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I am wondering whether unique numbers in $\pmod m$ can have the same multiplicative inverse where $m$ isn't prime (I have proved that if $m$ is prime, this is impossible). I feel intuitively that this is impossible because it would somehow imply that these 2 numbers with the same inverse are congruent $\pmod m$ and not unique. However, I have been struggling to show this. The crux of the matter seems to be whether if I have $c \equiv c \pmod m$ where $c$ is any integer, and $a \not\equiv b \pmod m $, is $ca \not\equiv cb \pmod m$ necessarily the case? This relates to my question because $c$ could be the shared inverse, and $a$ and $b$ the unique numbers in modulo $m$. I would appreciate how to show this.

Princess Mia
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    The usual proof of unique inverses works in rings (with $1$), not just fields (or if these terms are unfamiliar, with any modulus, not just prime ones). In particular, suppose $a$ has two inverses $b$ and $c$, and try computing $bac$ in two ways, using associativity. – Theo Bendit Aug 30 '23 at 04:49
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    You're asking if $c$ is cancellable modulo $m$. By the dupe: $c$ is cancellable iff $c$ is invertible iff $c$ is coprime $m,$ see the proofs there. – Bill Dubuque Aug 30 '23 at 04:56

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The inverse of the inverse of $x$ is $x$, so you can't ever have different things with the same inverse. I suspect your question is not really about inverses, though: in the integers mod $m$, only those residues coprime to $m$ form a multiplicative group and have inverses. The others do not.

For example, in the integers mod $6$ we have $2 \not\equiv 4 \pmod 6$ but $3 \times 2 \equiv 3 \times 4 \equiv 0 \pmod 6$.

Robert Israel
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