Simple yet interesting.

[Trurl]

 

The problem is it is Sqrt[(x^3*pnp^3)/(x^3*pnp+x)-pnp

Where pnp is the constant SemiPrime. 2564855351 in this case

 

x is the value on the axis. Where y on the graph equals zero, x approaches the smaller SemiPrime.

I believe in your graphs y is variable.

 

I am working on testing the program. But if you take

 

p = 2564855351;
x = 3;
Monitor[While[x <= p, If[(Sqrt[p^3/(p*x^2 + x)] - p) < 0.5, Print[x];
   Break[];];
  x += 2;];

and display x when it breaks it should be within range of the smaller Prime factor.

I know testing sounds simple enough but my computers and software are aging. Software upgrades faster than I do. All my software is scattered across 5 computers. And I don’t think Win7 and XP systems are safe on the internet. But that is just my computer maintenance problems. Hope this helps.

[Trurl]

 

 

p = 2564855351;
x = 3;
Monitor[While[x <= p, If[(Sqrt[p^3/(p*x^2 + x)] - p) < 0.5, Print[x];
   Break[];];
  x += 2;];

 

Syntax error. Should read:

 

p = 2564855351;
x = 3;
Monitor[While[x <= p, If[x*(Sqrt[p^3/(p*x^2 + x)]) - p) < 0.5, Print[x];
   Break[];];
  x += 2;];

That is why it wasn’t working. It was y, the second Prime factor. But we would have no way of knowing when y was reached. Multiple y by x to get p. Subtract p and the y of graph equal to zero. That y intercept is where x approaches the value of the smaller Prime factor.

[Ghideon]

On 1/24/2024 at 9:06 PM, Trurl said:

and display x when it breaks it should be within range of the smaller Prime factor.

What does "within range" mean?

23 hours ago, Trurl said:

That y intercept is where x approaches the value of the smaller Prime factor.

What do you mean by "approaches"? 

[Trurl]

Well leaving out the x was a happy accident. It just shows again x times y = p and we are just using algebra.

4 hours ago, Ghideon said:

What does "within range" mean?

That is why the second part of the program used division. The first algorithm found where x value of the small factor approached zero value at y. I put the value of y at p minus x times y so it is easier to find.

I could have said x at factor and y equals pnp.

I said within range because there is some error with x. It is close but within error.

4 hours ago, Ghideon said:

What do you mean by "approaches”?

The equation is too complex to solve for x even though it has x as the only unknown. I say approaches because I am plugging in a test value of x to find y. That is why I graphed it.

I know it is not pretty math and is more of a hack. But using test values you know the area of the factors because the further away the y value on the graph is from zero the further away you are from the factor.

[Ghideon]

Your descriptions do not match program code you have posted. What program do you try to describe?

[Trurl]

p = 2564855351;
x = 3;
Monitor[While[x <= p, If[(Sqrt[p^3/(p*x^2 + x)] - p) < 0.5, Print[x];
   Break[];];
  x += 2;];
 If[x <= p, While[x <= p, If[Divisible[p, x], Print[x];
     Break[];];
    x += 2;];], x]

 

First function corrected:

p = 2564855351;
x = 3;
Monitor[While[x <= p, If[x*(Sqrt[p^3/(p*x^2 + x)]) - p) < 0.5, Print[x];
   Break[];];
  x += 2;

 

Same as all my recent descriptions.

 

 

 

 

 

 

 

[Ghideon]

Your code does not seem syntactically correct. 

12 hours ago, Trurl said:

p = 2564855351;
x = 3;
Monitor[While[x <= p, If[(Sqrt[p^3/(p*x^2 + x)] - p) < 0.5, Print[x];
   Break[];];
  x += 2;];
 If[x <= p, While[x <= p, If[Divisible[p, x], Print[x];
     Break[];];
    x += 2;];], x]

 

First function corrected:

p = 2564855351;
x = 3;
Monitor[While[x <= p, If[x*(Sqrt[p^3/(p*x^2 + x)]) - p) < 0.5, Print[x];
   Break[];];
  x += 2;

 

Same as all my recent descriptions.

 

[Trurl]

p = 2564855351;
x = 3;
Monitor[While[x <= p, If[(x*(Sqrt[p^3/(p*x^2 + x)]) - p) < 0.5, Print[x];
   Break[];];
  x += 2;

 

Missing parentheses, but the logic is there.

The language is Mathematica. I collaborated with another Mathematica coder. I’m not familiar with the “Monitor” class.

But this class is the representation of everything I have posted. If it is the syntax I will have to consult more Mathematica coders.

Attention all coders!!!

[Trurl]

This code works!!!!!!!

