A special curve $\left( \frac { x^{ 2 }+1 }{ 1-x^{ 2 } } \right)$. Let $a,x\in \mathbb{Q}$ and $n\in\mathbb{N}$
For $0<x<1$, $a>0$ and $n>1$, show that
$$\ \frac { x^{ 2 }+1 }{ 1-x^{ 2 } }   \neq { a }^{ n }$$.
For example, for $x = 0.3$, then
$\frac {109 }{ 91 }$ can not be shown as $a^2$ or $a^3$ etc.
 A: Assume on the contrary that $\dfrac{1+x^2}{1-x^2}=a^n$ for some $x, a \in \mathbb{Q}, n \in \mathbb{N}, 0<x<1, a>0, n>1$. Write $x=\frac{s}{t}, s<t, s, t \in \mathbb{Z}^+, \gcd(s, t)=1$, so $\dfrac{t^2+s^2}{t^2-s^2}=a^n$
We consider 2 cases:
Case 1: $s, t$ have different parities. 
Then $\gcd(t^2+s^2, t^2-s^2)=1$, so we have $t^2+s^2=b^n, t^2-s^2=c^n, a=\frac{b}{c}$ for some $b, c \in \mathbb{Z}^+$. Thus $c^n=(t-s)(t+s)$, so since $\gcd(t-s, t+s)=1$, we have $t-s=d^n, t+s=e^n, c=de$ for some $d, e \in \mathbb{Z}^+$. Thus $d^{2n}+e^{2n}=(t-s)^2+(t+s)^2=2(t^2+s^2)=2b^n$. 
If $n \geq 3$, by results of Darmon and Merel, we have no non-trivial primitive solutions to the equation $x^n+y^n=2z^n$ (where trivial means $|xyz| \leq 1$) Clearly $\gcd(d, e)=1$, so $(x, y, z)=(d^2, e^2, b)$ is a primitive solution to the Fermat-like equation $x^n+y^n=2z^n$. It must thus be trivial, so $|d^2e^2b| \leq 1$, so $d=e=b=1$ since $d, e, b$ are positive integers. This implies that $2s=e^n-d^n=0$, a contradiction.
Therefore $n=2$. Now $t^2+s^2=b^2, t^2-s^2=c^2$, so $t^4-s^4=(t^2-s^2)(t^2+s^2)=(bc)^2$. We proceed via infinite descent (by comparing the size of $b$) to show that there are no positive integer solutions $(t, s, b, c)$ satisfying both equations simultaneously, where $t, s$ are of different parities and $\gcd(t, s)=1$. 
Note that if $t^2=s^2+c^2 \not \equiv 0 \pmod{4}$, so $t$ cannot be even, so $s$ is even. Using the parameterization of primitive Pythagorean triplets we get $t^2=k^2+l^2, s=2kl, c^2=k^2-l^2$, where $k, l$ are of different parities and $\gcd(k, l)=1$. Then $(k, l, t, c)$ is also a solution of the 2 equations, and we have $t<b$. We have thus obtained a smaller positive integer solution, and this leads (via infinite descent) to a contradiction. 
Case 2: $s, t$ are of the same parity.
Since $\gcd(s, t)=1$, $s, t$ are both odd. We have $\gcd(t^2+s^2, t^2-s^2)=2$, so $t^2+s^2=2b^n, t^2-s^2=2c^n$ for some $b, c \in \mathbb{Z}^+$. If $n$ is even, then we may take $n$ to be $2$, since we can replace $b, c$ with $b^{\frac{n}{2}}, c^{\frac{n}{2}}$ respectively. We thus have $t^2+s^2=2b^2, t^2-s^2=2c^2$, so $t^2=b^2+c^2, s^2=b^2-c^2$. Note that $\gcd(b, c)=1$ and $b, c$ are of different parities (since $s, t$ are odd), so applying the result from Case 1 (With a variable change $(t, s, b, c) \to (b, c, t, s)$), we get a contradiction. 
Therefore $n$ is odd. Thus $2b^n=(t^2-s^2)+2s^2=2c^n+2s^2$, so $b^n+(-c)^n=s^2$. If $n \geq 5$, then again by results of Darmon and Merel, we have no non-trivial primitive solutions to the equation $x^n+y^n=z^2$. Clearly $\gcd(b, c)=1$, so $(x, y, z)=(b, -c, s)$ is a primitive solution to the equation $x^n+y^n=z^2$. It must thus be trivial, so $|bcs| \leq 1$, so $b=c=s=1$ since $b, c, s$ are positive integers. However this implies that $t^2=s^2+2c^n=3$, a contradiction.
Therefore $n=3$, and we have $t^2+s^2=2b^3, t^2-s^2=2c^3$. We have $2c^3=(t-s)(t+s)$, so we have $\{(t-s), (t+s)\}=\{2d^3, e^3\}$ for some $d, e \in \mathbb{Z}^+$. Notice that it does not matter which of $t+s, t-s$ is $2d^3$ and which is $e^3$, since we will only consider $4b^3=2(s^2+t^2)=(t-s)^2+(t+s)^2=4d^6+e^6$. Note that $\gcd(t-s, t+s)=2$, so $e$ is even and $d$ is odd. This also implies that $\gcd(d, e)=1$. 
Write $e=2f, f \in \mathbb{Z}^+, \gcd(d, f)=1$. Then $4b^3=4d^6+64f^6$, so $b^3=d^6+16f^6$, so $b^3+(-d^2)^3=2(2f^2)^3$. Since $\gcd(-d^2, 2f^2)=1$ and $|2bd^2f^2|>1$, $(x, y, z)=(b, -d^2, 2f^2)$ is a non-trivial primitive solution to the equation $x^3+y^3=2z^2$, which again by results of Darmon and Merel (as in Case 1) leads to a contradiction.
Conclusion: We have obtained a contradiction in both cases, so the equation $\dfrac{1+x^2}{1-x^2}=a^n$ has no solutions for $x, a \in \mathbb{Q}, n \in \mathbb{N}, 0<x<1, a>0, n>1$.
