"Prove the following theorem using cases:
‘No perfect square is of the form 4k + 3, where k is an integer.’"
So this is how I've gone about solving it (simplified):
Suppose n is some even/odd integer. That is, n =2m/2m+1, so n^2 is equal to... etc.
For both cases, assume for a contradiction that n^2 is of the form 4k+3 where k is an integer.
Then n^2=4k+3=2m^2 if n is even, and
n^2=4k+3=4m^2+4m+1 if n is odd.
By rearranging to get the value of k, I show that k is not an integer thus violating my original assumption, thus no perfect squares are of the form 4k+3.
So my question is this: Is it sufficient to say that since k is not an integer, then we have achieved our contradiction? Is this sufficient to prove the theorem if I've achieved this contradiction for even and odd numbers?