Given coprime $a,b,c$ where $c\gg ab$ we know there are integers $0
On coprime integers.
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modular-arithmetic
analytic-number-theory
1 Answers
0
Accepting the implicit assumption that $a,b>1$, not only are $d,e,f$ not all equal, no two of them are equal to each other.
Define $g\equiv a^{-1} \bmod c$ and $h\equiv b^{-1} \bmod c$
From the definitions we can divide through appropriately to find:
$da\equiv 1\bmod bc$
$e\equiv 1\bmod c$
$fb\equiv 1\bmod ac$
So
$d\equiv a^{-1}\bmod bc$
$f\equiv b^{-1}\bmod ac$
This also means that $d\equiv g \bmod c$ and $f \equiv h \bmod c$. We know from the coprime condition that these numbers are distinct and $\not\equiv 1$, so $d,e,f$ are all distinct values $\bmod c$.