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Given $a,b,c,d,e,f\in\mathbb{Z}$, I'd like to find all $x\in\mathbb{Q}$ such that $$ \sqrt{\frac{ax^2+bx+c}{dx^2+ex+f}}\in\mathbb{Q}. $$ How would one approach such a problem?

As an explicit example, what are the solutions in the case where $(a,b,c,d,e,f)=(1,2,0,1,1,1)$?

Dustin G. Mixon
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    Generally speaking, the rationality of the numerator and denominator can be treated separately, then those results can be brought together. Since $x\in \Bbb Q$, if there are such solutions individually for both quadratics then there will be combined solutions based on an LCM of sorts... – abiessu Jul 01 '21 at 13:38
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    @abiessu - The numerator and denominator need not be rational for their fraction to be rational, right? – Dustin G. Mixon Jul 01 '21 at 13:40
  • True, giving additional solution possibilities. – abiessu Jul 01 '21 at 13:41
  • Let $\sqrt{Q_1(x)/Q_2(x)}=q\in\mathbb{Q}$ so $\frac{Q_1(x)}{Q_2(x)}=q^2$ and hence you have a quadratic (or lower) $Q(x)=0$ with rational solution $x=\dots$ iff ... – user10354138 Jul 01 '21 at 14:14

3 Answers3

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This is by no means a complete answer, but just the straightforwrd attempt-

Let $$\frac{ax^2+bx+c}{dx^2+ex+f}=\frac{m^2}{n^2}$$ Then $$(an^2-dm^2)x^2+(bn^2-em^2)x+(cn^2-fm^2)=0$$ which implies $$x=\frac{-(bn^2-em^2)\pm\sqrt{(bn^2-em^2)^2-4(an^2-dm^2)(cn^2-fn^2)}}{2(an^2-dm^2)}$$ To make $x$ a rational number, we want the expression $$E=\{(bn^2-em^2)^2-4(an^2-dm^2)(cn^2-fn^2)\}$$ to be a perfect square.

In your example, $E$ reduces to $\{(2n^2-m^2)^2-4(n^2-m^2)m^2\}$. As we can see here, the solutions either have $m=\pm n$ or $m=0$ or $n=0$. So, chances of a non trivial solution seem to be quite thin.

But, I tweaked your explicit example a little to $(a,b,c,d,e,f)=(1,2,1,1,1,0)$. In this case however, e have a very nice solution. We can see that \begin{align*} E&=(2n^2-m^2)^2-4(n^2-m^2)m^2\\ &=\{n^2+(n^2-m^2)\}^2-4(n^2-m^2)n^2\\ &=\{n^2-(n^2-m^2)\}^2\\ &=m^4 \end{align*} So, for this particular choice of $(a,b,c,d,e,f)$, all rational values of $x$ will give a rational square root.

Sayan Dutta
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Here's a solution due to Boris Alexeev.

We want $x\in\mathbb{Q}$ such that $\sqrt{p(x)/q(x)}\in\mathbb{Q}$. Suppose the roots of $p$ and $q$ are all distinct, since otherwise the problem is easier. Multiplying by $q(x)\in\mathbb{Q}$, we equivalently want $\sqrt{g(x)}\in\mathbb{Q}$, where $g(x):=p(x)q(x)$. That is, we want $(x,y)\in\mathbb{Q}^2$ such that $g(x)=y^2$. Since $g$ has all distinct roots by assumption, this is equivalent to finding rational points on a certain elliptic curve (see this treatment, for example). The desired solutions can therefore be obtained with the help of Magma or Sage.

Dustin G. Mixon
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The given equation is equvalent to:

$\frac{ax^2+bx+c}{dx^2+ex+f}=\frac{m^2}{n^2}$

We take:

$(a,b,c)=(1,-10,2)$

$(d,e,f)=(1,-8,-5)$

Hence we have:

$x^2-10x+2=m^2 ----(1)$

$x^2-8x-5=n^2 ---(2)$

Solving equation (1) &(2) we get:

$m^2-n^2=(7-2x)$

$(m+n)(m-n)=(7-2x)*(1)$

We take:

$m+n=(7-2x)$

$m-n=(1)$

Hence, $(m,n)=((4-x),(3-x))$

Substituting (m,n) in equations (1) & (2) we get:

$x=-7$

Hence we get:

$(m,n)=(11,10)$

$Hence (m/n)=(11/10)$

Samuel
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