4 ms·
> consider Z/3Z = {0, 1, 2}, all of whom have multiplicative inverses. 0 has no multiplicative inverse. 1 and 2 = -1 are their own inverse.
by tromp 4y ago
> consider Z/3Z = {0, 1, 2}, all of whom have multiplicative inverses.
0 has no multiplicative inverse. 1 and 2 = -1 are their own inverse.
- vitus 4y agoAccurate. I'm not sure why I wrote they all have inverses, or the point I was trying to make -- phi(3) = 2, so your provided equation already demonstrates that for all even e, n^e = n^0 = 1 mod m (which is obviously not a bijection for n > 2). It may have worked better with phi(7) = 6: it's not immediately obvious that f(n) = n^3 mod 7 is not a bijection for n in Z/7Z, but easy to demonstrate nonetheless: 2^3 = 1 mod 7 = 1^3. (That said, I think you hinted at the requirement of coprimality with phi when you pointed to "1 or 3 mod 4" instead of just saying "odd numbers" for 10.) I think my follow-on point re: bases with non-square powers mapping multiple elements to zero is still correct and not covered by your explanation, though.