Pythagoras numbers and fermats last theorem I am reading "What Is Mathematics? An Elementary Approach to Ideas and Methods"
And I am stuck here, I don't get it. I have posted a screen shot underlining what my doubt is..
I dont get it when the author says while the pythagoras theorem is : $a^2 + b^2 = c^2$
and then he says $x=a/c$ and $y=b/c$ 
and then the equation should be according to me , $ax+by=c$.. right??
but the author writes y^2 = (1-x)(1+x)
Which I think may have came from something like
$x^2 + y^2 = 1$
$y^2= 1^2 + x^2$
$y^2= (1+x)(1-x)$     (since $a^2 + b^2 = (a+b)(a-b)$ )
but I dont get it where did that x^2 + y^2 = 1 came from ?? 
Is it that author assumed that x=a/c and y=b/c and then stoped talking about pythagoras theorem and started talking on x^2 + y^2 = 1???
and then further he introduces a number t.. i don't get it how it got converted into y=t(1+x) and (1-x)=ty ???
Can please someone help?
also , while i was writing the question it clicked me that if:
 x=a/c (i.e. opposite upon hypotenuse means x is sin ) 
 y=b/c (i.e. adjacent upon hypotenuse means y is cos)
therefore x^2 + y^2 = 1 (sin^2 + cos^2 = 1)
but then if i assume i am right about the sin cos thing then how come the last step is derived ?? (the one in blue color underline and box)
also that as per me x is sin , but the formula in that book is of cos2x? when t=tanx??
so m lil confused .. if you want any more clean way of me asking my doubt then do tell me i will rephrase the entire question .. :)

 A: For the first part of your question: the author divides both sides of the Pythagorean equation by $c^2$. This yields: $\frac{a^2}{c^2} + \frac{b^2}{c^2} = 1$. Now, he defines $x = a/c$ and $y = b/c$. The equation can than be rewritten to: $x^2 + y^2 = 1$. Substracting $x^2$ from both sides yields $y^2 = 1 - x^2$, or equivalent: $y^2 = (1-x)(1+x)$. 
He then devides both sides by $y$ and $(1+x)$. This gives $y/(1+x) = (1-x)/y$. His next step is to say that the expression is actually $t = t$, where $t$ is a rational. So, $ y/(1+x) = t \rightarrow y =t(1+x)$ and, in the same way, $(1-x)/y = t \rightarrow (1-x) = ty$. 
Working this out gives the two equations $tx - y = -t$ and $x + ty = 1$. Now you have two unknowns in two equations. 
This can be easily solved:
$tx - y = -t \rightarrow x = \frac{y-t}{t}$. Substitute this into the second equation. That gives us$\frac{y-t}{t} + ty = 1 \rightarrow (y-t) + t^2y = t$. Now, solve for y:
$y + t^2y = 2t \rightarrow y(1+t^2) = 2t \rightarrow y = \frac{2t}{1 + t^2}$. 
Substitute this value for y back into the first to get the expression for $x$.
A: You replaced one of the two $a$s making up $a^2$ with $cx$. Try replacing both (and similarly for the two $b$'s in $b^2$ with $cy$)
