The goal of this exercise is to demonstrate that RSA decryption works for any $x \in \mathbb{Z}_n$, where $n = pq$ is the product of two primes $p$ and $q$, and $\mathbb{Z}_n$ denotes the integers modulo $n$. Specifically, we want to show that $D_{(n,d)}(E_{(n,e)}(x)) = x^{ed} \mod n = x$ holds, including for cases where $x \notin \mathbb{Z}_n^*$, meaning when $x$ is not coprime to $n$.

Case 1: When $x \in \mathbb{Z}_n^*$

This case has already been proven in standard RSA theory, but I'll briefly summarize for context. When $x \in \mathbb{Z}_n^*$, $x$ is coprime to $n$, meaning $\gcd(x, n) = 1$. In this case, $ed \equiv 1 \mod \varphi(n)$, where $\varphi(n) = (p-1)(q-1)$ is Euler’s totient function. Since $ed \equiv 1 \mod \varphi(n)$, this implies:

$x^{ed} \equiv x \mod n$

Thus, for $x \in \mathbb{Z}_n^*$, RSA decryption works as expected.

Case 2: When $x \notin \mathbb{Z}_n^*$

Now we consider the case where $x \in \mathbb{Z}_n$ but $x \notin \mathbb{Z}_n^*$, meaning that $x$ shares a common factor with $n$, either $p$, $q$, or both. In this case, we need to use the Chinese Remainder Theorem (CRT) to prove that decryption still works.

By the CRT, we need to show that $x^{ed} \equiv x \mod p$ and $x^{ed} \equiv x \mod q$. We will handle these two cases separately.

Sub-case 1: $x \equiv 0 \mod p$

If $x \equiv 0 \mod p$, then clearly $x^{ed} \equiv 0^{ed} \equiv 0 \mod p$. Thus:

$x^{ed} \equiv x \mod p$

Sub-case 2: $x \equiv 0 \mod q$

Similarly, if $x \equiv 0 \mod q$, then $x^{ed} \equiv 0^{ed} \equiv 0 \mod q$. Thus:

$x^{ed} \equiv x \mod q$

Sub-case 3: $x \not\equiv 0 \mod p$ and $x \not\equiv 0 \mod q$

In this scenario, $x$ is non-zero modulo both $p$ and $q$, but it is not necessarily coprime to either. In this case, the exponents $ed \equiv 1 \mod (p-1)$ and $ed \equiv 1 \mod (q-1)$ still hold, which implies:

$x^{ed} \equiv x \mod p \quad \text{and} \quad x^{ed} \equiv x \mod q$

Conclusion by the Chinese Remainder Theorem

By the CRT, since we have $x^{ed} \equiv x \mod p$ and $x^{ed} \equiv x \mod q$, it follows that:

$x^{ed} \equiv x \mod n$

Thus, RSA decryption works for any $x \in \mathbb{Z}_n$, even when $x \notin \mathbb{Z}_n^*$.

Summary:

We have shown that for any $x \in \mathbb{Z}_n$, $D_{(n,d)}(E_{(n,e)}(x)) = x$. This includes the case where $x \notin \mathbb{Z}_n^*$, using the Chinese Remainder Theorem to prove the result modulo $p$ and $q$.

Would you like further clarification or details on any part of this?

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Here are some related questions to deepen understanding:

1. How does the Chinese Remainder Theorem work in general, and why is it useful in RSA cryptography?
2. What are the conditions under which RSA encryption might fail, and how can they be mitigated?
3. How does Euler's Theorem play a role in proving RSA decryption for $x \in \mathbb{Z}_n^*$?
4. What happens when $x = p$ or $x = q$ in RSA decryption?
5. How does the choice of $e$ and $d$ affect the security of RSA?

Tip: In RSA, selecting large primes $p$ and $q$ is crucial for security, as factoring $n = pq$ is computationally hard.