If the coefficients of a quadratic equation are odd numbers, show that it cannot have rational roots

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 $$ax^2+bx=c=0$$?

• $x^2+3x+2=0$ has rational solutions with $a$ and $b$ odd so you can't discount $c$... Jul 9 '20 at 16:30
• @MartinHansen, if we take $x^2+3x+4=0$, then it fails. wouldn't a general information be more useful Jul 9 '20 at 16:34
• I was just giving you the first line of what's now in mr_e_man's answer... Jul 9 '20 at 16:50

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.

• This is so much simpler than the accepted answer. Jul 10 '20 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)

• 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. Jul 10 '20 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

$$1^2=1$$

$$3^2=9=8\cdot1+1\equiv1$$

$$5^2=25=8\cdot3+1\equiv1$$

$$7^2=49=8\cdot6+1\equiv1.$$

And any odd number times $$4$$ is

$$4\cdot1=4$$

$$4\cdot3=12=8\cdot1+4\equiv4$$

$$4\cdot5=20=8\cdot2+4\equiv4$$

$$4\cdot7=28=8\cdot3+4\equiv4.$$

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

$$b^2-4ac\equiv1-4=-3=8\cdot(-1)+5\equiv5\not\equiv1$$

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

$$=4q^2+1+4q-4(4pr+2p+2r+1)$$

$$=4k-3$$

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

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

$$k-m^2-m=1$$

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

• 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. Aug 5 '20 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.