I've been thinking about this

>>7466

>what do you do to find the lookup x for c that gives you na or (n-1)

>find the lookup x

>as opposed to the actual x

I'm thinking maybe the x we're looking for isn't actually the correct x. Maybe it's a value that we can substitute for x to find na or (n-1) as one of the other values. This lookup x would probably need to be in our list of knowns (or stuff we can find that we haven't paid attention to). More evidence for it not actually being the correct x is the thing I've been saying about the cell in (e,1) where a[t]=na being irrelevant if you find the correct x for it since it's the same x as the solution record. If we're looking for a different x value that gives us na in a different way (probably based on (an)(a(n-1))), we would still be using (e,1) and (f,1), and we would also still be looking for an x value related in some way to na, so it would still fit all the criteria.

So if we had a different, incorrect x value, and we applied it to our knowns, how would that produce na, (n-1) or (an)(a(n-1))? Well, if we are keeping our knowns the same, that would mean we'd find it as a, b or n. This doesn't seem to really make any sense, though. If it turned up as a or b, c would need to be divisible by it, which, in the case of semiprimes, isn't possible. If it turned up as n, a and b would need to change, which, again, in the case of semiprimes, isn't possible. So this means we're probably applying it to numbers that aren't directly related to our given c.

I think maybe we're looking for the x value from a cell where a=aa and b=n(n-1), or something similar to that. Maybe, since n can be either greater than or smaller than a, and since b always has to be greater than a, whether the n we're looking for is greater or smaller than the a we're looking for determines whether we're looking for na or (n-1), since VQC mentioned both.

>>7528

If this applies to semiprimes, a will be odd. That means aa will be odd. In order to be a valid difference-of-two-squares cell, a and b have to have the same parity. That means n(n-1) would have to be odd. But since it's a number multiplied by itself minus 1, whether n is odd or even, the other one will have the opposite parity. When a number is multiplied by another number with the opposite parity, you get an even number. That means, in this case, a would be odd and b would be even. That means it doesn't exist in the grid. I still think we're looking for an incorrect x value that somehow gets applied to a known (or multiple) and produces either (an)(a(n-1)), na or (n-1) somehow. But I guess it's not going to be in the cell I was thinking of.