For odd integers $a,b,c$ line $ax+by+c=0$ cannot intersect parabola $y=x^2$ in a rational point(where both abscissa and ordinate are rational numbers.)

We need to solve the equation of the line $ax+by+c=0$ with the parabola $y=x^2$.

Put $y=x^2$ in the equation $ax+by+c=0$,we get $bx^2+ax+c=0$,

Now we need to check whether the discriminant of the quadratic equation is a perfect square or not.

Discriminant $=a^2-4bc$

As $a,b,c$ are odd integers,let $a=2k_1+1,b=2k_2+1,c=2k_3+1$

Discriminant $=a^2-4bc=(2k_1+1)^2-4(2k_2+1)(2k_3+1)$

But I do not know whether $4k_1^2+4k_1-16k_2k_3-8k_2-8k_3-3$ is a perfect square or not

  • 1
    $\begingroup$ For there to be rational roots then there has to exist two factors of $bc$ which adds to give $a$, e.g. $14x^2 +31x + 15 = 14x^2 + 21x + 10x + 15 = 7x(2x + 3) + 5(2x + 3) = (7x + 5)(2x + 3)$. But if $b$ and $c$ are both odd, then $bc$ is also odd, which means that any two factors much be odd--but two odds add to give an even number. $\endgroup$ – Jared Jan 20 '16 at 2:48

Proof by contradiction:

Assume that $bx^2 + ax + c = 0$ has rational roots and $a$, $b$, and $c$ are odd. This means that there exists four integers, $k_1$, $k_2$, $k_3$, and $k_4$ such that:

\begin{align} bx^2 + ax + c =&\ (k_1x + k_2)(k_3x + k_4)\\ =&\ k_1k_3x^2 + (k_1k_4 + k_2k_3)x + k_2k_4 \end{align}

Because $b$ and $c$ must be odd, we can assume that $k_1$, $k_2$, $k_3$, and $k_4$ must all be odd (since $k_1k_3 = b$ must be odd and $k_2k_4 = c$ must be odd). If all of those are odd, then $k_1k_4$ is odd and $k_2k_3$ is odd and therefore $k_1k_4 + k_2k_3 = a$ must be even since an odd plus an odd results in an even number.

This is a contradiction to the assumption that $a$ is odd, and thus it cannot be the case that $a$, $b$, and $c$ are odd and you have a rational intersection.


Suppose that $a,b,c$ are odd, $bx^2+ax+c=0$ and $x$ is rational. Let $x=p/q$ where $p,q$ are integers with no common factor. Then $$bp^2+apq+cq^2=0\ .$$ Reading the equation modulo $2$, we have $a\equiv b\equiv c\equiv 1$; also $p^2\equiv p$ and $q^2\equiv q$; hence $$p+pq+q\equiv0\ .$$ Therefore $$(p+1)(q+1)\equiv1\ ;$$ this implies $$p+1\equiv q+1\equiv1$$ and so $$p\equiv q\equiv0\ .$$ That is, $p$ and $q$ are both even, which is a contradiction since we assumed they have no common factor.


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