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# x - Faculty of Computer Science - Technische Universität Dresden

x - Faculty of Computer Science - Technische Universität Dresden

## Proof

Proof (1) c • d � 1 (mod �(n)) � �k� Z : c • d - 1 = k • �(n) � �k� Z : c • d = k • �(n) + 1 Therefore m c • d � m k • �(n) +1 (mod n) Using the Theorem of Fermat �m� Z n*: m �(n) � 1 (mod n) it follows for all m coprime to p m p-1 � 1 (mod p) Because p-1 is a factor of �(n), it holds m k • �(n) +1 �p m k • (p-1)(q-1) +1 �p m • (m p-1 k • (q-1) ) �p m 1 1 209

Proof (2) Holds, of course, for m � p 0. So we have it for all m � Z p. Same argumentation for q gives m k • �(n) +1 �q m Because congruence holds relating to p as well as q, according to the CRA, it holds relating to p • q = n. Therefore, for all m � Z n m c • d � m k • �(n) +1 � m (mod n) Attention: There is (until now ?) no proof RSA is easy to break � to factor is easy 210

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