Question 479798
Set up 3 cases for a:
a = 3n
a = 3n+1
a = 3n+2
where n represents any integer
Case 1:
{{{a^2 = (3n)^2 = 9n^2}}}
Divide by 3
{{{9n^2/3 = 3n^2 +0}}}
Remainder is 0 for all n
Case 2:
{{{a^2 = (3n+1)^2 = 9n^2+6n+1}}}
Divide by 3
{{{(9n^2+6n+1)/3 = 3n^2 +2n +(1/3)}}}
Remainder is 1 for all n
Case 3:
{{{a^2 = (3n+2)^2 = 9n^2+12n+4}}}
Divide by 3
{{{(9n^2+12n+4)/3 = 3n^2 +4n +(4/3) = 3n^2 +4n +1 + (1/3)}}}
Remainder is 1 for all n
Therefore a^2 = 0 or 1 mod 3 for all a (integers)