or rather the result won't be used as an answer it self. We won't be able to solve it, but it should help us.

So I'm also looking into the idea of reversing his crumbs. We know that we're supposed to use this as a shortcut to end up with the answer. We're supposed to get an offset from D[t] - d, which means an offset to what and by working our way backwards, how can we see this offset in use?