Re: JSH: Key in the lock, factoring technique

On Feb 9, 5:32 pm, JSH <jst...@xxxxxxxxx> wrote:
On Feb 9, 12:54 pm, marcus_b <marcus_bruck...@xxxxxxxxx> wrote:



On Feb 9, 1:31 pm, JSH <jst...@xxxxxxxxx> wrote:

So guess what? Big new thing is that I came up with this guessing
technique to factor, which is a key in the lock technique, where the
guesses have to be correct or you don't get all integers.

Trouble is, and why I'm still putting out the posts, the math people
are either quiet or still arguing with me, which in this case means
showing complete contempt for your knowledge of basic algebra:

The four equations are:

(f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = r_1 + k_1*p_1

(g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = r_2 + k_2*p_2

and

(f_1 + c_1*p_1) = (g_1 + d_1*p_2)

(f_2 + c_2*p_1) = (g_2 + d_2*p_2)

Note however that if you substitute the right
sides of equations 3 and 4 into equation 1, you
get exactly equation 2. The 4 equations are
not independent. You really have only 3 equations
in 4 unknowns.

Which means in general there are infinitely
many solutions. Of course you are only interested
in solutions which are all-integers. So how do you
find those solutions?

Algebra says that with 4 equations and 4 unknowns you can solve for
the variables unless some of your equations are equal to each other.


No, algebra does NOT say that. Consider the system of 3
equations:

x + y - z = 2
x + 2y - 3z = 4
2x + 3y - 4z = 6

Three unknowns, three different equations. No two equal.
None is a multiple of one of the others. But the system is linearly
dependent: the third equation is the sum of the first two equations.
The third equation does not add any new information to the
first two. Effectively you have two equations in 3
unknowns. There are infinitely many solutions (x, y, z).
In fact, in this case there are infinitely many INTEGER
solutions. Here are 3 of them:

x = -1, y = 4, z = 1
x = -2, y = 6, z = 2
x = -3, y = 8, z = 3

I bet you could find a few more.

Like

x + 2y = z

and

2x + 4y = 2z

so you can't eliminate a variable with those two as they are
equivalent.


As usual, your example is too simplistic and it misleads
you.

Degrees of freedom are about differing equations.

(f_1 + c_1*p_1) = (g_1 + d_1*p_2)

and

(f_2 + c_2*p_1) = (g_2 + d_2*p_2)

are conditionally true in that they are not always true for any values
you might select.

And with all integers you have a more selective condition still.

So, like if T = 119 and p_1 = 11, which has 7 and 17 as factors and
YOU pick f_1 = 1 mod 11, then f_2 = 9 mod 11, right?

But are there any integer values for all the f's, g's, c's and d's
such that those equations above can be true?

Yes. There are as that is equivalent to the trivial factorization.

Now then, what if you pick f_1 = 2 mod 11? Then f_2 = 10 mod 11.

Are there any integer solutions then?

That information MEANS SOMETHING and if the algebra can't distinguish
between it, then what good is it?

You're on a physics newsgroup.


Ooooh. Now I am worried.

If physics students don't know their degrees of freedom then they
don't belong in the field.


The implications are shocking.

So what are YOU doing here?

But crucially they may not know how desperate math people can be to
lie about even the most basic mathematics that has enough importance
that denial can destabilize the financial world.

Yes. It's that important physics people.


Ye have been warned!

We're talking about a result that can END entire nations if it's not
acknowledged. Even the US could go up in financial flames.


Financial flames! The worst kind!

These people are so far gone that they would destroy this world to
save their lies.

They'd let you die, and anyone else, but worse, they'd let you be
wrong, and believe you knew things that were not true.


We are evil incarnate. Professional liars.

They are anti-knowledge. A parasitic sub-species of humanity that
took over mathematics.


(Grunt, whine, scratch, faaaart!) Me not like Extreme Math!

Marcus.

James Harris

.

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