On coprime integers.

Question

Given coprime $a,b,c$ where $c\gg ab$ we know there are integers $0<d,e,f<abc$ such that $da^2\equiv a\bmod abc$, $eab\equiv ab\bmod abc$ and $fb^2\equiv b\bmod abc$ holds. Is $0<d=e=f<abc$ possible?

Answer 1

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 $\bmod c$.