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