Why $s(1-s)$ numbers which are squares in a field are written $\frac{u^2}{1+u^2}$? Trying to exercise in Math again after years of other activities, I need a little help on this :
Let $F$ be some arbitrary field with characteristic > 2.
Let $S \subseteq F$ be the set of numbers s such that $s(1-s)=a^2$ for some $a \in F$ , and $s \neq 1$
Let $U \subseteq F$ be the set of numbers of the form $\frac{u^2}{1+u^2}$ for some $u \in F$, and $u^2 \neq -1$
We must show that the $S = U$.
I can easily show that $U \subseteq S$
ut I'm stuck on showing that $S \subseteq U$

Imagine a straight-edged triangle, with hypotenuse of unit length. Then its height is the 
geometric mean of the two segments into which it splits the hypotenuse, $h=\sqrt{s(1-s)}$ . 
At the same time, the locus of all points in two-dimensional space which form a straight–
edged triangle with a given hypotenuse is the circle whose diameter that hypotenuse is. 
Now, imagine that the circle is centered at the origin. It follows that the height in question 
is $r\sin\theta$, where r is the radius, and $\theta$ the angle which the radius uniting the point to the 
origin forms with the hypotenuse. But $\sin\theta$ can also be written as $\sqrt{\dfrac{\tan^2\theta}{1+\tan^2\theta}}$ , since 
$1+\tan^2\theta=\dfrac1{\cos^2\theta}=\sec^2\theta$. See also Weierstrass substitution for more information.
