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To prove this, I think the Δ should =$k^2$ so I let a=2p-1, b=2q-1, c=2r-1, where p, q, r are all positive integers, then I calculated $ b^2-4ac$ which is $-16 p r + 8 p + 4 q^2 - 4 q + 8 r - 3$ and find it hard to prove that $-16 p r + 8 p + 4 q^2 - 4 q + 8 r - 3 ≠ k^2$ so how to prove Δ ≠ $k^2$ and is it possible to use method of contradiction ( let a root $x_0$= p/q and $gcd(p,q)=1$)

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  • $\begingroup$ Super short version, $x=p/q$ with $(p,q)=1$ implies $ap^2+bpq+cq^2=0$, which implies $p\mid c$, $q\mid a$ which are odd, so $p,q$ are odd, and so left side of the equation is sum of three odd numbers, hence odd, hence not $0$. $\endgroup$ – Sil Nov 19 '20 at 21:08
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Step 1: We show that it either has zero or two rational solutions. Assume then contrary, that is $x_1 \in \mathbb{Q}, x_2 \in \mathbb{R}\setminus\mathbb{Q}$. Then $x_1x_2 \in \mathbb{R}\setminus\mathbb{Q}$, or $-ac\in \mathbb{R}\setminus\mathbb{Q}$. Contradiction.

Step 2: Assume that is has two rational solutions. So it can be written as: \begin{align*} (x - \frac{n_1}{m_1})(x - \frac{n_2}{m_2}) &= 0\\ (m_1x - n_1)(m_2x - n_2) &= 0 \\ m_1m_2x^2 - (n_1m_2 + n_2m_1)x + n_1n_2 &= 0 \end{align*}

Now I will claim that we are done. Note that we can pick $n_i, m_i$ such that $\gcd(n_i, m_i) = 1$. We need the coefficients of $x$ to have the same parity. If $m_1$ is even then $n_2$ must be even which gives $n_1m_2 + n_2m_1$ odd. The symmetric argument applies when $m_2$ is even. Finally, if both $m$ are odd then $n$ are odd but now $(n_1m_2 + n_2m_1)$ is even so we reach the required contradiction.

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