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. 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 I can't get out of my head, only math. So, uh, you should check it out. :) 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 the graph of xy=1, scaled up by that area. You can 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. For areas that are products of integers, the y values of the portion of a curve between y=2 and y=1 can be represented by fractions where the numerator begins at (area+1)/2-1 and descends to 1, and the denominator is equal to x. If the fraction is reducable, then x divided by the reduced denominator is a factor of the number. For example, the fractions for the area 15 are:
The curve continues beyond area,1 with y<1. In the case of area=15, the fractions are 1/16, 2/17, 3/18, etc. The same rule applies: x divided by a reducable fraction's denomininator is a factor of the number. A deterministic solution to the factoring problem may not exist, but it does seem like some of the parts are there to be able to come up with some faster algorithms for finding those intersection points or reducable fractions. 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. 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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