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

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.

• $$c+a \equiv ab \pmod{p} \iff c \equiv a(b-1)\pmod{p}$$ – Daniel Fischer Apr 15 '14 at 15:02
• @NumberFour : I changed five instances of (\mathrm{mod}~p) to \pmod p. That is standard. (With more than one character included, you need braces, thus \pmod{23} yields $\pmod{23}$.) ${}\qquad{}$ – Michael Hardy Apr 15 '14 at 15:41

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