Score:0 Crypto

# |RSA| Is it normal for $\phi(n)$ to work as RSA modulus? 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: $$gcd(220, 7)\\ 220 = 7 * 31 + 3 \\ 7 = 3 * 2 + 1 \\$$ $$1 = 7 + 3(-2)\\ 1 = 7 + (220 + 7(-31))(-2)\\ 1 = 7(63) + 220(-2)\\$$  Nope. Could could you show your work?  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?  thanks for your response, I edited my post and added the numbers
Score:1 Crypto 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.  