(2) Prove that if a given integer n is a perfect square then either n=3q or n=3q+1 for some ineteger q.
n is a perfect square so can be written a^2, where a is an integer.
the conditions to show are that either:
a*a = 3q OR a*a = 3q + 1
this condition can be proven by showing that:
a*a is not equal to 2 (mod 3), or that
a*a =/= 3q + 2
where a and q are still integers.
by square rooting both sides,
a = sqrt(3q + 2)
we must now show that sqrt(3q + 2) is not an integer, for all values of q.
using induction, and testing the case for q=0,
sqrt 2 is not an integer.
now assume it is true for all q, and show it is true for q+1:
sqrt(3(q+1) + 2)) = sqrt(3q + 5)
now I will show that sqrt(3q+5) cannot be an integer...
*15 minutes, and a lot of messy weird work to do with roots of a quadratic, proof by contradiction and double negatives, later*
... well, that's that. I leave for someone else to continue in my messy footsteps.
(Q.E.D.) ..not quite