Number theory problem - contradiction

Question

In an algebraic proof (for my problem it doesn't matter which proof) I have a special setting:

$a,b,c \in \mathbb{Z}, \text{gcd}(a,c)=1,b<c \ \text{and} \ a \in \left\lbrace 1, \ldots , c\right\rbrace$

I tried to make a proof by contradiction and received the following equation $$ \frac{ab}{c}=d, d \in \mathbb{Z}.$$

Now I think this is unpossible because $\text{gcd}(a,c)=1$ and $b<c$ but I am not sure how I should argue exactly.

Answer 1

With your rather scarce data, you have a contradiction. If $\;p\;$ is a prime dividing $\;c\;$ , then:

$$\left(\frac{ab}c=d\iff ab=cd\right)\implies p\mid ab\;,\;\;p\nmid a\implies p\mid b $$

Taking the maximal power of each prime dividing $\;c\;$ we get the same as above, but then we'd have that all the powers of all the primes that divide $\;c\;$ also divide $\;b\;$ and then $\;b\ge c\;$ , contradiction.

Answer 2

By Euclid's Lemma $\, \gcd(c,a)=1,\, \ c\mid ab\,\Rightarrow\, c\mid b,\,$ which is a contradiction if $\,|b| < |c|.\,$ But you assume only that $\,b < c,\,$ so the contradiction fails if $\,b < 0,\,$ e.g. if $\, b = -c.\,$ But it works if $\,b\ge 0.$