The Curve of Constant Area - A Logue Aug 2016
I've been thinking about the Curve of Constant Area off and on for the past few years, and now I think you should too. I'm pretty bad at math, so for all I know it's some kind of crackpot illuminati time cube brainworm thing, like a song you can't get out of your head, only math. So, uh, you should really check it out and think about it a bit too. It's cool. Trust me. :)
The Curve of Constant Area is super-simple: Just take any number and call it the area of a square or a rectangle. Then, using an x,y axis like the kind you stared at in grade school for hours with +X going to the right and +Y going up, the Curve of Constant Area (let's call it the CCA for short) is the location of the upper right corner of all of the rectangles with that area, assuming the lower left corner is at 0,0 and the rectangles are "orthogonal" (not rotated at all). Here's what the graph for an area of 9 looks like:
So what's so interesting about that? Well, the first thing we can see is that the CCA is the same shape for any area, it's just scaled by that area. For a proof of that, you could think of the curve staying right where it is and scaling the graph underneath it such that x and y are equal to the square root of the area where the curve crosses the diagonal line x=y. The curve is also symmetrical around that line.
The thing that's most interesting though, at least to me, is related to how and when the curve passes through integer x,y intersections as it is scaled. For example, for an area that's a square of a prime number such as 4, 9, 25, 49, ... the CCA passes through three and only three x,y intersections. They are easy to find because they are always at x=1 y=area, x=area y=1, and finally the point where x and y both equal the square root of the area. This is true even if the area is the square of a prime number that's 400 digits long.
Now consider the curve for an area that is the product of two prime numbers. For example, 15, which is the product of the primes 3 and 5:
If we scale the curve such that the x underneath the point where the curve crosses the diagonal equals the square root of 15, then presto: the curve intersects x,y at the area's factors 3 and 5. Given nothing but the area, the curve of constant area "found" the factors instantly. That's interesting because it works even when the two factors are prime numbers that are each hundreds of digits long. Woot! Super hard factoring problem solved! Alas, not quite... Our visual pattern recognition is really the thing that "found" those prime factors and it was only able to do so because the graph is so small. If we obtain an area by multiplying two primes that are each hundreds of digits long, the graph will be too big to process visually, but the same thing will be happening: the curve will pass through integer x,y intersections at the factors and only at the factors.
Where I'm headed with this is wondering whether there's some clever way to use something like the CCA to find those points of intersection. Unlike modulo division, multiplying two prime factors doesn't throw any information away. That's the thing I haven't been able to get out of my head.
Maybe if we rotate the graph counter-clockwise by 45 degrees and stretch it such that the x-axis lines and the curve line are straight and only the y axis lines curve, then add a simultaneous equation where the y axis lines are straight... hmm.
A deterministic solution to the factoring problem may not exist, but it sure feels like all the parts are there to be able to come up with some fourth order equations that those intersection points pop out of. I think there's a good chance that someone has already figured out an approach to this problem that works, only they're not talking.
As of August 25 2016, google returns seven (7) results for "curve of constant area". I found that pretty surprising, and thought I should try to increase it to eight (8) since it's related to the majority of cryptography as we know it and probably to alien civilizations and Donald Trump's hair as well. If you've figured out how to factor products of two large prime numbers, let me know so I can stop thinking about it.
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