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Theory of Relative Squares - Printable Version

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RE: Theory of Relative Squares - AJMansfield - 02-14-2014

If you could find a really efficient way to calculate [Image: gif.latex?%28n+1%29%5E2] given [Image: gif.latex?m%3Dn%5E2] for some [Image: gif.latex?n%5Cin%5CBbb%7BN%7D], there is a way to use that to crack an RSA encryption.

RE: Theory of Relative Squares - AFtExploision - 02-14-2014

(02-14-2014, 03:02 AM)AJMansfield Wrote: If you could find a really efficient way to calculate [Image: gif.latex?%28n+1%29%5E2] given [Image: gif.latex?m%3Dn%5E2] for some [Image: gif.latex?n%5Cin%5CBbb%7BN%7D], there is a way to use that to crack an RSA encryption.
what

RE: Theory of Relative Squares - Darkroom - 02-14-2014

all (n+1)^2 is an addition and a multiplication literally its super fast even for huge bit sizes

RE: Theory of Relative Squares - Iceglade - 02-14-2014

And if you don't like the composition then just do n^2+2n+1... but I can't imagine why you'd want that.

RE: Theory of Relative Squares - AJMansfield - 02-14-2014

What I mean, is that if all you knew was [Image: gif.latex?%5Cinline%20m], normally you have to take its square root (a tedious, [Image: gif.latex?%5Cinline%20O%28M%28n%29%29] process, where the current best for [Image: gif.latex?%5Cinline%20M%28n%29%3Dn%5Clog...5E*n%29%7D]) to find [Image: gif.latex?%5Cinline%20n], and then you can just do [Image: gif.latex?%5Cinline%20m+2n+1] to get [Image: gif.latex?%5Cinline%20%28n+1%29%5E2]. But if you could figure out how to get [Image: gif.latex?%5Cinline%20%28n+1%29%5E2] from [Image: gif.latex?%5Cinline%20n%5E2] without needing to calculate [Image: gif.latex?%5Cinline%20n], you could crack RSA using the difference-of-squares method much more quickly than existing methods.

Basically, find a way of computing a square root of a perfect square that is significantly faster than multiplying two numbers of the same lemgth, and you win. (Not that I expect this to ever happen.)

RE: Theory of Relative Squares - Iceglade - 02-15-2014

Yeah, I'm not sure how that would be done if it was at all. It would be an interesting thing to look into at some point though, I guess.

RE: Theory of Relative Squares - Dcentrics - 02-21-2014

* Dcentrics stares in disbelief

RE: Theory of Relative Squares - CrazyPyroEagle - 06-08-2016

(a + b)^2 - a^2 = a^2 + 2ab + b^2 - a^2 = b^2 + 2ab

It's not as simple in different powers though.

(a + b)^3 - a^3 = a^3 + 3ba^2 + 3ab^2 + b^3 - a^3 = 3ba^2 + 3ab^2 + b^3

RE: Theory of Relative Squares - Chibill - 06-08-2016

Please don't bump threads older than a year.

RE: Theory of Relative Squares - LordDecapo - 08-13-2016

It was bumped already...
But I'll add this here
N^2 = (N-1)^2 + (N-1) + N