Is it secure if I disclose an element equals 1 modulo p in Zn?

Bob

11/15/23, 9:27 AM

Let $n = pq$, $p,q$ are two large primes, then $\mathbb{Z}_n^*\cong \mathbb{Z}_p^* \times \mathbb{Z}_q^*$. We disclose $n$ and keep $p, q$ secret. Is it secure if we disclose a random element $a$:

$a\in \mathbb{Z}_n^*$, $a = 1 \mod p$

That is, to disclose a random chosen element in $\langle 1\rangle \times \mathbb{Z}_q^*$ ? How to prove it?

0 + 5

finite-field

factoring

Daniel S

11/15/23, 9:29 AM

No, it's not secure. May I ask how the question arises?

Dattier

11/15/23, 9:34 AM

No, because $p=\gcd(n, a-1)$

Bob

11/15/23, 9:37 AM

Oh, gcd(kp ,n) = p. What about a choose from $<g>\times \mathbb{Z_q}^*$, where $<g>$ is a small subgroup of $\mathbb{Z}_p^*$?

Daniel S

11/15/23, 9:41 AM

Nope. If we know the subgroup order $s$ we just compute $\mathrm{GCD}(a^s-1,n)$ otherwise $\mathrm{GCD}(a^{k!}-1,n)$ is likely to capture $p$ for moderate sized $k$. see the $p-1$ factoring method.

Dattier

11/15/23, 9:52 AM

@Bob if $card(<g>) >\sqrt p$, you can disclose $g$.

Elon Musk

I sit in a Tesla and translated this thread with Ai:

EN: Is it secure if I disclose an element equals 1 modulo p in Zn?

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