$ax^2+ax+b$ has rational roots $\Rightarrow a$ square, for coprime $a,b\in\Bbb Z,\ a>0$

Question

This is something that came up as part of another problem that I was trying to solve.

My theory is that if a quadratic equation with integer coefficients has coefficients that are equal for the $x^2$ term and the $x$ term and has rational roots this will imply that the leading and second coefficient are square numbers, i.e.

If $a>0$ and $b$ are coprime integers and $ax^2+ax+b=0$ has rational roots, then $a$ is a perfect square.

I've tried looking at the discriminant and also subbing in $\frac{p}{q}$ and thought about taking mods but so far am a bit stuck.

Edit: as noted in comments, we must assume $a>0$, and $a$ and $b$ are coprime (otherwise we could multiply the quadratic by any non-square number to create counter examples).

Answer 1

Edit: I had missed that when $a$ is even, a slight change is needed. Thanks to commenters for pointing this out.

For $x$ to be rational, the discriminant $a^2-4ab$ must be a square number; say it's $c^2$. Adding $(2b)^2$ to both sides, $$(a-2b)^2=(2b)^2+c^2$$

which is a Pythagorean triple. Per the question $a$ and $b$ are coprime, and if $a$ is odd, the triple is primitive. So we know (Euclid) that

$$a-2b=m^2+n^2;\quad 2b=2mn;\quad c=m^2-n^2$$

From the first two of these, $a=m^2+n^2+2mn=(m+n)^2$ so $a$ is indeed a square number, as conjectured.

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If $a$ is even, then $b$ must be odd ($a$ and $b$ are coprime). Say $a=2a'$, where $a'$ and $b$ are coprime. Now

$$(2a'-2b)^2=(2b)^2+c^2$$

so $c$ must be even; $c=2c'$ and

$$(a'-b)^2=b^2+c'^2$$

Recall $a'$ and $b$ are coprime; so this triple is primitive. In this case, since $b$ is odd, we must have

$$a'-b=m^2+n^2;\quad b=m^2-n^2;\quad c'=2mn$$

so $a'=2m^2$ and $a=4m^2=(2m)^2$, so again the conjecture holds.

Answer 2

${{\dfrac{\!\!\!-b}{\color{#c00}a}} =\overbrace{ x^2\!+\!x}^{x\ =\ \large \frac{p}q\!\!} = \dfrac{p^2\!+pq}{\color{#c00}{q^2}}}\:\!$ thus $\, \color{#c00}{a = q^2}\,$ by $\overbrace{\style{font-family:inherit;}{\text{both fractions reduced}}}^{\large \begin{align}\rm wlog \ \ &(q,\:\!p)\,=\,1,\ \rm so\\ (q,\,p^2+pq)\,=\,&(q,\:\!p^2)=1\\ \end{align}}\!,\, a>0.\ $ $\bf\small QED$

Answer 3

If $\,\color{#0a0}{2\nmid a}\,$ the square discriminant has $\color{#0a0}{\rm coprime}$ factors $\:\!a,\, a\!-\!\color{#0a0}4\:\!b\:\!$ thus both factors must be squares. Else $\,\color{#c00}{2\,|\,a,\ {\rm so}\,\ 4\,|\,a}\,$ by parity (cf. below) so same idea as above applies after cancelling $\:\!4^2$.

Answer 4

Another way. Suppose $a=t^2r$ where $r\gt1$ is square-free.The discriminant is $$a^2-4ab=a(a-4b)=\square$$ then $a-4b=s^2r$ so we have $$4b=r(t^2-s^2)$$ Since $b$ cannot divide any prime factor of $r$ because $(a,b)=1$ and $r$ is not square we get $r=2$. Then $$2b=(t+s)(t-s)$$ If $t$ and $s$ have same parity then $2b$ is multiple of $4$ and $b$ is even so $(a,b)\gt1$, contradiction. And if $t$ and $s$ have distinct parity then $2b$ is odd, contradiction. Consequently $r=1$ and $a$ is square.

Answer 5

Since equation $ax^2+ax+b=0$ has rational roots, then we can factor it as follow: $$a\left(x+\frac{r}{s}\right)\left(x+\frac{u}{v}\right)=0, \quad r,s,u,v\in\mathbb{Z},(r,s)=1$$ Without loss of generality, let $v=s$, and observing that (expand LHS and compare with the primal): $$\frac{u}{v}=\frac{s-r}{s}$$ $$ax^2+ax+a\cdot\frac{r(s-r)}{s^2}=0$$ Due to $$a\cdot\frac{r(s-r)}{s^2}=b,\quad a,b\in\mathbb{Z}, (a,b)=1$$ then $$(r(s-r), s^2)=1$$ which means $\frac{r(s-r)}{s^2}=\frac{b}a$ is an irreducible fraction.

Thus, $a=s^2>0$ is a square.

Answer 6

$\overbrace{a\!\neq\! n^2\Rightarrow a \!=\! j\:\! p^{\:\!\color{#0af}{2\:\!i+1}},\, \color{#0a0}{p\nmid j}}^{a\ \text{non $\square\:\!$ so has a prime $\:\!p\:\!$ to $\rm\color{#0af}{odd}$ power}}\!\!\!.\,$ Let $X\! =\! p^i x\,$ so $\!\overbrace{a \,x^2\,+\,a\,x + b \!=\! f(x)}^{\textstyle \color{#c00}pj X^2\! +\! \color{#c00}pj\:\!p^i\! X^{\vphantom{|^|}}\!\! +\! b \!=\! g(X)}\!.\,$ If $\,x_1\!\in\! \Bbb Q\,$ is a root of $\:\!f\:\!$ then $\,X_1\!\in\!\Bbb Q\,$ is a root of $\:\!g,\,\:\!$ contra $\,g^{\vphantom{|^|}}\,$ is $\,\underbrace{{\color{#0a0}{\rm reverse}\ \color{#c00}{\rm Eisenstein}}}_{\large \color{#0a0}{p\ \nmid\ j}\:\!;\,\ p\ \nmid\ b\ {\rm by}\ (a,b)=1}\:\!$ so irreducible.