Check if true : Atleast one of the integers, a, b, c must be even

Question

Suppose a, b, c are integers such that the equation $ ax^2 + bx + c =. 0 $ has a rational root.
Check if true : Atleast one of the integers, a, b, c must be even. I know for rational roots $ b^2 - 4ac $ must be a perfect square.
So let $ b^2 - 4ac = p^2 $ for some $ p \epsilon Q $ .
How to proceed now ?

Answer 1

It is easy to see that $p$ is an integer. Note that

If $n$ is odd, then $n^2\equiv 1\pmod 8$.

Suppose $b$ is odd. Since $b^2-4ac$ is odd, $p$ must be odd. Then $4ac=b^2-p^2\equiv 0\pmod 8$. This implies $ac$ is even. Hence, one of $a,c$ is even.

Answer 2

We have $p \in \mathbb{Z}$, as if square of rational is integer, then the rational is an integer.

So $(b - p) (b + p) = 4ac$. If $a$ and $c$ are odd, then $(b - p) (b + p) \equiv 4 \pmod 8$.

It means if $b$ is odd then $p$ is odd. If $b \equiv p \pmod 4$, then $b - p \equiv 0 \pmod 4$ and $b + p$ is even, so $(b - p) \cdot (b + p) \equiv 0 \pmod 8$. Similarly if $b \not \equiv p \pmod 4$ but both are odd, then $b + p \equiv 0 \pmod 4$ and $b - p$ is even, so in this case $(b - p) (b + p) \equiv 0 \pmod 8$ too.

Answer 3

Suppose not. i.e. all coef's of $f(x)$ are odd. By the Rational Root Test if it has reduced root $\,r = e/d\,$ then $\,d\,$ divides the odd lead coef $c,\,$ so $d$ is odd, so $\bmod 2\!:\,\ r \equiv e/d\equiv e/1\,$ is well-defined.

Hence $\bmod 2\!:\,\ a,b,c\equiv1\,\Rightarrow\, f(x) \equiv x^2+x+1\,\Rightarrow\, f(0)\equiv 1\equiv f(1)$

Thus $f(x)$ takes only odd values. In particular $\,f(r)\neq 0,\,$ contradiction.

Remark $ $ This is the quadratic case of the following

Parity Root Test $\ $ A polynomial $\rm\:P(x)\:$ with integer coefficients has no integer roots
when its constant coefficient $\,\rm P(0)\,$ and coefficient sum $\,\rm P(1)\,$ are both odd.

Proof $\ $ The test verifies that $\rm\ P(0) \equiv P(1)\equiv 1\ \ (mod\ 2),\ $ i.e. that $\rm\:P(x)\:$ has no roots mod $2$, hence it has no integer roots. $\quad$

See here for further discussion.