new asymmetric crypto algorithm V2

[remy.au...@gmail.com]

Hello

I do not say that this proposal is mathematically safe and reliable
I propose a scheme where factoring the public integers
does not break the asymmetric crypto algorithm ,thank you for any advice


Alice

p1=24443
p2=24469
publicalice=p1*101
publicalice=publicalice+2+2626-p2*100
publicalice%p2=2
publicalice%p1=28

Alice constructs an integer such that p1,p2 >sqrt(publicalice)
and publicalice%p1 and publicalice%p2 very small


Bob
bob only uses integers smaller than sqrt(publicalice)
sqrt(publicalice)=156....
msg=123
k=106
publicbob=publicalice*k+msg


Alice uses these private keys p1 p2 to calculate k and Bob's msg
k=(publicbob%p1-publicbob%p2)/(publicalice%p1-publicalice%p2)
(publicbob%p1)-k*28=123
(publicbob%p2)-k*2=123


question
After how many tests or trials can Eve find two close prime numbers with small values


cdl remy

[remy.au...@gmail.com]

oops I mixed up my mind and forgot one step
pour info
publicbob/publicalice-2*3*5*7*11*13*17*19*23*29*31*37*41*43*47*53*59*12345 =52030467668663799592676352


*************start***********
Alice
publicalice=101*31+1
publicalice%29=1
publicalice%31=2
**********
Bob
sqrt(publicalice)=55....
k=4
msg=12
publicbob=publicalice*k+msg+2*3*5*7*11*13*17*19*23*29*31*37*41*43*47*53*59*123456798

***************
Alice
publicbob%31-publicbob%29=4
publicbob%31-4*2=msg=12
****end*********

I am not yet at the calculation method, I am rather on the attacks and the robustness of the crypto system
thank you for any contribution, idea, attack or opinion

cdl remy

[remy.au...@gmail.com]

following several requests in private uh not a lot
bon bref

https://www.cjoint.com/doc/21_07/KGplCMzESMJ_Cryptographie-Asymetrique.pdf

don't forget the author's note thanks for any help

cdl remy

[711 Spooky Mart]

403 Forbidden
nginx/1.16.0

--
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█░░░░░░░██░░█░░░░██░░███░░░░██░░███ spooky mart
██████░░██░░███░░██░░█████░░██░░███ always open
██████░░██░░███░░██░░█████░░██░░███ stay spooky
██████░░██░░█░░░░██░░░░█░░░░██░░░░█ https://bitmessage.org
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███████████████████████████████████

[Chris M. Thomasson]

On 7/15/2021 1:47 PM, 711 Spooky Mart wrote:
> On 7/15/21 6:30 AM, remy.au...@gmail.com wrote:
>>
>> following several requests in private uh not a lot
>> bon bref
>>
>> https://www.cjoint.com/doc/21_07/KGplCMzESMJ_Cryptographie-Asymetrique.pdf
>>
>> don't forget the author's note thanks for any help

> 403 Forbidden
> nginx/1.16.0
>

I can see the pdf: https://i.ibb.co/bXkVm41/remy.png

[remy]

>
> 403 Forbidden
> nginx/1.16.0
>

ok http://remyaumeunier.chez-alice.fr/tmp/CryptographieAsymetrique.pdf

thank you for any feedback or even unfavorable opinion

cdl remy

--
http://remyaumeunier.chez-alice.fr/
toujours autant dyslexique

[711 Spooky Mart]

On 7/16/21 2:30 AM, remy wrote:
>
>>
>> 403 Forbidden
>> nginx/1.16.0
>>
>
> ok http://remyaumeunier.chez-alice.fr/tmp/CryptographieAsymetrique.pdf
>
> thank you for any feedback or even unfavorable opinion
>
> cdl remy
>

The paper on second link was accessible. Something about the SSL cert on
the first site was making my browser refuse to proceed.

I also cannot see how knowing the primes would recover the private
numbers. At first glance instinct says you are on to something here.

Have you tried mapping sequential and random inputs to orthogonal
squares or matrices with filters? That would be a good way to test for
low-entropy patterns. It would also be a good attack to render results
in your final paper.

I also see ways to possibly gussy it up.

If you combine modular exponentiation into it, and/or fraction the prime
banach space multiplications, it will likely be a formidable trap door.
Each of those additional operations increases a combinatorial search space.

For instance with x as a secret number:

if 2*3*5*7 is converted to (x/2)*(x/3)*(x/5)*(x/7) we have fractional
banach space. We can also have {x(1)...x(n)} with coded ordering to each
slot in the series for even more combinatorial complexity for the
attacker. Coded ordering of the fraction values would be a bit more
tricky in software implementations but certainly worth the resistance to
attack.

