If the coefficients of a quadratic equation $$ax^2+bx+c=0$$ are all odd numbers, show that the equation will not have rational solutions.

I am also not sure if I should consider $c$ as a coefficient of $x^0$, suppose if I take that $c$ is also odd,

then $$b^2-4ac $$ will be odd. But that $-b$(odd), in the quadratic formula will cancel out the oddness of $\sqrt{b^2-4ac}$, in case if it is a perfect square. If it's not a perfect square then the root is irrational.

If I take $c$ to be even, even then the same argument runs but we noticed that when we take $c$ odd, we get that when discriminant is perfect square, so that means the question asks for $c$ not to be an coefficient.

Final question: Is this right to take $c$ as one of the coefficients of the equation $ax^2+bx=c=0$?

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    $\begingroup$ $x^2+3x+2=0$ has rational solutions with $a$ and $b$ odd so you can't discount $c$... $\endgroup$ Jul 9, 2020 at 16:30
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    $\begingroup$ @MartinHansen, if we take $x^2+3x+4=0$, then it fails. wouldn't a general information be more useful $\endgroup$
    – Arjun
    Jul 9, 2020 at 16:34
  • $\begingroup$ I was just giving you the first line of what's now in mr_e_man's answer... $\endgroup$ Jul 9, 2020 at 16:50

5 Answers 5


If the quadratic has rational roots, it can be expressed in the form $$ ax^2+bx+c = (Ax+B)(Cx+D) $$ for integers A, B, C, and D. Expanding and matching, we see that $$ a=AC\qquad b=AD+BC\qquad c=BD $$ For $a$ to be odd, we require $A$ and $C$ to both be odd. Similarly, for $c$ to be odd, we require both $B$ and $D$ to be odd. However, if all of $A$, $B$, $C$, and $D$ are odd, then $AD+BC$ must be even, and thus $b$ must be even.

Thus, to have rational roots, all the coefficients cannot be odd at the same time.

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    $\begingroup$ This is so much simpler than the accepted answer. $\endgroup$ Jul 10, 2020 at 15:07

Let the quadratic be $f(x) = ax^2+bx+c$ where $a, b, c \equiv 1 \pmod{2}$. By the Rational Root Theorem, if $\frac{p}{q}$ is a root of the quadratic in its lowest terms, then $p | c$ and $q | a$. Since $a$ and $c$ are odd, then both $p$ and $q$ must be odd. Then, we have $$f(\frac{p}{q}) = a\cdot \frac{p^2}{q^2}+b\cdot \frac{p}{q}+c = \frac{ap^2+bpq+cq^2}{q^2}.$$

However, we have that $a, b, c, p,$ and $q$ are all odd, so then $ap^2+bpq+cq^2$ is also odd, which means we cannot have $f(\frac{p}{q}) = 0$ by contradiction. Therefore, the quadratic $f(x)$ cannot have any rational roots.

(Partly derived from AoPS Algebra 2 textbook)

  • $\begingroup$ You don’t need the rational root theorem; at most one of $p$ and $q$ is even. In either case, $ap^2+bpq+cq^2$ is odd; if they’re both odd, also $ap^2+bpq+q^2$ is odd. Of course this is an abridged form of the RRT. $\endgroup$
    – egreg
    Jul 10, 2020 at 8:52

Take $a=1,b=3,c=2$ to get the rational solutions $-2,-1$. So the statement is false unless $c$ is also required to be odd.

Now consider squares modulo $8$. Any odd number has the form $8n+1$, $8n+3$, $8n+5$, or $8n+7$ (these are abbreviated as $\equiv1,3,5,7\bmod8$). So an odd number squared is





And any odd number times $4$ is





Therefore, if $a,b,c$ are all odd, then $ac$ is also odd, and


so $b^2-4ac$ cannot be a square.


Let $a=2p+1, b=2q+1, c=2r+1$, where $p,q,r$ are some integers

Then $b^2-4ac=(2q+1)^2-4(2p+1)(2r+1)$



where $k=q^2+q-4pr-2p-2r$, an even integer

So $b^2-4ac$ is an odd number.So if it is square of some integer , then that integer is odd.

Let $4k-3=(2m+1)^2=4m^2+4m+1$

$\Rightarrow 4(k-m^2-m)=4$


$k=m(m+1)+1$ an odd integer, a contradiction

  • $\begingroup$ You can shorten this. You prove that $b^2-4ac=4k-3$ where $k$ is even, and deduce that $b^2-4ac$ is odd. So $b^2-4ac=5\mod 8$ (or $\ne 1\mod 8$) so it is not an odd square. $\endgroup$
    – Rosie F
    Aug 5, 2020 at 6:31

Every odd square is 1, modulo 8.

The square root of an integer $n$ is either an integer (times $i$, if $n<0$) or irrational.

$a, b$ and $c$ are all odd. So, modulo 8, $b^2=1$, $4ac=4$, and $D=b^2-4ac=5$. Thus $D$ is not a square. But $D$ is an integer, so $\sqrt{D}$ is irrational, so the quadratic's roots are not rational.


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