I have read multiple websites to see what would happen with RSA when $\phi(n)$ and $e$ are not relatively prime, but no one has discussed it.
Can someone explain how to decrypt $c$ if $\phi(n)$ and $e$ aren't relatively prime?
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This is the exact dupe : In RSA encryption, why does the public exponent (usually, 'e') have to be coprime with $\varphi(n)$? – kelalaka Aug 27 '20 at 08:52
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2Actually, the https://crypto.stackexchange.com/questions/81949/how-to-compute-m-value-from-rsa-if-phin-is-not-relative-prime-with-the-e/81966#81966 answer addresses the specific question Yolo asked – poncho Aug 27 '20 at 13:32
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This answer was moved here since it fits that other question well. – fgrieu Aug 27 '20 at 16:48
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@poncho The answer was moved, you could copy above comment there if you want, as the mods seemingly cannot. – Maarten Bodewes Aug 27 '20 at 18:46
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Regarding Can someone explain how to decrypt $c$ if $n$ is prime, and $\phi(n)$ and $e$ aren't relatively prime?: have you checked poncho's answer which has a large intersection with your case? – fgrieu Aug 27 '20 at 21:21
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When $n$ is prime than it is not RSA. It is Pohlig-Hellman Cipher. I've roll-back your edit. If you really consider $n$ is prime, then please read about Pohlig-Hellman Cipher. In RSA, we need at least two distinct primes... – kelalaka Aug 27 '20 at 21:47