# Optimal Integer Factorization and Quantum Theory of Numbers

### Optimal Integer Factorization and Quantum Theory of Numbers

"I_C

We begin far apart with simple ideas and with little knowledge of the path before us,

But we shall meet at the middle and handshake our grand triumph."
--David Cole (aka primework123)

Stage 0:

Keywords: Decomposition of n = p * q, primes are the atoms of the integers, base-10 and the finite field, $$Z_{2 }$$, are sub-atomic levels of the integers,...

Relevant Keywords: Carry-bit/number input, Residue bit/number, Carry-bit/number output, Optimization and Synchronization (or Handshaking), ...

If we let s, t$$\in$$ $$Z_{10 }$$ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, the product, s* t, can be easily computed as follows:

Carry-bit input or $$c_{i0 }$$ = 0 (always);

Residue bit or $$r_{0 }$$ = s * t (mod 10);

Carry-bit output or $$c_{o0 }$$ = (s * t - $$r_{0 }$$)/10 ;

Simple Example: Let s = 4 and s = t, then we have the following results:

$$c_{i0 }$$ = 0 (always);

$$r_{0 }$$ = s * t (mod 10) = 12 (mod 10) = 2;

$$c_{o0 }$$ = (s * t - $$r_{0 }$$)/10 = ( 12 - 2 ) /10 = 10/10 =1;

So, s * t = 12.

... End of Stage 0.

David Cole
(aka primework123)
Please support my research work at: https://www.gofundme.com/david_cole
Thank you! Thank Lord GOD!
primework123

Posts: 22
Joined: Sat Oct 31, 2015 2:07 pm
Reputation: 0

### Re: Optimal Integer Factorization and Quantum Theory of Numb

"I_C

We begin far apart with simple ideas and with little knowledge of the path before us,

But we shall meet at the middle and handshake our grand triumph."
--David Cole (aka primework123)

Stage 0:

Keywords: Prime decomposition of n = p * q, primes are the atoms of the integers, base-10 and the finite field, $$Z_{2 }$$, are sub-atomic levels of the positive integers, Complexity and Change,...

Relevant Keywords: Carry-bit/number input, Residue bit/number, Carry-bit/number output, Optimization and Synchronization (or Handshaking), ...

If we let s, t$$\in$$ $$Z_{10 }$$ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, the product, s* t, can be easily computed as follows:

Carry-bit input or $$c_{i0 }$$ = 0 (always);

Residue bit or $$r_{0 }$$ = s * t (mod 10);

Carry-bit output or $$c_{o0 }$$ = (s * t - $$r_{0 }$$)/10 ;

Simple Example: Let s = 4 and s = t, then we have the following results:

$$c_{i0 }$$ = 0 (always);

$$r_{0 }$$ = s * t (mod 10) = 12 (mod 10) = 2;

$$c_{o0 }$$ = (s * t - $$r_{0 }$$)/10 = ( 12 - 2 ) /10 = 10/10 =1;

So, s * t = 12.

... End of Stage 0.

David Cole
(aka primework123)
Please support my research work at: https://www.gofundme.com/david_cole
Thank you! Thank Lord GOD!

primework123

Posts: 22
Joined: Sat Oct 31, 2015 2:07 pm
Reputation: 0

### Re: Optimal Integer Factorization and Quantum Theory of Numb

Hey primework123, I like your previous post.

Suppose, we are given a positive integer, n, which is the product of two large primes, p and q, how do you factor n?

In base-10, we can let p = $$\sum_{i=0}^{j}a_{i}*10^{i}$$, and q = $$\sum_{i=0}^{k}b_{i}*10^{i}$$ where

$$a_{i}, b_{i}\in Z_{10 }$$={0,1,2,3,4,5,6,7,8,9}. Furthermore, we assume $$500\le j \le k$$, and $$p \ne q$$.
Guest

### Re: Optimal Integer Factorization and Quantum Theory of Numb

Guest wrote:Hey primework123, I like your previous post.

Suppose, we are given a positive integer, n, which is the product of two large primes, p and q, how do you factor n?

In base-10, we can let p = $$\sum_{i=0}^{j}a_{i}*10^{i}$$, and q = $$\sum_{i=0}^{k}b_{i}*10^{i}$$ where

$$a_{i}, b_{i}\in Z_{10 }$$={0,1,2,3,4,5,6,7,8,9}. Furthermore, we assume $$500\le j \le k$$, and $$p \ne q$$.
Guest

### Re: Optimal Integer Factorization and Quantum Theory of Numb

Guest wrote:
Guest wrote:Hey primework123, I like your previous post.