With exponentiation:

(2^x mod n)*(3^x mod n)*(5^x mod n)*(7^x mod n) we mix in the DLP
problem, which gets hidden by the banach space and the modular addition
of your initial scheme.

or exponentiation + banach factoring x/(2^y mod n)*x/(3^y mod n)*x/(5^y
mod n)*x/(7^y mod n)

Even with all that it should still be faster than asymmetric elliptic
curves, and manageable key size will have very high bit security. What
you have is worth exploring further.

I am off on a journey and when I get back I will take a crack at it and
let you know if I discover anything. I hope you keep cracking.

[remy]

>
> With exponentiation:
>
> (2^x mod n)*(3^x mod n)*(5^x mod n)*(7^x mod n) we mix in the DLP
> problem, which gets hidden by the banach space and the modular addition
> of your initial scheme.
>

don't you have a numerical example to fix the ideas,because ...

alice
...
publickekAlice


Bob
...
publickeyBob


alice
...=msg

[Daniel]

El 14/7/21 a les 15:56, remy.au...@gmail.com ha escrit:

> Hello
>
> I do not say that this proposal is mathematically safe and reliable
> I propose a scheme where factoring the public integers
> does not break the asymmetric crypto algorithm ,thank you for any advice
>
>
> Alice
>
> p1=24443
> p2=24469
> publicalice=p1*101
> publicalice=publicalice+2+2626-p2*100
> publicalice%p2=2
> publicalice%p1=28
>
> Alice constructs an integer such that p1,p2 >sqrt(publicalice)
> and publicalice%p1 and publicalice%p2 very small
>
>
> Bob
> bob only uses integers smaller than sqrt(publicalice)
> sqrt(publicalice)=156....
> msg=123
> k=106
> publicbob=publicalice*k+msg
>

Your method of presentation is cryptic. Anyway

k=floor(publicbob/publicalice), as msg<publicalice. Your method for
adding noise I see neither clear if it's useful, you're constrained to
very small primes.

>
> Alice uses these private keys p1 p2 to calculate k and Bob's msg
> k=(publicbob%p1-publicbob%p2)/(publicalice%p1-publicalice%p2)
> (publicbob%p1)-k*28=123
> (publicbob%p2)-k*2=123
>
>
> question
> After how many tests or trials can Eve find two close prime numbers with small values
>
>
> cdl remy
>

Cheers
Daniel

[Daniel]

El 14/7/21 a les 17:42, remy.au...@gmail.com ha escrit:

> oops I mixed up my mind and forgot one step
> pour info
> publicbob/publicalice-2*3*5*7*11*13*17*19*23*29*31*37*41*43*47*53*59*12345 =52030467668663799592676352
>
>
> *************start***********
> Alice
> publicalice=101*31+1
> publicalice%29=1
> publicalice%31=2

publicalice%31=1, not 2, but let set this fact aside.

> **********
> Bob
> sqrt(publicalice)=55....
> k=4
> msg=12
> publicbob=publicalice*k+msg+2*3*5*7*11*13*17*19*23*29*31*37*41*43*47*53*59*123456798
>
> ***************
> Alice
> publicbob%31-publicbob%29=4
> publicbob%31-4*2=msg=12
> ****end*********

so k=4 and 2 is publicalice%31. Can work. But the size of the msg is
limited to the size of the biggest prime in your noise. Bob doesn't know
this limit so it must add all small primes up to some value. publicbob
can grow exceedingly in size and the value of msg be still small.

>
> I am not yet at the calculation method, I am rather on the attacks and the robustness of the crypto system
> thank you for any contribution, idea, attack or opinion
>
> cdl remy
>

Cheers
Daniel

[remy]

it must be imagined that the prime numbers are large in the order of
+200 bits I could always define the application domain a little later
the main difficulty for me is not really there,

it is necessary to be able to generate noise but thank you for the remark

>>
>> I am not yet at the calculation method, I am rather on the attacks and
>> the robustness of the crypto system
>> thank you for any contribution, idea, attack or opinion
>>
>> cdl remy
>>
>
> Cheers
> Daniel

[remy.au...@gmail.com]

after this little weekend I think that the use of the factorial is not reasonable because
p<q pgcd(fact(sqrt(p*q).toInt()), p*q)= p

Sorry for the notations, but the objective is above all to be understood
so I suggest you come and tiptoe in a space reserved for the free and adventurous mind

I don't know how to do it but I would like

n1!=n2
(n1)mod(2,3,5,7,11)={......}
(n2)mod(2,3,5,7,11,....,pn)={......}

(11)mod(2,3,5,7,11,13,17)={1,2,1,4,0,11,11}
(31)mod(2,3,5,7,11,13,17,23,29,31,37}={1,1,1,3,9,5,14,8,2,0,31}
11 (operating mode to be defined ) 31= n2

(n2)mod((2,3,5,7,11,13,17,23,29,31,37}=( (1,2,1,4) ,(9,5,14,8,2,....)}

The idea is simple, I want to replace a part of the signature
of an integer by another one without having to use the factorial.

