I started a thread and stickied it here. >>5774
>one of the four types of solution (odd x+n with either even or odd e), then quickly for another and then the other two.
I've graphed four different graphs for the different parities of (x+n) and e (so (odd, even), (even, odd), (odd, odd) and (even, even)). They're the first four pictures here. (x+n) is the x axis and e is the y axis. The numbers in these grids are first the co-ordinates ((x+n), e) and then the third number is f.
As you can see in the (x+n) odd e even and (x+n) even e odd pictures, most of the f values are squares. So these are our infinite sets where (x+n)(x+n) = f. They only ever occur when (x+n) and e have opposite parity. As you'll also probably see (sorry if one or two of them are blurry), there are some other non-linear f values scattered around in weird places. I can't tell right away why that is. You'll also see they just make a big (visual) square and can infinitely and uniformly stretch downwards or sideways. It's an infinite set of infinite sets, each with an increasing amount of noise. If you figure out the noise, you win.
Looking at the pictures where (x+n) and e have the same parity, these contain the finite sets. Some parts of the finite sets are contained within the grids where (x+n) and e have different parity like I explained above as those "f values scattered around in weird places", so they aren't singled out completely, but none of the f values are square/infinite unless (x+n) and e have different parity. So that's quite a useful piece of information. Another thing from these is that since they show the finite series you can see them increasing in length as y increases. I've shown that a little more in the fifth picture. It increases at an increasing rate (so from (x+n)=3 it's 1, 5, 11, 19, 29 etc, meaning +4, +6, +8, +10). As you might also see if the numbers aren't too small is that in both cases where (x+n) and e have the same parity, most of the values of f follow strict patterns, but there's also noise here. So if there wasn't noise, this would be the pattern: from the (visually) top value of e downwards (in even even it's from 0, and in odd odd it's from 3 since when e=1 (x+n) is even according to the other grid where that's the case), the value of f decreases by 2 whenever e increases by 4, and as (x+n) increases by 2 and e stays the same, f increases by an increasing amount (so if you look at the even even picture, at {4, 0, 7} and {6, 0, 17}, as (x+n) increases by 2, f first increases by 10, then, if there wasn't noise, it would be 31 next, meaning an increase of 14, then an increase of 18 to 49, so it's increasing by 10, 14, 18, 22 etc, meaning the increase increases by 4 each time).
Hopefully that wasn't too much ramble. If there's anything to take away from it it's that (x+n)(x+n) = f when (x+n) and e have opposite parity.
Have you done this on any other examples or any other divisor fs? It looks like some degree of symmetry could come from that.