Date: 11/08/2001 at 09:39:07 From: Chaim Lodish Subject: Modular arithmetic This is something my class has been working on, but we seem stuck. For any integer a, a^4 is congruent to 0 or 1 (mod 5). We get to the point where we let a = 2k and a = 2k+1 for both even and odd cases, but are unable to get further to say that 16k^4 is congruent to 0 or 1 mod 5. We were able to work with the odd number case, and found at least ten k's to be a factor of five. So it is just the even case that is getting us stuck. Many thanks, Chaim
Date: 11/08/2001 at 09:50:45 From: Doctor Paul Subject: Re: Modular arithmetic Why don't you look at these five possibilities: a = 0 mod 5 a = 1 mod 5 a = 2 mod 5 a = 3 mod 5 a = 4 mod 5 If a = 0 mod 5, a^4 = 0 mod 5 if a = 1 mod 5, a^4 = 1 mod 5 if a = 2 mod 5, a^4 = 16 = 1 mod 5 if a = 3 mod 5, a^4 = 81 = 1 mod 5 if a = 4 mod 5, a^4 = 256 = 1 mod 5 - Doctor Paul, The Math Forum http://mathforum.org/dr.math/
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