0
$\begingroup$

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 $=a^2-4bc=(2k_1+1)^2-4(2k_2+1)(2k_3+1)$
$=4k_1^2+4k_1-16k_2k_3-8k_2-8k_3-3$

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

$\endgroup$
  • 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
2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.