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Given a key $e = 3$ (the public exponent for RSA) and a group of numbers, say $Z_n^*$ (the multiplicative group of $n = pq$, where $p$ and $q$ are primes as requested by RSA).

I can find the decryption key: $d = e^{-1}$

Is it correct to say than any member $a$ in $Z_n^*$ has a source such that: $source = a ^ d$?

Edit: suppose we have gcd(3,phi(n)) = 1 as requested by RSA

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  • $\begingroup$ Question not clear for me. Are you asking; does every ciphertext has a unique plaintext? $\endgroup$ – kelalaka Dec 18 '18 at 13:46
  • $\begingroup$ @kelalaka What i mean is given a member of some group in Zn, and a key e, if i am free to choose my plaintext p, then no matter what member x in Zn i am given i can always generate the source of x, right ? I just need to perform x ^ (e^-1) and i get the source ? $\endgroup$ – caffein Dec 18 '18 at 15:12
  • $\begingroup$ So that basically means than given a specific key i can cover all of Zn: i just take a member in Zn, create the decryption key, decrypt and get the source and therefore i can generate a source for any member in Zn given a specific key, right ? $\endgroup$ – caffein Dec 18 '18 at 15:13
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If you're asking whether the power map $x\mapsto x^a \pmod n$ is injective, it suffices to take $a$ with $1 < a < \phi(n)$ and $\gcd(a,\phi(n)) = 1$.

Edit: I couldn't complete the answer earlier. As mentioned in the comments $gcd(a,n)$ is enough, but for computational efficiency one might as well take $a \pmod n$ as the exponent. In the RSA setup, $n$ is square free thus the function is bijective.

To see this, note that after all $e,d$ have similar roles and if one of them, say $e$, satisfies $\gcd(e,n)=1,$ then there is an integer $f$ such that $ed+fn=1$ (by the fact that they are inverses) thus demonstrating $gcd(d,n)=1.$)

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    $\begingroup$ Actually, the condition $1 < a < \phi(n)$ is not actually required; it is injective iff $\gcd(a,\phi(n)) = 1$ $\endgroup$ – poncho Dec 18 '18 at 15:50
  • $\begingroup$ Further, if in addition to $\gcd(a,\phi(n)) = 1$ it holds that $n$ is square-free [that is the product of distinct prime(s)], then $x\mapsto x^a \bmod n$ is a bijection over $\Bbb Z_n$, in addition to being a bijection over $\Bbb Z_n^*$. $\endgroup$ – fgrieu Dec 18 '18 at 17:17

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