Thread: Product of integers that are relatively prime to m

Originally Posted by I like Serena
Starting somewhere, what happened to $b_1$ and $b_{\phi(m)}$?
$b_1$ is always $1$ and $b_{\phi(m)}$ is always equal to $m-1$.

2. Originally Posted by evinda
$b_1$ is always $1$ and $b_{\phi(m)}$ is always equal to $m-1$.
Ah, okay. Perhaps it'd be useful to mention that.
In particular it means that whatever $m$ is, 1 and -1 will always be their own inverse, and they will from a pair that are additive inverses.
(Assuming $m>2$.)

Otherwise, it seems your proof is correct.

Originally Posted by I like Serena
Ah, okay. Perhaps it'd be useful to mention that.
Yes, I agree...

Originally Posted by I like Serena
In particular it means that whatever $m$ is, 1 and -1 will always be their own inverse, and they will from a pair that are additive inverses.
(Assuming $m>2$.)

Originally Posted by I like Serena

Otherwise, it seems your proof is correct.
Nice... Thank you!!!