Hello there.

I heard that protection against cracking opend keis, based on

impossibility of

breaking of bignumbers to its simple co-multiplicators.

Hm. ...beg your pardon for english mathematical lexic,

I sure know it in Ukrainian only :).

Well knowm algorythm :

..................

for (current=2;current<MAX_RANGE;current++)

{

simple=1;

for (ch=0;((ch<base_pnt) && simple);ch++)

{

simple=0!=(current%base[ch]);

}

if (simple)

{

base[base_pnt++]=current;

steps[current-last_simple]++;

last_simple=current;

}

.......................

theoretycally allow to find it but in a practice

requaire a lot of memory and a lot of time...

Even not "a lot of" but "fantastical".

Let's look for enother way...

For every positiv integer number (C)

exist at least two positiv integer numbers (A,B)

which for exist equivalence C=A*B;

Sure, if number C is simple one of such two numbers equal to C other to

"1". :)

But in other case (when C is not simple) co-multiplicators (A and B)

lies between 1 and C exceptionaly.

So.

Lets define that exist two integer positive co-multiplicators

of integer positiv number. And shall take that of such co-multiplicators

from all possible which difference between lower as

in any other possible case. (Example for 54 it's 9 and 6 , but not 18

and 3)

Let name just definition def. #1.

Again.

A number is simple when its co-multiplicators which allow to def #1 and

one of them equal to 1 and other one equal. to such number,

If it isn't then we could divide number per it's co-multiplicators,

and its co-mutiplicators for co-multiplicators again and again untill

we won't get a set of simple co-multiplicators.

Rest, is a simple :) but not so , to find a co-multiplicators defined

in def.#1

In simplest case, we could try to get square root from the examinating

number.

And if it was an integer number then it number is equalt

to both of co-multiplicators (A and B)

sqrt(C) is integer

A=B=sqrt(C)

C=A*B

In other case we could try to use such formula:

sqr(X)=sqr(a) + (X+a)(X-a).

Really let's calculate a square (D) of some integer (E)

neares but bigger as number (C),

C >= D

D = sqr(E) (1)

find a difference (between D and C), and try to find square root (F)

from it.

Then plus such square root (F) to number E which was a source for square

(D)

biger to examinating number (C),

D=sqr(F)+C

D=sqr(F)+(E+F)(E-F)

C=(E+F)(E-F)

A=(E+F)

B=(E-F)

C=AB

sqr(F)=D-E

F=sqrt(D-E)

Really, it works.

Let call number F "corrector".

But if number F is not integer, what's then ?

In such case:

A=(E+G)

B=(E+H)

C=(E+G)+(E+H)

D=sqr(E)

D>=C (but D neares to C)

D=I+C

Where G and H are "correctors".

How to find such correctors ?

Good question... :)

Finally I do not know, but ... have some thinks about it.

We could write number I by the same formule which we used for a C :

sqr(X)=sqr(a)+(X+a)(X-a)

after second apling of such formula we get somthin like:

sqr(X)=sqr(Sqr(b)+(c+b)(c+d))+(e+(Sqr(b)+(c+b))(c+d)(e-(Sqr(b)+(c+b)(c+d))

after convertion it into multy-member we got somthing longer as I could

write

in 80-characters line. :)

(And perhaps we have to

look for sqr(b) number equal

to integer in 4-th degree (not sure that translate this term true).)

But I hope that formula sould get difficalter and difficalter,

untill it would allow to unite it in expression and calculate

"correctrs".

I hope so, but not sure.

That's all.

Have you any thinks ?

Good bye there.

--Alternative.

P.S. I do not subscribed for this news group, as such for any other

Or call me in Kyiv (aka Kiev) 550-8285 (ask for Misha)

but I speak english worse as write. :)