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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.

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    $\begingroup$ $$c+a \equiv ab \pmod{p} \iff c \equiv a(b-1)\pmod{p}$$ $\endgroup$ – Daniel Fischer Apr 15 '14 at 15:02
  • $\begingroup$ @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 $\pmod{23}$.) ${}\qquad{}$ $\endgroup$ – Michael Hardy Apr 15 '14 at 15:41
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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.

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Hint: justify and/or prove the following

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

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