# Let a, b and c be odd positive integers . Show that the quadratic equation 𝑎𝑥^2+𝑏𝑥+𝑐=0 has no rational solution. [duplicate]

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 $$gcd(p,q)=1$$)

• 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 $0$. – Sil Nov 19 '20 at 21:08

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.