I don't know if it is possible ?

the only thing I know how to do for the moment is to impose a value
n-(n)mod(p)+val ,imposing 2 values is already a tinkering so

an idea or opinion ?

cdl remy

[remy]

Le 19/07/2021 à 10:04, remy.au...@gmail.com a écrit :
> after this little weekend I think that the use of the factorial is not reasonable because
> p<q pgcd(fact(sqrt(p*q).toInt()), p*q)= p

as it is useless to hope to calculate such a factorial or primorial
because the size of the integer is too big and I wish to be able to use
large prime numbers I exclude this solution to create some noise

[remy]

Le 16/07/2021 à 09:30, remy a écrit :
>
>>
>> 403 Forbidden
>> nginx/1.16.0
>>
>
> ok http://remyaumeunier.chez-alice.fr/tmp/CryptographieAsymetrique.pdf
>
> thank you for any feedback or even unfavorable opinion
>
> cdl remy
>

I might have a solution,I did not test it,a simple reflection with a
loud voice
publicalice=13
msg=2
k=5
n=(13*k)+msg;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;n%23;n%29;n%31;
1
2
0
2
10
0
14
8
19
7
3


n=(13+52)+msg;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;n%23;n%29;n%31;
1
2
0
2
10
0
14
8
19
7
3


so on a part of the set an addition behaves like a multiplication
so if bob makes

n=(13+528948954894894)+msg;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;n%23;n%29;n%31;

alice doesn't change anything in the algo, it calculates a k and then
deducts the msg, right?

[remy]

Le 16/07/2021 à 09:30, remy a écrit :

> ok http://remyaumeunier.chez-alice.fr/tmp/CryptographieAsymetrique.pdf




n=5;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;
1
2
0
5
5
5
5
5
n=23;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;
1
2
3
2
1
10
6
4
n=173;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;
1
2
3
5
8
4
3
2
n=2273;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;
1
2
3
5
7
11
12
12
n=452723;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;
1
2
3
5
7
11
13
10
n=6578843;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;
1
2
3
5
7
11
13
17

n=1+2*2+2*3*3+2*3*5*5+2*3*5*7*10+0+2*3*5*7*11*13*15+2*3*5*7*11*13*17*12;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;
1
2
3
5
7
11
13
17


n=6578843-5;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;
0
0
3
0
2
6
8
12
n=6578843-23;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;
0
0
0
3
6
1
7
13
n=6578843-173;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;
0
0
0
0
10
7
10
15
n=6578843-2273;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;
0
0
0
0
0
0
1
5
n=6578843-452723;n%2;n%3;n%5;n%7;n%11;n%13;n%17;n%19;
0
0
0
0
0
0
0
7


(6578843-452723)/(2*3*5*7*11*13*17)=12


first conclusion
all integers are composable in sum of primoriels

and that the primoriel is smaller than expected

pb how to calculate n %(31..101 ...)

[remy]

Le 27/07/2021 à 11:53, remy a écrit :
>  pb how to calculate n %(31..101 ...)

finally, it is as often much simpler to predict
I decompose n into a sum such that

n=x(2*3*5*7*11*13*17....p)+r; x<p

so.

n%2=r%2
n%3=r%3
n%5=r%5
n%7=r%7
n%11=r%11
n%13=r%13
n%17=r%17

...

n%p=r%p

for values greater than p and smaller than r

x*( (2*3*5*7*11*13*17....p)%(p+x))+r%(p+x)

attention the decomposition in sum and unique but
several different integers can have the same r

I think now I have everything I need to set up the crypto system
don't hesitate to walk past me.

cdl remy

[remy]

Le 27/07/2021 à 11:53, remy a écrit :

> Le 16/07/2021 à 09:30, remy a écrit :
>> ok http://remyaumeunier.chez-alice.fr/tmp/CryptographieAsymetrique.pdf
>

as often these much simpler than expected

I decompose n into a sum such that
n=x(2*3*5*7*11*13*17....p)+r; x<p

so.

n%2=r%2
n%3=r%3
n%5=r%5
n%7=r%7
n%11=r%11
n%13=r%13
n%17=r%17

...

n%p=r%p

for values greater than p and smaller than n

x*( (2*3*5*7*11*13*17....p)%(p+x))+r%(p+x)

attention the decomposition in sum and unique but
several different integers can have the same r

I think now I have everything I need to set up the crypto system
don't hesitate to walk past me.

cdl remy