Working Mathematica Code:

 

Clear[x,p];



p=2564855351;
x=3;


Monitor[While[x<p, If[(x*(Sqrt[p^3/(p*x^2+x)])-p)<0.5, Print[x];
   Break[];];
  x+=2;];
 If [x <= p, While[x<p, If[Divisible[p,x],
     Print[x];
     Break[];];
    x +=2;];], x]
3
41227

 

[Trurl]

Ok, I haven’t been here in 3 months. I was walking around and read the writing on the wall. And it made me think I should be working on mechanics instead of such abstract ideas of factoring SemiPrimes.

 

 

But my work was not without reward. But of course graphs can appear deceiving. Often graphs show a limited view. (Like in the old days of CAD when you couldn’t revolve the view in 360 degrees. You had to change the angle of the view by typing in coordinates).

 

 

The question is, “Is this graph deceiving?”

 

 

 

 

 

Clear[x, pnp]

pnp = 2564855351

Show[
 
 
 
 
 Plot[(( (((pnp^2/x) + x^2)) / x) /pnp), {x, 0, 60000}]
 
 
 ]










Plot[((pnp - (Sqrt[(x^2 * pnp^4 + 2 *pnp*x^5) + x^8])/
     pnp^4 - (1 - (x^2/(2*pnp))) *(pnp^2/x^2))), {x, 40000, 60000}]










Plot[(1/((pnp / (((pnp^2/x) + x^2)) / x))), {x, 0, 60000}]

 

[KJW]

27 minutes ago, Trurl said:

I should be working on mechanics instead of such abstract ideas of factoring SemiPrimes.

But my work was not without reward.

As a result of this thread, I did some exploration of factoring semiprimes. I even came up with a solution of sorts, but this didn't seem to be any more efficient than brute force.

 

[Trurl]

You are right about the brute force. I don’t know how useful it is when the smaller factor is occurring close to zero anyway.

 

But what is given me a glimmer of hope is the first graph of my last post cross the x axis at around 41227. That’s where it appears to cross but computer generated graphics can be deceiving.

 

If anyone could show where y on the graph equal zero that is the x intercept of that fist graph of the last post, it would be well appreciated.

[Trurl]

On 4/20/2024 at 11:25 PM, KJW said:

As a result of this thread, I did some exploration of factoring semiprimes. I even came up with a solution of sorts, but this didn't seem to be any more efficient than brute force.

I also wanted to say I’m happy to hear this post encouraged you to research SemiPrimes. That is what it is all about. I was going to be an adult educator until the professor failed me. But that is another story.

 

I also hope that if you take the inverse of the dozens of equations that it will make the graph better for visual inspection.

 

I also would be interested in what you are working on. Maybe you will share in the forums someday.

 

BTW does anyone know what the picture of the writing on the wall is?

[Trurl]

Clear[x, pnp] pnp = 2564855351 Show[ Plot[(( (((pnp^2/x) + x^2)) / x) /pnp), {x, 0, 60000}]

 

Graph this equation. It is the inverse.

It should make it easier to find the SemiPrime. It will decrease number of trails, but on small numbers will still be more time than brute force. It will however give you a graphical understanding of the smaller SemiPrime. Remember 2 unknowns should be impossible.

[Trurl]

105951

105950

 

 

105909

105910

105911

 

 

 

 

Ok on an odometer when there are at least 3 matching numbers say 555 or two sets of repeating numbers, say 5500, then we will try to predict when each occur chronologically on the odometer.

 

 

Sure you could just fill in the numbers with the corresponding numbers. But imagine these numbers were related to Prime distribution. The pattern is harder to see with the odometer as it moves linearly.

 

 

 

 

Imagine you have a regular odometer. It goes zero through nine. So in the single-digit place of the odometer (the start of counting), you note that 3, 5, 7 are Prime. So now you go to the tens-digit. And note that 3+1 or 3+3 or 3+5 or 3+7 are eliminated as Prime. To get another Prime number you would have to add an even number to 3, 5, or 7.

 

Of course adding an even number doesn’t always result in a Prime. This is just a graphical representation of a sieve. But just as repeating numbers in the odometer are hard to predict linearly throughout the revolutions of the odometer, are we not doing the same thing when looking for patterns in Prime numbers?

 

I study Prime numbers because I like finding patterns. Patterns can confuse or look like they could be there, but patterns are what we see in math. I might have stretched the truth when I claimed RSA was in trouble, but that one-way function is why I started all this math in the first place. The odometer sieve would require a lot of calculation. Again it would only show what is going on. It is the same with my graphs. I can only estimate where the semi-Prime factor is on a graph. Graphing 128-bit numbers and analyzing the graph is a challenge, but necessary to prove as the curve on the graph approaches zero x approaches the smaller Prime factor.

 

That is where I believe if the graph holds true, for larger N’s, it will be superior to brute force.

 

But as the graph becomes larger and more difficult to evaluate so does choosing an N. I have read that if N becomes too large, the enciphering of the message becomes too cumbersome.

 

In a future post I will be sending a graph. But It is important for anyone reading this to share if they had any results with the graph. I have tried to make the graph more useful (less test values) by inverting the equation.