SOLUTION: Prove that a squared = 0 or 1 mod 3 for all a∈ Integers

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Question 479798: Prove that a squared = 0 or 1 mod 3 for all a∈ Integers
Answer by ccs2011(207) About Me  (Show Source):
You can put this solution on YOUR website!
Set up 3 cases for a:
a = 3n
a = 3n+1
a = 3n+2
where n represents any integer
Case 1:
a%5E2+=+%283n%29%5E2+=+9n%5E2
Divide by 3
9n%5E2%2F3+=+3n%5E2+%2B0
Remainder is 0 for all n
Case 2:
a%5E2+=+%283n%2B1%29%5E2+=+9n%5E2%2B6n%2B1
Divide by 3
%289n%5E2%2B6n%2B1%29%2F3+=+3n%5E2+%2B2n+%2B%281%2F3%29
Remainder is 1 for all n
Case 3:
a%5E2+=+%283n%2B2%29%5E2+=+9n%5E2%2B12n%2B4
Divide by 3
%289n%5E2%2B12n%2B4%29%2F3+=+3n%5E2+%2B4n+%2B%284%2F3%29+=+3n%5E2+%2B4n+%2B1+%2B+%281%2F3%29
Remainder is 1 for all n
Therefore a^2 = 0 or 1 mod 3 for all a (integers)