Finding a general formula for two squares whose sum = 2 Presumably, a formula for u & v that generates all the possible u & v such that $u^2 + v^2 = 2$.
With the limitation that the problem must be solved without using roots or fractional powers assuming that u & v are only required to be real numbers. 
I'm not exactly sure where to start. 

As $(1-t^2)^2+(2t)^2=(1+t^2)^2$, we can parametrize $a^2+b^2=1$ by letting $a=\frac{1-t^2}{1+t^2}$, $b=\frac{2t}{1+t^2}$. 
Now to get to $u^2+v^2=2$ without introducing $\sqrt 2$, just take $u=a+b$, $v=a-b$ (which makes $u^2+v^2=a^2+2ab+b^2+a^2-2ab+b^2=2(a^2+b^)=2$), so take
$$\tag1 u = \frac{1-t^2+2t}{1+t^2},\qquad v = \frac{1-t^2-2t}{1+t^2}.$$
Note that as $|t|$ runs from $0$ to $\infty$, $a$ rund from $1$ (inclusinve) to $-1$ (exclusive). And negating $t$ keeps $a$ while negating $b$. Thus $(a,b)$ above really runs through all of $S^1$ except the single point $(-1,0)$. Accordingly, $(1)$ parametrizes all points with $u^2+v^2=2$ with the single exception $(-1,-1)$. In order to really catch all pairs, you can either accept $t=\infty$ as parameter value, or use the double angle formulas 
$\sin 2x=2\sin x\cos x$, $\cos2x=\cos^2x-\sin^2x$ to run around the circle (almost) twice. That is, we'd take $c=2ab$, $d=b^2-a^2$ and then $u=c+d$, $v=c-d$.
