# Simple yet interesting.

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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.

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

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.

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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"?

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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.

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

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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.

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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.

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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!!!

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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]
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