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A lot of questions about this have unsatisfying answers that either argues how unsafe RSA is when $p=q$ or points out that $\phi(n) \neq (p-1)(q-1)$ for $p=q$.

However, I'd like to know why the RSA fails, i.e why the Decipher $R$ is not equal to the Message $M$ when $p=q$, regardless of safety concerns and assuming the public and private keys are just blindly computed in modulo $(p-1)(q-1)$.

Jean Lille
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    Isn't it enough to say that $\phi(n) \neq (p-1)(q-1)$? Try encoding and deciphering with $p=q=5$. – Ben Grossmann Dec 08 '15 at 15:30
  • Expanding on @Omnomnomnom's comment, $\phi(pq)=(p-1)(q-1)$ for $p\neq q$ (the RSA case), but $\phi(p^2)=p(p-1)$, which is different. Hence Fermat's Little Theorem will not help in the same way. – vadim123 Dec 08 '15 at 15:33

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