If $c \not\equiv 0 \pmod p$ then $\forall a \not\equiv 0\ \exists b \not\equiv 1$ so $c+a\equiv ab \pmod p$

Question

Im looking for a correct argumentation of why the folowing holds, any help would be great:

For $p$ prime, if $c \not\equiv 0 \pmod p$ then $\forall a \not\equiv 0 \pmod p ~\exists b \not\equiv 1 \pmod p$ such that $c+a\equiv ab \pmod p$

I presume this could be shown easily, but Im looking at least for hint how to start.

Answer 1

Hint $\ {\rm mod}\ p\!:\ a\not\equiv 0\,$ so, by Bezout, $\,a^{-1}$ exists, so $\ ab\equiv a+c\overset{\times\ a^{-1}}\Rightarrow b\equiv 1 + a^{-1}c,\ $ hence $\,b\equiv 1\,$ implies $\ a^{-1}c\equiv 0\,\overset{\times\ a}\Rightarrow\ c\equiv 0\, $ contra hypothesis.

Remark $\ $ Alternatively, employ $\ n x \equiv k \ $ has a unique solution if $\,n\,$ is coprime to the modulus, i.e. use the existence and uniqueness of fractions with denominator coprime to the modulus.

Answer 2

Hint: justify and/or prove the following

$$c+a=ab\pmod p\implies b=a^{-1}(a+c)=1+a^{-1}c\pmod p$$