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# Encryption ( RSA)

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Suppose Fred sees your RSA signaure on m1 and m2, (i.e., he sees (md1 mod n) and (md2 mod n).
How does he compute the signature on each of mj1 mod n (for positive integer j), m&#8722;1
1 mod n,
m1 × m2 mod n, and in general mj1 × mk2 mod n (for arbitrary j and k)?

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#### Solution Preview

Multiplicative property of RSA
The RSA signature scheme (as well as the encryption method) has the following multiplicative property, sometimes referred to as the homomorphic Property.
If s1 =md1 mod n and s2 = md2 mod n are signatures on messages m1 and m2, respectively (or more properly on messages with redundancy added),
then s = s1s2 mod n has the property that s = (m1m2)d mod n.
If m = m1m2 has the proper redundancy (i.e., m MR),
then s will be a valid signature for it. Hence, it is important that the redundancy function R is not multiplicative, i.e., for essentially all pairs a, b M, R(a · b) = R(a)R(b). ...

#### Solution Summary

Encryption (RSA) is examined in this solution.

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