This screenshot gives an equation that will solve for the 2 Semi-Prime factors that make up 2564855351. Yes I know they are already known. It is for proof of concept. Where y equals zero x is the smaller factor of the equation. We graph to reveal (test) this.

I don't understand the error perfectly, but at 41227, the graph should be zero. Mathematically you can plug and chug values and test, but I believe 41227 can be isolated rather fast. So the method is useful and is a step in the right direction for Prime factorization.

Can anyone guide me on how to draw this graph to match what the numbers say is the graph? The graph looks like a barcode and not a one to one graph. It is the same graph I used earlier with a factors of error. In that graph y equaled 0.38 where x was 41227.

If I did not explain this enough, tell me and I will explain further. Thanks.

Not sure what you mean by

The problem is that more than just the Prime factor come close. I would have to test from 50,000 to 41,000

eq1 = ((x*Sqrt[pnp^3/(pnp*x^2 + x)] + x^4/(pnp^2 + x)) - pnp)
eq2 = (Sqrt[((((x^2*pnp^4 + 2*pnp*x^5) + x^8)/pnp^4)*pnp^2/x^2)] - pnp)

Plot[{eq1, eq2}, {x, 500, 10000},
 PlotRange -> All,
 PlotLabels -> "Expressions"]

FindRoot[eq1 == eq2, {x, 5000}]
(* {x -> 5064.549946} *)

The curves intersect at ~5064.55

Please edit your response and add the correct equations in a form that can be copied and pasted into a notebook.

Why do you think

The should be equal around 41000.

In the domain over which you are plotting the expressions, they are different by 9 orders of magnitude.

ClearAll[x, pnp];
eq1 = ((x*Sqrt[pnp^3/(pnp*x^2 + x)] + x^4/(pnp^2 + x)) - pnp)
eq2 = pnp - (Sqrt[((x^2*pnp^4 + 2*pnp*x^5) + x^8)/pnp^4] - (1 - x^2/(2*pnp)))

pnp = 2564855352;
Plot[{eq1, eq2}, {x, 10000, 50000},
 PlotRange -> All,
 PlotLabels -> "Expressions"]

By looking at the two expressions it is clear that they can only be close in magnitude for very large values of x.

Plot[{eq1, eq2}, {x, 9000000, 10000000},
 PlotRange -> All,
 PlotLabels -> "Expressions"]

FindRoot[eq1 == eq2, {x, 10000000}, WorkingPrecision -> 10]
(* {x -> 9.583792130*10^6} *)

Which is not ~41000.

WL is case-sensitive, this does nothing and returns unevaluated. Should be Clear.

clear[x, pnp]

This is what I attempted to do. The problem is the graphs don't equal exactly. The should be equal around 41000.

Well, this is a mess. You're not using the correct symbols.

• Mathematica is case-sensitive
• << should probably be <
• the [Dash] should probalby be -

Please get the code to be at least syntactically correct before asking us to debug it.

g[N_] := (N^2/x + x^2)/N*x == 2564855351

I want the value of x when y=2564855351 (That is y is G[x] and y equals N of the SemiPrime.

I thought the equation would give an estimate of a factored SemiPrime.

But the numbers aren’t exact. There is error. I thought I could graph 2 graphs and the intersection would be a good estimate. But the smallest error in N=N cause the computer to not be able to give me the estimate of x, where y equals N.

I want a feature that will tell me the point 2 graphs intersect.

It seems simple enough, but I don’t know of any way to draw the graph without changing the range. But with numbers of hundred digits, Mathematica draws the graph in seconds, but finding one specific point in a million is difficult.

I guess it does draw the graph in real-time, but I want views in different ranges over the graph.

I could put y=0 in the domain, but with large numbers, the scale of the graph is off.

I don’t know any solution except maybe writing a loop and comparing the equations.

I was just hoping there is something simple I haven’t learned yet.