|RSA| Is it normal for $\phi(n)$ to work as RSA modulus?

cn flag

So I was casually practicing RSA on paper for an exam, I did the whole process I wrote bellow, and when I tried the encryption and decryption I got distracted and instead of doing $m^e \mod n$

I did $m^e \mod {\phi(n)}$ and both the decryption and encryption worked. Is this normal?

Here are the numbers: $$ p = 11\\ q = 23\\ n = (p\cdot q) = (7 \cdot 23) = 253\\ \phi(n) = (p-1) \cdot (q-1) = 220\\ e = 7\\ d = 63 \\ $$ I got d using the Extended Euclidean Algorithm: $$ gcd(220, 7)\\ 220 = 7 * 31 + 3 \\ 7 = 3 * 2 + 1 \\ $$ $$ 1 = 7 + 3(-2)\\ 1 = 7 + (220 + 7(-31))(-2)\\ 1 = 7(63) + 220(-2)\\ $$

kelalaka avatar
in flag
Nope. Could could you show your work?
DannyNiu avatar
vu flag
I don't think it's normal. It's probably the numbers you used just happen to work, so can you show us the numbers you used?
frog avatar
cn flag
thanks for your response, I edited my post and added the numbers
ru flag

In general a pair of RSA decryption exponents calculated in this way for a modulus $N$ will also work for any modulus $M$ that satisfies $\lambda(M)|\phi(N)$ where $\lambda$ is the Carmichael function.

In your example $M=\phi(N)=220$ and $\lambda(M)=\mathrm{lcm}(\phi(4),\phi(5),\phi(11))=\mathrm{lcm}(2,4,10)=20$ does indeed divide $\phi(N)=220$. This set of circumstances was helped by the fact that 11 divides $\phi(23)$. In general if the RSA modulus $pq$ has $p|q-1$ then $\phi(p)|\phi(\phi(q))$ and this helps a great deal.

This sort of phenomenon is less likely to happen when $p-1$ and $q-1$ have large prime divisors. However, it should be possible to construct other examples by choosing $p$ and $q$ where $p-1$ and $q-1$ are not divisible by any large primes, but are divisible by all small prime powers.


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