<s>[INST] If $\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$ for real $t$ and $0
$$\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$$ 
  where $0<x<1$ and $t\in\mathbb{R}$. Prove that $x=1/2$. 
It is evident that $x=1/2$ satisfies the above equation. Please help.
 [/INST] Well, define $y \in \mathbb{R}$ such that $x=1-y$ (noting that also $0<y<1$), we get
$$\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$$
$$\implies \frac{t^2-(1-y)^2}{(t^2+(1-y)^2)^2}+\frac{y^2-t^2}{(y^2+t^2)^2}=0$$
Note that the first equation implies 
$$\frac{t^2-x^2}{(t^2+x^2)^2}=\frac{t^2-(1-x)^2}{((1-x)^2+t^2)^2}$$
While the second implies 
$$\frac{t^2-(1-y)^2}{(t^2+(1-y)^2)^2}=\frac{t^2-y^2}{(y^2+t^2)^2}$$
So, in fact, the two equations are equivalent, which means $y=x \implies 1-x=x \implies x=1/2$
Edit: I figured out that what I proved was if $a$ is a solution (for $x$) then $1-a$ is also a solution. This means that $x=1/2$ for sure is a solution for all $t$. 
But when I tried graphing the equation on desmos and added a slider for $t$, I get that as $t$ ranges over all reals (or $t^2$ ranges over positive reals), $x$ and $1-x$ vary with it. For example, $t^2 = \frac{1}{6}\left(\sqrt{601} - 13\right) \implies x \in \{-2,1/2,3\}$
Note that $x=1/2$ is the general solution, and $x=3$ is a specific solution and if $3$ is a solution then $1-3=-2$ is also a solution. 
If we want $x=4$ as a solution we have $x=1-4=-3$ is also a solution and $t^2=\frac{1}{6}\left(\sqrt{2353} - 25\right)$ satisfies the equation. So, there's no point of solving for a specific $x$ if $x$ just varies depending on $t^2$ over the reals (and of course $1-x$ accordingly), but for all $t$, $x=1/2$ is certainly a solution.
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