Is this an accurate proof that no perfect square is of the form $4k+3$? ($k$ an integer) 
A positive integer $n$ is a perfect square. Prove that it cannot be of the form $4k+3$, where $k$ is an integer.

I tried to prove this by proof by contradiction: if $n$ is a perfect square, then its square root, say $x$, is an integer. Suppose $n$ is of the form $4k+3$. Then $$x^2= 4k+3$$ which we can also write as $$x^2\equiv3 \mod 4$$ However, this congruence has no solutions. Therefore our initial assumption that $n$ is of the form $4k+3$ was false and $n$ cannot be of the form $4k+3$. 
I know there's something missing or wrong in this proof but I don't know what. Any help would be appreciated.
 A: For even $\sqrt n=2m$, $n=(2m)^2=4m^2\equiv\color{blue}0\mod 4$.
For odd $\sqrt n=2m+1$, $n=(2m+1)^2=4m^2+4m+1\equiv\color{blue}1\mod 4$.
So you never achieve congruency to $\color{blue}3$, as there are no other cases.
As a byproduct, this also establishes that $4k+2$ is never a perfect square.
A: $4k+3$ is an odd number.
A square of an even number is even.
So the only candidates to have a square of the form $4k +3$ are 
$4a +1$ and $4a+3$.
Now $(4a+1)(4a+1) = 16a^2+8a+1$ and that is of the "form" $4k+1$.
Also $(4a+3)(4a+3) = 16 a^2 + 24a + 9 = 16 a^2 + 24 a + 8 + 1$ is of the form $4k+1$.
QED.
A: Assume to the contrary that  there is integer $\ k$ such that $\ 4k+ 3$ is a perfect square
Let $\ x^2= 4k+3$
$\ 4k+3 = 2(2k+2)+1 \implies x^2$ is odd which follows $\ x$ is also odd
Then  
Let $\ x= 2a+1$ for some integer $\ a$
Now we can write that  \begin{align} &x^2  = 4k+3\\
\implies &(2a+1)^2 =4k+3\\   
\implies &4a^2+4a+1 =      4k+3 \\
\implies &4a^2+4a-4k =       2 \\
\implies &4(a^2+a+k) =        2 \\
\implies &2(a^2+a+k)=        1 \\  
\end{align}     which is impossible since (a^2+a+k) is integer qed
