-1
$\begingroup$

Why is the Rabin mapping: $f_i(x_i)=x_i^2 \bmod N_i$ not a Permutation over $\mathbb{Z}_N^*$?

$\endgroup$
2
$\begingroup$

Why is the Rabin mapping: $f_i(x_i)=x_i^2 \bmod N_i$ not a Permutation over $\mathbb{Z}_N^*$?

Because it maps distinct inputs to the same output.

For example, $f_i(1) = f_i(N_i - 1)$, with both being 1

$\endgroup$
  • $\begingroup$ or, general case: $f_i(j) = f_i(N-j)$ $\endgroup$ – kelalaka Dec 10 '18 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.