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A somewhat more general proof involves the use of the Chinese remainder theorem and the fact that 4 is divisible by the order of the multiplicative group of bot
by contravariant 9y ago
A somewhat more general proof involves the use of the Chinese remainder theorem and the fact that 4 is divisible by the order of the multiplicative group of both integers mod 2 and mod 5. Therefore x^5 = x mod 2 and x^5 = x mod 5 for all x, and as a consequence x^5 = x mod 10 for all x.
This construction seems to fail if the base isn't square free. In particular in base 4 the number 2 has as powers 2, 10, 20, 100 etc.
- knappa 9y agoThat's not true for x=2 : 2^4 = 6 mod 10 Also, 4 does not divide the order of the multiplicative group of Z/2Z. i.e. 4 does not divide 1. (Also, the claim was that x^5 = x mod 10 not x^5=1, but I assume that's a typo.)
- contravariant 9y agoHmm yes I suppose I was a bit sloppy, although I got the gist of it right somehow. I guess the problem here is that x^4 = 1 mod 2 is false if x = 0 mod 2, but x^5 = x mod 2 does hold for all numbers. I've fixed the problems you mentioned, thanks.