Proof by Exhaustion - Is this sufficient?

[urza4315]

Assignment question!
"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?

[dudiobugtron]

You don't need to do all of that for even integers, you just have to show that 4k+3 is odd.

[Jplus]

To put it differently: yes, what you did is more than sufficient.

[DavCrav]

Alternatively, notice that the operation of taking remainders when dividing by 4 "commutes" with multiplication, in the sense that if I take the remainder of xy, it is equal to the remainder of x times the remainder of y (possibly after you take remainders again, as 3*3=9, which should be 1).

Consequently, the fact that 0^2=0, 1^2=1, 2^2=4-> 0 (take remainders) and 3^2=9->1, means any square is of the form 4k or 4k+1.

(This is an example of something called modular arithmetic.)

[PM 2Ring]

Now show that the sum of two perfect squares cannot be of the form 4k + 3.