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

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.
• It's a field, for all nonzero $u$ there is $u^{-1}$ such that $u^{-1}u=1\pmod p.$ – kodlu Jun 4 at 9:04
• 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. – kub0x Jun 4 at 14:15
• Yes, that's exactly what it means since $c-c'$ is nonzero if and only if $c\neq c'$ in $Z_p$. – kodlu Jun 4 at 21:43