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.
 A: HINT
$$\text{We know that all odd squares are of the form $8k+1$ (Why?)} \tag{$\star$}$$
Use $(\star)$, to prove what you want. Move your mouse over the gray area for the complete solution.

 First note that $b^2-4ac$ is odd, if $a,b,c$ are all odd. Hence, if it is a square, it has to be a square of an odd number. Since $a,c$ are odd, we have $a=2M_a+1$ and $c= 2M_c+1$. Hence, we get $$4ac = 4(2M_a+1)(2M_b+1) = 16M_a M_b + 8(M_a + M_b) + 4 \equiv 4 \pmod8$$ Also, from $(\star)$, $b^2 \equiv 1 \pmod8$. Hence, $$b^2-4ac \equiv 5 \pmod8$$ contradicting $(\star)$.

A: Suppose $b^2-4ac$ is a square. Since $a,b,c$ are odd, $b^2-4ac$ must be an odd square. Hence $x=\frac{-b+\sqrt{b^2-4ac}}{2}$ is an integer. Since $x^2+bx+ac=0$, therefore $x^2+bx+ac\equiv 0\,(mod\,2)$. By FLT $x^2\equiv  x\,(mod\,2)$. Hence $(1+b)x+ac\equiv 0\,(mod\,2)$. Since $2|1+b$. Thus $0+ac\equiv 0\,(mod\,2)$ (contradiction)
A: $$a,b,c=\text{odd}\quad\iff\quad a=2A+1\quad;\quad b=2B+1\quad;\quad c=2C+1$$ $$\Delta=b^2-4ac=n^2\quad\iff\quad(2B+1)^2-4\,(2A+1)(2C+1)=n^2$$ $$(4B^2+4B+1)-4\,(4AC+2A+2C+1)=n^2$$ $$4\,\Big[(B^2+B)-(4AC+2A+2C+1)\Big]+1=n^2$$ $$\iff n=2k+1\iff n^2=4k^2+4k+1\iff$$ $$\underbrace{\underbrace{B(B+1)}_{even}-\underbrace{2\,(2AC+A+C)}_{even}-1}_{odd}=\underbrace{k(k+1)}_{even}$$ Contradiction !
A: 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 $q_1p_2+q_2p_1=-b\Rightarrow\Leftarrow.$ 
A: $\begin{align}
b^2 - 4ac &= n^2 \rightarrow \text{ ...$n$ must be odd} \\
(b - n)(b + n) &= 4k \rightarrow \text{ ...product of 2 even numbers} \\
b-n=2 \land b+n &=2k \rightarrow \text{ ...because $k$ is odd} \\
2n+2 &= 2k \rightarrow\\
 n &= k - 1 \text { ...contradicting $n$ and $k$ are odd.}
\end{align}$ 
A: 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.
