Yeah okay Jan
So as it turns out, the smoothness of N-n as you add more small prime factors to c (using q) actually decreases (i.e. the highest prime factor goes up). Weren't we told this whole time that smoothness would increase?
>When is the remainder of (DPN + XPN) a square?
If e is ever a square it means that c is the sum of two squares. If (d+n)+(x+n) is the sum of two squares, either (d+n) and (x+n) are squares or are equivalent to the sum of two squares when added together (i.e. 16+9 = 15+10). (d+n)(d+n)-(x+n)(x+n)=c, and (d+n)-(x+n)=a. d+n=i and x+n=j, so (d+n)+(x+n) = b. If (d+n)+(x+n) have a remainder of a square and are therefore the sum of two squares, b will be the sum of two squares. a won't necessarily be the sum of two squares though. That's all I can deduce logically at the moment. Here are some examples. I'm not seeing anything obviously useful.
>What about the remainder for (DPN + XPN) of d or e?
d' or e' for (d+n)+(x+n) would have a remainder of a square if they themselves are the sum of two squares. Odd sums of two squares are only divisible by other odd sums of two squares, so if d' is odd and has a remainder of a square, (d+n)+(x+n) has to be the square of a sum of two squares (i.e. (4+9)(4+9)=169, d'=(4+9)), since d' is its square root. Since d' is the sum of two squares, (d+n)+(x+n) will also be the sum of two squares, since a square is always the sum of two squares at the very least as itself plus 0 squared. With evens I don't think there was an equivalent pattern, so I don't know about that. I don't know about when e' would be a sum of two squares either.
I interpreted the "remainder" as e, and treated (d+n)+(x+n) as c to find its e value. In other words, it's a bunch of examples of (d+n)+(x+n) values for which ((d+n)+(x+n)) - (floor(sqrt((d+n)+(x+n)))floor(sqrt((d+n)+(x+n)))) or c'-(d'd') is a square.