fathom fjord
Hello, I’m a bit stumped on writing up a proof for question 13. Note that Exercise 4.6.12 is just question 12. Mainly, I’d like some help with improving the reasoning in my proof sketch:
I understand that since a is a unit mod n and -a is also a unit mod n then this implies that the set of all units mod n, say S, is equivalent to -S (or is a permutation of S). Furthermore, since S is the set of all units mod n, and -a is also a unit mod n, then S must also include all additive inverses -a mod n. The idea here is to now group each unit in the sum with their additive inverse such that the total sum equals 0. However, I’m struggling to articulate why I can group up each term with its additive inverse without one number being left alone.
It was suggested to me that I follow the proof of Wilson’s theorem and how pairing terms was handled there, but I can’t seem to understand how to tie it into this proof. For example, the only case in which a number would be on its own is if $a \equiv -a (\mod n) \implies 2a \equiv 0 (\mod n) \implies n \mid 2a \implies n \mid 2$. This implies that when n = 1 or n = 2, then there exists a number which has its own additive inverse.
Will this help me to prove why I am able to pair terms with their inverses ‘nicely’, or is there a better solution that I’m missing? Any help is appreciated!
tacit bramble
summer crag
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keen merlin
