If we have a point in a field $c$. Can we get another value $c^{'}$ such that $\left(c^{\prime}-c\right)$ is invertible in $Z_p$ ?


In $\mathbb{Z}_p$ all points $c\neq 0,$ are invertible.

Choosing any $c'\neq c$ will give you $c-c' \neq 0,$ thus it will be invertible.

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  • $\begingroup$ Are you sure? How you define invertible? $\endgroup$ – Heba Mohsen Jun 4 '19 at 7:58
  • $\begingroup$ It's a field, for all nonzero $u$ there is $u^{-1}$ such that $u^{-1}u=1\pmod p.$ $\endgroup$ – kodlu Jun 4 '19 at 9:04
  • $\begingroup$ Every element in the field is coprime with $p$ since it's prime. That means that every element is an unit and it has a multiplicative inverse in the group $Z{_p}^*$. If the modulus wasn't coprime then $c'-c$ could be a non unit depending on the result of the substraction. $\endgroup$ – kub0x Jun 4 '19 at 14:15
  • $\begingroup$ is this means that the result of (c'-c) should be invertible? $\endgroup$ – Heba Mohsen Jun 4 '19 at 17:33
  • $\begingroup$ Yes, that's exactly what it means since $c-c'$ is nonzero if and only if $c\neq c'$ in $Z_p$. $\endgroup$ – kodlu Jun 4 '19 at 21:43

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