RSA: How Euler's Theorem is used? I'm trying to understand the working of RSA algorithm. I am getting confused in the decryption part.
I'm assuming 
$$n = pq$$
$$m = \phi(n) = (p - 1)(q - 1)$$
E is the encryption key $\gcd(\phi(n), E) = 1$
D is the decryption key, and 
$DE = 1 \mod \phi(n)$
$x$ is the plain text
Encryption works as ($y = x^E \mod n$) and decryption works as ($x = y^D \mod n$)
The explanation for why the decryption works is that since $DE = 1 + k\phi(n)$,
$$y^D = x^{ED} = x^{1 + k \phi(n)} = x(x^{\phi(n)})^k = x \mod n$$
The reason why last expression works is $x^{\phi(n)} =  1 \mod n$ ?
According to Eulers theorem this is true only if $x \text{ and }\phi(n)$ are coprimes. But $x$ is only restricted to be  $0 < x < n$  and $\phi(n) < n$. So $x$ should be chosen to be coprime with $\phi(n)$?
Help me clear out the confusion!
 A: Even if the plaintext $x$ is not pairwise coprime with $p$ or $q$, RSA still works as advertised. Here is why:
$p$ and $q$ are prime, so $x$ is a multiple of either $p$ or $q$, given the restriction that $x < pq$.
Assume that $x \equiv 0 \pmod p$. If it is congruent to $0$ mod $q$ the below still applies, just switch the name assigned to the two primes.
$x^k \equiv 0 \pmod p$ for all $k > 0$, i.e $x^k \equiv x \pmod p$.
$$
\begin{align*}
x^{1+ z \phi(n)} & \equiv x^{1+ z \phi(p) \phi(q) } \\
&\equiv x^1 \cdot x^{\phi(q) \phi(p) z} \\
&\equiv x \pmod q
\end{align*}
$$
Combining both equations with the Chinese Remainder Theorem yields $x$, the plaintext.
A: First, the statement should read $x^{\phi(n)}\equiv1\pmod{n}$, not modulo $\phi(n)$. You are right that this assumes that $x$ and $n$ are coprime. Given that $p,q$ are very large primes, the fraction of possible $x$ that is not coprime with $n$ is exceedingly small: $\frac1p+\frac1q-\frac1n$. In fact the security of the method is based on the smallness of this number: if there were any reasonable chance of finding a number $m$ not coprime with $n$ by picking a random number between $0$ and $n$, then one could compute $\gcd(m,n)\in\{p,q\}$ using it, and factor $n$. But formally, the test of coprimality of the plain text $x$ with $n$ should be done by the encoder, just like the test you already assumed that $x\neq0$. In the very unlikely event that coprimality fails, one must add some noise to the plain text.
A: Aside from it being unlikely that $\gcd(x,n) \neq 1$, note that the only possibilities are $\gcd(x,n) \in \{1,p,q,n\}$.  Therefore, if $x$ and $n$ are not coprime, the one can decipher the text anyways (since one then knows either $p$ or $q$ and can easily find the other factor, and hence $n$).  
In other words, if $x \in \mathbb{Z}/ n \mathbb{Z}$ is not a unit, we know that
$$
x^{ED} \equiv x \pmod{n} \Longleftrightarrow
\left\{\begin{array}{ccc}
x^{ED} \equiv x \pmod{p}
\\
x^{ED} \equiv x \pmod{q} 
\end{array} \right\}.
$$
Hope this helps.