Suppose, we are given a positive integer, n, which is the product of two large primes, p and q, how do you factor n?

In base-10, we can let p = $$\sum_{i=0}^{j}a_{i}*10^{i}$$, and q = $$\sum_{i=0}^{k}b_{i}*10^{i}$$ where

$$a_{i}, b_{i}\in Z_{10 }$$={0,1,2,3,4,5,6,7,8,9}. Furthermore, we assume $$500\le j \le k$$, and $$p \ne q$$.

So, n = $$\sum_{i=0}^{l}r_{i}*10^{i}$$ where $$r_{0} \in {1, 3, 7, 9}.$$
Guest

### Re: Optimal Integer Factorization and Quantum Theory of Numb

Guest wrote:
Guest wrote:
Guest wrote:Hey primework123, I like your previous post.

Suppose, we are given a positive integer, n, which is the product of two large primes, p and q, how do you factor n?

In base-10, we can let p = $$\sum_{i=0}^{j}a_{i}*10^{i}$$, and q = $$\sum_{i=0}^{k}b_{i}*10^{i}$$ where

$$a_{i}, b_{i}\in Z_{10 }$$={0,1,2,3,4,5,6,7,8,9}. Furthermore, we assume $$500\le j \le k$$, and $$p \ne q$$.

So, n = $$\sum_{i=0}^{l}r_{i}*10^{i}$$ where $$r_{0} \in$$ {1, 3, 7, 9}.

primework123,

So, given n as a product two large, unknown, and distinct prime numbers, p and q, we need to solve a system of equations to find

$$a_{i}$$ and $$b_{i}$$ for all i, $$0 \le i \le j$$ and $$0 \le i \le k$$, respectively, in order to construct the unknown prime factors, p and q.

But, how do optimization and handshaking helps us to solve the system of equations efficiently (in polynomial time or less)?
Guest

### Re: Optimal Integer Factorization and Quantum Theory of Numb

Hey primework123, I like your previous post.

Suppose, we are given a positive integer, n, which is the product of two large primes, p and q, how do you factor n?

In base-10, we can let p = $$\sum_{i=0}^{j}a_{i}*10^{i}$$, and q = $$\sum_{i=0}^{k}b_{i}*10^{i}$$ where

$$a_{i}, b_{i}\in Z_{10 }$$={0,1,2,3,4,5,6,7,8,9}. Furthermore, we assume $$500\le j \le k$$, and $$p \ne q$$.[/quote][/quote]

So, n = $$\sum_{i=0}^{l}r_{i}*10^{i}$$ where $$r_{0} \in$$ {1, 3, 7, 9}.[/quote]

primework123,

So, given n as a product two large, unknown, and distinct prime numbers, p and q, we need to solve a system of equations to find

$$a_{i}$$ and $$b_{i}$$ for all i, $$0 \le i \le j$$ and $$0 \le i \le k$$, respectively, in order to construct the unknown prime factors, p and q.

But, how do optimization and handshaking helps us to solve the system of equations efficiently (in polynomial time or less)?
_________________________________________
Hi,

Yes, you're on the right path to a solution which was discovered back in the fall of 2010. Keep up the good work, and you'll sure to find the proper solution. I have given enough hints...

Best wishes,

David Cole
(aka primework123)
Keep the faith (effort and hope) and keep an open mind.

http://biblia.com/verseoftheday/image/Ro8.28

primework123

Posts: 22
Joined: Sat Oct 31, 2015 2:07 pm
Reputation: 0

### Re: Optimal Integer Factorization and Quantum Theory of Numb

Howdy,

If n =41989, and we know it to be a product two distinct primes, p and q, what are they?

Furthermore, n = $$\sum_{i=0}^{4 }*r_{i}*10^i$$
Guest

### Re: Optimal Integer Factorization and Quantum Theory of Numb

Guest wrote:Howdy,

If n =41989, and we know it to be a product two distinct primes, p and q, what are they?

Furthermore, n = $$\sum_{i=0}^{4 }r_{i}*10^i$$
, then

$$r_{0}$$ = 9, $$r_{1}$$ = 8, $$r_{2}$$ = 9, $$r_{3}$$ = 1, and $$r_{4}$$ = 4.

We assume, p = $$\sum_{i=0}^{2}a_{i}*10^i$$ and q = $$\sum_{i=0}^{2 }b_{i}*10^i$$ since the square root of n is roughly a three-digit number.
Guest

### Re: Optimal Integer Factorization and Quantum Theory of Numb

Guest wrote:
Guest wrote:Howdy,

If n =41989, and we know it to be a product two distinct primes, p and q, what are they?

