No rational has a square of the form $\frac 1t-t$ where $ t $ is a rational in $(0,1)$

Question

Let $ t\in (0,1)\cap \Bbb Q$.

Prove that there is no rational $ a $ satisfying $$a^2=\frac 1t -t$$

I tried to start from the quadratic $ t^2+a^2t-1=0$ with $ a^4+4 $ as discriminant but i couldn't find any idea. thanks in advance.

Answer 1

Assume $a$ is rational, i.e.,

$$a = \frac{p}{q}, \; p, q \in \mathbb{Z}, \; q \neq 0, \; \gcd(p,q) = 1 \tag{1}\label{eq1A}$$

With $t$ being rational, the discriminant, i.e., $a^4 + 4$, must be the square of a rational. Thus, for some $s,w \in \mathbb{Z}, \; w \neq 0, \; \gcd(s,w) = 1$,

$$\begin{equation}\begin{aligned} \left(\frac{p}{q}\right)^4 + 4 & = \left(\frac{s}{w}\right)^2 \\ \frac{p^4 + 4q^4}{q^4} & = \frac{s^2}{w^2} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Since both the left & right side fractions are in lowest terms, this means the numerators & denominators are equal to each other. In particular,

$$\begin{equation}\begin{aligned} p^4 + 4q^4 & = s^2 \\ (p^2)^2 + (2q^2)^2 & = (s)^2 \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

If $p$ is odd, then $(p^2, 2q^2, s)$ forms a primitive Pythagorean triple. Thus, as the linked Wikipedia article states, for some $m$ and $n$ coprime and both not odd,

$$p^2 = m^2 - n^2 \tag{4}\label{eq4A}$$

$$2q^2 = 2mn \implies q^2 = mn \tag{5}\label{eq5A}$$

Since $m$ and $n$ are coprime, then \eqref{eq5A} shows that $m$ and $n$ must each be perfect squares, i.e., there are $y, z \in \mathbb{N}$ where $m = y^2$ and $n = z^2$. Using this in \eqref{eq4A} gives

$$p^2 = y^4 - z^4 \tag{6}\label{eq6A}$$

However, as proven in x^4-y^4=z^2 has no solutions in positive integers, there are no solutions in positive integers for \eqref{eq6A}. Since none of the values can be $0$ (e.g. $p$ is odd, and $z = 0 \implies n = 0 \implies q = 0$, but \eqref{eq1A} states $q \neq 0$), this means there are no solutions.

The other case for \eqref{eq3A} is if $p$ is even, i.e., $p = 2x$ for some $x \in \mathbb{N}$, so the left side is even, which means $s = 2r$ for some $r \in \mathbb{N}$. Thus, \eqref{eq3A} becomes

$$\begin{equation}\begin{aligned} 16x^4 + 4q^4 & = 4r^2 \\ 4x^4 + q^4 & = r^2 \\ (q^2)^2 + (2x^2)^2 & = (r)^2 \end{aligned}\end{equation}\tag{7}\label{eq7A}$$

This is now of the same form as \eqref{eq3A}, but with $q$ definitely being odd, so repeating the same arguments as earlier (but with the explanation for \eqref{eq6A} not having a solution being somewhat different, e.g., $z = 0 \implies x = p = 0 \implies t = \pm 1$) results in, once again, that there is no solution.

Since both parity cases for $p$ in \eqref{eq3A} shows there's no solutions, the assumption in \eqref{eq1A}, i.e., $a$ is rational, must be false.