Our task is simple. We are going to completely break the entire RSA cryptosystem! We're bigshots now.
But wait, how are we going to just crack RSA? With the Virtual Quantum Computer! Quick rundown to follow.
The main mathematical task that must be achieved to break RSA is to be able to factorize any integer instantly. The VQC is a promising way to do that.
What is the VQC?
The virtual quantum computer (VQC) is a grid made of infinite yet constructable sets that follow a known pattern. Like a quantum spreadsheet.
The grid is the superposition. The collapse of that superposition will be two input parameters, d and e which can be calculated easily for all integers, c, where c is the difference of two squares.
So, when the integers that are the difference of two squares are arranged into the grid and their corresponding properties are shown, a pattern emerges that shows calculation instead of searching is possible.
Legend
The map's legend is {e:n:d:x:a:b}, where d is the result of removing the largest square from c AKA the square root,
e is the remainder,
n is what you add to d to be exactly halfway between a and b,
and x is what you add to a to make d.
c is any number that is the difference of two squares, so odd numbers are included.
na and nb for any c can be found n places apart in the cell at (e,1).
Rules of the grid: global rules
Each cell of the grid (e,n) has infinite elements or ZERO elements.
Each cell with one value has infinite elements, since every element can make a new one.
By induction, a cell only needs one value to make infinite values, that's part of the power of this and is why it is a virtual quantum computer as a whole.
The t variable is what will allow you to walk across these infinite elements.
If a grid cell has elements, all elements are constructable from a finite set of root elements.
The grid is indexed using e, n, and t, where e is the rows, n is the columns, and t is the specific element in the cell-group.
Thus, only three variables are required to identify an element: e, n and t.
All products of odd numbers and all products of pairs of even numbers are the difference of two squares.
The x-intercept of the line that goes through the point containing the factors of c is (a + 1).
(1, 1) - the key
The values of a and b at 1,1 are related to the length of the longest side in right angled triangles.
The values here can be used to create the entire grid.
The values here determine the values of the rows to the left and right, which determine the values of the whole column.
(f, 1) is an interesting cell.
Columns
Each cell at n=1 contains the roots of products in the column.
If c is a prime number, it will appear in one column exactly once.
If c is the product of two prime numbers that do not equal eachother, c will appear in two cells of one column.
All products (integers) c that are the sum of two squares appear (only) in columns where e=0,1,4,9,16,25β¦
All factors in a column are factors of the elements of the first cell in their column.
All Fermat primes (except) 3 appear in column one.
(e, 1)
If a number at position t has a factor s, then s is a factor at (t+s), (t+2s) and so on for a at (e,1).
Also, if a number at position t has a factor s at (e+1), then s is a factor at (s+1-t), (2s+1-t), etc for a at (e,1).
Rows
(1, n)
The cells in row one where n=1 have a relationship with the cells 2n to the right and 2n to the left.
Each "a" from the first row equals na because xx+e = 2na and na is half of that. That's BIG part of the KEY
Each element in a cell can be generated by moving up (t-1 = x-2) or down (t+1 = x+2). Other variables can be generated from x.
Every single factor of any value of a in the first row will be referred to as s.
Useful Equations and Notation
ab = c
dd + e = c
(d + n)(d + n)-(x + n)(x + n) = c
a + 2x + 2n = b
a = d - x
d = a + x
d = floor_sqrt(c)
e = c - (dd)
b = c / a
n = the difference between the square root d and the larger of the two squares
n = ((a + b) / 2) - d
d + n = number that is exactly halfway between a and b
d + n = i
x = d - a
x = (floor_sqrt(( (d+n)*(d+n) - c))) - n
x + n = j
f = e - 2d + 1
t = the variable that lets you traverse the infinite elements in for a given (e, n) that has values.
if (e is even) t = (x + 2) / 2
if (e is odd) t = (x + 1) / 2