Prove that if $a, b, c$ are positive odd integers, then $b^2 - 4ac$ cannot be a perfect square. Prove that if $a, b, c$ are positive odd integers, then $b^2 - 4ac$ cannot be a perfect square.

What I have done:

This has to either be done with contradiction or contraposition, I was thinking contradiction more likely.
If $b^2-4ac$ was a perfect square then the polynomial $ax^2+bx+c$  would have some rationals $\frac {p_1}{q_1}, \frac {p_2}{q_2}$ as roots($\frac{p_i}{q_i}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$). Therefore $(q_1x-p_1)(q_2x-p_2)=ax^2+bx+c$. 
So $q_1,q_2,p_1,p_2$ are odd integers (since $q_1q_2=a,p_1p_2=c$) and $q_1p_2+q_2p_1=-b\Rightarrow\Leftarrow.$
If we let $a$, $b$, and $c$ be $2x+1$, $2y+1$, and $2z+1$ respectively, We will get a general from of a multiple of $8$ minus $3$.
Since one portion of our from can have an $8$ factored out and the other portion be $4$ multiplied by the product of consecutive integers which is $4$ multiplied by an even number, giving us a multiple of $8$.
So our expression is $-3 \pmod 8$ or $5 \pmod 8$, but note that any number squared is $0, 1, \text { or } 4 \pmod 8$.
This is because any number is $0, 1, 2, 3, \dots \text { or } 7 \pmod 8$. Squaring all those values $\pmod 8$ will satisfy the statement.
And so we are done.