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In the RSA problem, picking a message $x \in \mathbb{Z}_N \setminus \mathbb{Z}_N^*$ implies factorizing $N$. Since factorization with respect to the standard RSA generator is hard assuming the RSA problem is hard, it is likely that selecting $x \in \mathbb{Z}_N \setminus \mathbb{Z}_N^*$ is hard. Thus, one may ask the question:

Given $x$ uniformly distributed over $\mathbb{Z}_N$ what is the probability that $x \notin \mathbb{Z}_N^*?$

with the hope that it is negligible. So I write:

$P[x \notin \mathbb{Z}_N^*] = 1 - \frac{\phi(N)}{N} = \frac{1}{p}+\frac{1}{q} - \frac{1}{pq}$

However, how does one prove this is negligible?

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  • $\begingroup$ Hint: $p$ and $q$ both are large, including much larger than $4^k$ for $k$-bit security. Note: since you hypothesized that $N=p\,q$ with primes $p$ and $q$ such that $\phi(N)=(p-1)(q-1)$, you must have $p\ne q$; and then it is no longer necessary to restrict to $\Bbb Z_N^*$ for RSA; see Does RSA work for any message $m$. $\endgroup$ – fgrieu Feb 1 at 16:32
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    $\begingroup$ "one tries to avoid to pick a message $x \in \mathbb Z_N$" Whatever you meant to say here, it's probably wrong. $\endgroup$ – fkraiem Feb 1 at 16:45
  • $\begingroup$ @fkraiem Probably $x\in \mathbb{Z}_N\setminus \mathbb{Z}_N^*$. Though, yes you do not try to avoid it, you just need that it doesn't happen except with negligible probability. $\endgroup$ – Maeher Feb 1 at 16:56
  • $\begingroup$ @Maeher Well, clearly if you can produce such an $x$, you can factor $N$, so it should go without saying that it is infeasible. ;) $\endgroup$ – fkraiem Feb 1 at 17:02
  • $\begingroup$ Basically, "one tries to avoid" makes it sound like "you could do it if you wanted to, but it's better not to", while in fact you just can't. $\endgroup$ – fkraiem Feb 1 at 17:09
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With fgrieu's idea:

If we consider the generator working on a security parameter of $n$ bits (meaning resistance to $\mathcal O(2^n)$ computational effort), then each of the prime $p$ and $q$ must be at least (about) $n$-bit, otherwise trial division would factor the public modulus. Then, $N=p\,q$ with $p$ and $q$ distinct primes at least $2^n$ implies $P[x \notin \mathbb{Z}_N^*]=1-\frac{(p-1)(q-1)}{p\,q}<\frac1p+\frac1q<\frac2{2^n}$, that is $\mathcal O(2^{-n})$ which is negligible.

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