Trurl Posted November 24 Posted November 24 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. -2
Trurl Posted yesterday at 12:19 AM Author Posted yesterday at 12:19 AM 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.
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