In what cases will RSA not work?

Question

I know that there are cases when RSA will not work like when the number to feed into the system is greater than the modulus. I was wondering if there were any other cases when RSA won't work

I looked all over and tried rephrasing the question in google many times

Edit: I am looking for what cases exist with a valid n, e and d used properly will not encrypt and decrypt correctly eg: message <= n

This contains some related information http://crypto.stackexchange.com/questions/1004/does-rsa-work-for-any-message-m — mikeazo, Jul 05 '15 at 22:10
@SEJPM 0 and 1 actually encrypt and decrypt just fine. Not securely, but yes correctly. — cpast, Jul 06 '15 at 00:02

score 4 · Accepted Answer · answered Jul 06 '15 at 06:09

It should be proven in any presentation of RSA that, with a correct public modulus $N$, public exponent $e$ and private exponent $d$, all integers $m \in \{0,1,\dots,N-1\}$ satisfy $$\left(m^e\bmod N\right)^d\bmod N = m.$$ So it is only possible for a number to "not encrypt or decrypt correctly" when it is not in $\{0,1,\dots,N-1\}$. Moreover, this necessary condition is obviously also sufficient, because the result of a $\bmod N$ operation will be in this set, so if $m$ is outside it, it can't possibly equal the result of the above computation.

Search the forum for 'unconcealed messages'. There are a minimum of 9 messages which result as the original message (quine). — Carl Knox, Nov 05 '17 at 18:53

In what cases will RSA not work?

1 Answers1