I'm trying to understand a proof in my textbook, which is about uniformity of functions over finite fields (GF(2^n)). It's about finding the maximum number of solutions for an equation containing a certain function f(x) over GF(2^n).
Consider $c,d \in \mathbb{F}_{2^n}$ and $f: \mathbb{F}_{2^n} \mapsto \mathbb{F}_{2^n} | f(x) = x^{-1}$
It starts with the equation:
$x^{-1} + (x+c)^{-1} = d$
for which the number of solutions for any given c and d is needed. After some transforming we have:
$\frac{c}{x} + \frac{1}{\frac{x}{c}+1} = c*d$
So far so good, but now, the authors just substitute:
$cd = d$ and $\frac{x}{c}=x$
While I can still understand the first one (since you also could rewrite it as something like $cd = e$, because e [or that 'new' d] is just a permutation of the domain $F_{2^n}$), the latter beats me. How may I know I don't change the number of solutions to that equation, when I effectively set $c=1$?
Another proof does an equal trick. It has:
$a^d\left(\left(\frac{x}{a}+1\right)^d + \left(\frac{x}{a}\right)^d \right) = b$ , with a and b as constants of $F_{2^n}$ and then it states, that the following has the same number of solutions for x:
$(x+1)^d + (x)^d = \frac{b}{a^d}$
and later even:
$(x+1)^d + (x)^d = b$
Again, they seem to just liquidate $\frac{x}{a}$ and I just don't get why this is possible.
UPDATE: The first proof ends in this:
$dx^2 + dx + 1 = 0$
In the next step it's suddenly just:
$x^2 + x + 1 = 0$, which is quietly assumed to have same number of solutions in $\mathbb{F}_{2^n}$. Just because it's a quadratic equation, I can leave out every parameter? Next step, after multiplying with $x(x+1):\\$ $x^4+x=0$
which would look different, if we hadn't just removed the 'd's.