crypto technique

Karl Lui Barrus klbarrus at owlnet.rice.edu
Wed Oct 20 17:42:39 PDT 1993


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Another possible problem with the technique is that the multiple
solutions are all valid.

For example, with two nestings and a = 11/2, b = 8, c = 25/2, d = 3, P = 16

I obtain g = 903 + 4675/4 x + 3025/8 x^2 mod 16
         g' = 7 + 3/4 x + 81/8 x^2 mod 16

where the g' is obtained from g by reducing the coefficients mod 16.

Solving the resulting equations yields two solutions:

a = 11/2, b = 8, c = 25/2, d = 3 (what I chose)

a = 31/2, b = 6, c = 17/2, d = 2

Plugging in the second solution:
 
         h = 359 + 6851/4 x + 16337/8 x^2 mod 16
         h' = 7 + 3/4 x + 81/8 x^2 mod 16

Notice that h' equals g'!

So the other solution can be used to form the same polynomial (which
we already saw doesn't encrypt uniquely).

Can this other solution be used for decryption as well?  I'd check but
I've REALLY got to go study now :-)

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