Originally Posted by HallsofIvy
If x is not divisible by 3 then it can be written as x= 3k+1 or 3k+2 for some integer k.
If y is not divisible by 3 then it can be written as y= 3j+1 or 3j+2 for some integer k.
That gives 4 cases to consider:
1) x= 3k+1 and y= 3j+1.
2) x= 3k+1 and y= 3j+2.
3) x= 3k+2 and y= 3j+1.
4) x= 3k+2 and y= 3j+2.
Calculate for each of those cases.
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Since +/- 0 = 0 are not cases 2 and 3 equivalent?
Since 2 = -1 and -1*-1 = 1 , (-1)^2 = (+1)^2 are not cases 1 and 4 equivalent?
Even better you can rewrite each term of an expression mod n before evaluating the expression mod n. Thus each of the terms can be rewriten mod 3 by substituting 0 for 3n and 1 for -1^2 (or 2^2) to get the equivalent expressions 1-1 = 0 mod 3 which is clearly correct.