A new type of equation to meet several equation descriptions. Can you solve this using known techniques?

[Trurl]

y = (((pnp^2/ x ) + x^2) / pnp)

 

pnp = x * y

 

(((((pnp^2 / x) + x^2)) / x) / pnp) where y = 0

 

 

If all of these are true in the factors we wish to find, x and y, is there a limit; a range; that could be computed that said if x is this big then y is that big? It wouldn’t be a differential equation that solves a spring. But how do I find and x that is true by testing if y is also true in these 4 constraints?

 

It is a simple idea, but what is the math that completes it? I know that where y on the graph equals zero x has the value approximate to the smaller factor. I have an equation that will tell me y factor knowing x. If you move x larger y gets smaller. Move x smaller y gets larger. There is only a certain range that will prove this true. Combine that with all the other constraints you have and equation that solves a polynomial.

[Trurl]

Clear[x, y, g, pnp]

pnp = 2211282552952966643528108525502623092761208950247001539441374831\
912882294140\
   2001986512729726569746599085900330031400051170742204560859276357953\
757185954\
   2988389587092292384910067030341246205457845664136645406842143612930\
176940208\
   46391065875914794251435144458199;

x = 1.13056560621865239372901234269585839625544`15.653559774527023*^100

y = (((pnp^2/x) + x^2)/pnp)

g = x*y

N[y]



Out[75]= 1.130565606218652*10^100

Out[76]= 1.955908211597676*10^159

Out[77]= 2.211282552952967*10^259

Out[78]= 1.95591*10^159

 

 

This is my estimate of RSA-206. Not yet factored. Crunch away!

 

Trying to quit number crushing. Just wanted to see if this is significant.

 

 

 

[Trurl]

On 12/20/2024 at 7:19 PM, Trurl said:

Out[75]= 1.130565606218652*10^100

Out[76]= 1.955908211597676*10^159

Out[77]= 2.211282552952967*10^259

Out[78]= 1.95591*10^159

So Out [77] is the semiPrime. The RSA-260 number not factored.

Out [76] is the larger Prime factor.

Out [75] is the smaller Prime factor. It was found by finding what the x-value of the graph is while the y-value of the graph equals zero.

So we know that this x-value of the graph can be no larger than where the y-value equals zero. And we plug that x-value into the equation y = (((pnp^2/x) + x^2)/pnp) to find that y in the equation is smaller than the x value in the equation. So we switch x and y of the equation. To find Out [75].

So we know that the SemiPrime factor x is no less than Out [75]. And the larger we test for x of the equation, y of the equation must be smaller then Out [76].

So now we crunch numbers by division, increasing in incrementation and test those numbers from Out [75] until we find the smaller Prime factor, x.

[Trurl]

Has anyone looked at this.

I am saying the smaller Prime factor is between 1.0 * 10^90 and 1.0 * 10^130.

I know that is a large amount of numbers.

Does it make it more solvable? If you already knew the range how did you solve for it?

If you already knew that it was in this range it isn’t useful. But if it is useful it could be a way to find meaning in random polynomials.