In RSA Crypto System we choose p and q such that they are distinct primes. Calculate n=pq and phi=(p-1)(q-1). Then e is chosen for public key (n,e) s.t. gcd(e,phi)=1
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2$\begingroup$ What are your thoughts? What efforts have you made? We expect you to make a serious effort to solve your own question before asking here. For instance, have you tried trying a few examples? Have you tried walking through the proof of correctness for RSA and seeing if it still holds if $p=q$? If your question is so short that it fits in a one-line title and you find that you don't have anything else to say in the body of the question, that should often be a warning sign. $\endgroup$– D.W. ♦Commented Dec 18, 2015 at 18:34
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$\begingroup$ I did try to solve it! For a few cases, the public and the private key becomes the same, depending upon the choice of e. Rest I just think that it makes it easier to decrypt if the numbers are same. $\endgroup$– Nilaxi MakenCommented Dec 21, 2015 at 19:10
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2 Answers
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The security of RSA relies on the fact that the best known way to compute $\phi(n)$ is to prime factorize $n$. For $n=pq$, where $p$ and $q$ are large, distinct primes, this is very hard. If instead $n=p^2$, then one could quickly find $p$ by calculating a square root. Then one could calculate $\phi(p^2)=p^2-p$ and break the encryption completely.
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In addition to SBareS's answer, let me mention that the formula $\varphi(pq) = (p-1)(q-1)$ only works if $p \neq q$: $\varphi(p^2) = p(p-1)$. Therefore if $p = q$ then decryption wouldn't be the inverse of encryption (unless you use the correct formula for $\varphi(n)$.
answered Dec 18, 2015 at 22:18