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Suppose we have to find if $aX^2+bY^2+cZ^2=0$ has any rational points or not, where $a,\ b,\ c$ are non-zero integers. Define $bc=\alpha^2s, \ ca=\beta^2t,\ ab = \gamma^2u$, where $s,\ t,\ u$ are square free integers. Then $A=\dfrac{a\alpha}{\beta\gamma},\ B=\dfrac{b\beta}{\gamma\alpha},\ C=\dfrac{c\gamma}{\alpha\beta}$ are square free, pairwise relatively prime integers and $aX^2+bY^2+cZ^2=0$ has a rational point iff $AX^2+BY^2+CZ^2=0$ has a rational point.

Now the second part is fairly easy to prove by simple substitution of values of $A,\ B,\ C$, but I'm having trouble in proving that $A,\ B,\ C$ are square free integers.

My Approach:

Dividing $bc$ and $ca$, we get: $$\dfrac{bt}{as}=\dfrac{\alpha^2}{\beta^2}\implies \dfrac{\alpha}{\beta}=\dfrac{\sqrt{bt}}{\sqrt{as}}$$

So we get $A=\dfrac{a\sqrt{bt}}{\gamma\sqrt{as}}=\dfrac{\sqrt{abt}}{\gamma\sqrt{s}}$

Now from $ab$, $\gamma=\dfrac{\sqrt{ab}}{\sqrt{u}}$.

Hence $$A=\dfrac{\sqrt{u}\sqrt{t}}{\sqrt{s}}$$

Now how do I prove that $A$ is even an integer, let alone a square free one, since $s,\ t,\ u$ are given to be square free.

Please help

THANK YOU

Devansh Kamra
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1 Answers1

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The problem states that

$$bc = \alpha^2s, \; \; ca = \beta^2t, \; \; ab = \gamma^2u \tag{1}\label{eq1A}$$

Multiplying all $3$ values together, then taking the square root & dividing to get $\sqrt{stu}$ by itself gives

$$\begin{equation}\begin{aligned} (abc)^2 & = (\alpha\beta\gamma)^2(stu) \\ \lvert abc \rvert & = \lvert \alpha\beta\gamma \rvert \sqrt{stu} \\ \left\lvert \frac{abc}{\alpha\beta\gamma} \right\rvert & = \sqrt{stu} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Since a square root of an integer being a rational number means that integer must be a perfect square, this means there's a positive integer $k$ such that

$$stu = k^2 \tag{3}\label{eq3A}$$

Each prime factor of $s$ must divide $k$, so the RHS has two factors of that prime factor. Since $s$ is square-free, this means those extra prime factors must be in $t$ or $u$. Thus, for some integers $d$, $e$, $f$ and $g$, we have

$$s = de, \; \; t = df, \; \; u = eg \tag{4}\label{eq4A}$$

Also, $s \mid k \; \to \; k = sm$ for some non-zero integer $m$. Using this and \eqref{eq4A} in \eqref{eq3A} gives

$$\begin{equation}\begin{aligned} (de)(df)(eg) & = (sm)^2 \\ d^2(e^2)(fg) & = (de)^2(m^2) \\ fg & = m^2 \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

Since $t$ and $u$ are square-free, then \eqref{eq4A} shows $f$ and $g$ are also square-free. Thus, their product being a square means they must be equal to each other, i.e.,

$$f = g \tag{6}\label{eq6A}$$

Using \eqref{eq4A} and \eqref{eq6A} in your result for $A$ (note we could also get this from \eqref{eq2A} by dividing the LHS by $\frac{bc}{\alpha^2}$ and the RHS by $s$), along with adding $\pm$ since $A$ depends on the signs of various values (though, if $\alpha$, $\beta$ and $\gamma$ are positive, then $\operatorname{sign}(A) = \operatorname{sign}(a)$), gives

$$\begin{equation}\begin{aligned} A & = \pm \sqrt{\frac{ut}{s}} \\ & = \pm \sqrt{\frac{(ef)(df)}{de}} \\ & = \pm \sqrt{f^2} \\ & = \pm \lvert f \rvert \end{aligned}\end{equation}\tag{7}\label{eq7A}$$

Similarly, we can show (I'll leave determining the details to you) that

$$B = \pm\lvert e \rvert, \; \; C = \pm\lvert d \rvert \tag{8}\label{eq8A}$$

This confirms $A$, $B$ and $C$ are all integers and, from \eqref{eq4A}, they're all square-free. Also from \eqref{eq4A}, since $s = de$ is square-free, this means $\gcd(d,e) = 1$, plus $t = df$ being square-free means $\gcd(d,f) = 1$. Similarly, $u = ef$ with $u$ being square-free means $\gcd(e,f) = 1$. This shows $A$, $B$ and $C$ are also pairwise relatively prime to each other.

John Omielan
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