A Cryptosystem Using Public Invertible Coefficients (CUPIC) was proposed during March, 2011. It was found to be defective.
http://groups.google.com/group/sci.crypt/browse_thread/thread/90505f68e63da7f1#
This CUPIC was abandoned and a retraction was published :
Cryptosystem Using Private Inverse Coefficient : CUPIC
by Alan C. Folmsbee
March 8, 2011 Rev. 95
Research this:
How to make public the square TT,
but keep inverse of T SECRET!
Choose a padding number T, to be squared to become TT
Choose secret primes j and k
public modulus n = jk
publish n
publish Z ≡ TT mod n
Secretly calculate coefficient " T inverse " = " 1/T "
private 1/T mod j or
private 1/T mod k
m is message
c is ciphertext
ENCRYPT
c≡sqrt(Zmm) mod n
DECRYPT
m ≡ (1/T mod k) * (sqrt(cc) mod n)
ABSTRACT without modulus
$$$$$$$$$$$$$$$
c=sqrt(TTmm)
cc=TTmm
cc/TT = mm
m=sqrt(cc/TT)
m = 1/T * sqrt(cc) ...INVERSE COEFFICIENT 1/T
$$$$$$$$$$$$$$$
maybe...
m = sqrt(cc) mod n * 1/T mod k
Use the fact: "if a is a residue (mod n) then a is a residue (mod p^k)
for every prime power dividing n. From
http://en.wikipedia.org/wiki/Quadratic_residue "
.......................................................
Example non-modular versions of cryptosystems
CUPIC:
m is message
T INVERSE is secret
TT IS PUBLIC
c=sqrt(TTmm)
cc=TTmm
cc/TT = mm
m=sqrt(cc/TT)
m=sqrt(cc) * 1/T (T INVERSE IS SECRET)
EXAMPLE
T=3 m=11
c=sqrt(9*121) = sqrt 1089 = 33
m=sqrt(33*33/9) = 1/3 * sqrt (1089) = 11
.....................................
rabin
c =mm PUBLIC COMPOSITE MODULUS pq
m=sqrt c SECRET PRIME MODULI : p and q
......................................
rsa
c=m^e PUBLIC EXPONENT : e
m=c^d SECRET EXPONENT : d
.................................
CUPIC
c=sqrt mmTT PUBLIC PADDING TT
m=(sqrt cc) * 1/T SECRET FACTOR : T INVERSE
.............................
TT PADDING IS PUBLIC, T INVERSE IS PRIVATE
CUPIC example with tiny numbers REV 95
padding number T=11, to be squared to become TT=121
Choose secret primes j and k = 5, 7
public modulus n = jk =35
publish n=35
publish TT mod n ≡ Z ≡ 16
Secretly calculate coefficient " T inverse " = " 1/T "
private 1/T mod j or
private 1/T mod k
11 mod 7 ≡ 4
inv(4) mod 7 ≡ 2
because 2 x 4 = 8 ≡ 1 mod 7
m is message = 10
c is ciphertext
c ≡ sqrt(Zmm) mod n
c ≡ sqrt 1600 mod 35 ≡ 5
maybe...
m ≡ sqrt(cc) mod n * 1/T mod k
m= 5 * 2 = 10
YES!
...
If you cannot factor n=jk then it is hard to take TT mod n and
calculate T inverse mod k. Collaborators welcome.
...................................................
RETRACTION ...
Abandoning this attempt.
Bryan is right, so I am abandoning this attempt at a public key scheme
because the goals are contradictory where:
Z ≡ TT mod jk (goal 1) where no inverse of Z exists mod jk
FOR NON-EXISTENCE OF INVERSE OF Z:
Z ≡ 0 mod jk goal 2
so both goals cannot be accomplished together.
End of attempt.
...................................................
This CUPIC is abandoned. I am retracting all of my statements I made about this.
I am sorry. I was wrong.
Alan Folmsbee , March 21, 2011