Furthermore, n = $$\sum_{i=0}^{4 }r_{i}*10^i$$,

then $$r_{0}$$ = 9, $$r_{1}$$ = 8, $$r_{2}$$ = 9, $$r_{3}$$ = 1, and $$r_{4}$$ = 4.

We assume, p = $$\sum_{i=0}^{2}a_{i}*10^i$$ and q = $$\sum_{i=0}^{2 }b_{i}*10^i$$ since the square root of n is roughly a three-digit number.

So we have:
$$a_{2 }a_{1 }a_{0 }$$

x $$b_{2 }b_{1 }b_{0 }$$
______________________________

We must generate the right system of equations to solve the problem... good luck
Guest

### Re: Optimal Integer Factorization and Quantum Theory of Numb

Guest wrote:
Guest wrote:
Guest wrote:Howdy,

If n =41989, and we know it to be a product two distinct primes, p and q, what are they?

Furthermore, n = $$\sum_{i=0}^{4 }r_{i}*10^i$$,

then $$r_{0}$$ = 9, $$r_{1}$$ = 8, $$r_{2}$$ = 9, $$r_{3}$$ = 1, and $$r_{4}$$ = 4.

We assume, p = $$\sum_{i=0}^{2}a_{i}*10^i$$ and q = $$\sum_{i=0}^{2 }b_{i}*10^i$$ since the square root of n is roughly a three-digit number.

So we have:
$$a_{2 }a_{1 }a_{0 }$$

x $$b_{2 }b_{1 }b_{0 }$$
______________________________

We must generate the right system of equations to solve the problem... good luck
Guest

### Re: Optimal Integer Factorization and Quantum Theory of Numb

$$r_{10 }$$ = $$a_{0 }*b_{0}$$ (mod 10); $$c_{11}$$ = ($$a_{0}b_{0} - r_{10}$$)/10;
$$r_{11 }$$ = $$c_{11} + a_{1 }*b_{0}$$ (mod 10); $$c_{12}$$ = ($$c_{11} + a_{1 }*b_{0} - r_{11}$$)/10;
$$r_{12 }$$ = $$c_{12} + a_{2 }*b_{0}$$ (mod 10); $$c_{1}$$ = ($$c_{12} + a_{2 }*b_{0} - r_{12}$$)/10;

So, $$a_{2 }a_{1 }a_{0 }$$

X $$b_{0 }$$
__________________________________________
= $$c_{1}r_{12 }r_{11 }r_{10 }$$ (Stage 1)

$$a_{2 }a_{1 }a_{0 }$$

X $$b_{1 }$$
__________________________________________
= $$c_{2}r_{22 }r_{21 }r_{20 }$$ 0 (Stage 2)

$$a_{2 }a_{1 }a_{0 }$$

X $$b_{2 }$$
__________________________________________
= $$c_{3}r_{32 }r_{31 }r_{30 }$$ 0 0 (Stage 3)

However, the 'devil' is not knowing the details... Please complete the multiplication process (modulus-10) for stages, two and three. Next, combine (add) all stages to complete the multiplication process (modulus-10). And correct any mistakes too.
Guest

### Re: Optimal Integer Factorization and Quantum Theory of Numb

C=1
X=(N-1)/2
If X is even c=c+1, X/2
If X is odd X=N-X
When X=1 , N mod C = (p+q)-1, after just use fermat
Guest

### Re: Optimal Integer Factorization and Quantum Theory of Numb

Guest wrote:C=1
X=(N-1)/2
If X is even c=c+1, X/2
If X is odd X=N-X
When X=1 , N mod C = (p+q)-1, after just use fermat

Nope, that's not even close to being fast. It's exponential as hell.
Guest

Guest

### Re: Optimal Integer Factorization and Quantum Theory of Numb

Guest wrote:C=1
X=(N-1)/2
If X is even c=c+1, X/2
If X is odd X=N-X
When X=1 , N mod C = (p+q)-1, after just use fermat

N=33. X=(33-1)/2 = 16, C =1.
X(2) = 16/2 =8, C = 2.
X(3) = 8/2 = 4, C = 3.
X(4) = 4/2 = 2, C = 4.
X(5) = 2/2 = 1, C = 5.

33 mod 5 = 3, while p+q-1 = 11+3-1 = 13.

The method is wrong.
Guest