On page 20 of "Proofs from THE BOOK" (click here) a proof of representing number as sum of two square is given. The proof is due to Roger Heath-Brown (1971,appeared in 1984). The first part of the proof is -
I didn't get the below paragraph at all-
What do we get from the study of $f$ ? The main observation is that since $f$ maps the sets $T$ and $U$ to their complements, it also interchanges the elements in $T\setminus U$ with these in $U\setminus T$. That is, there is the same number of solutions in $U$ that are not in T as there are solutions in T that are not in U -so $T$ and $U$ have the same cardinality.
So my questions is-
How does $f$ interchanges the elements in $T\setminus U$ with these in $U\setminus T$?
Notice that $\,S\,$ is partitioned into four pairwise disjoint sets: $$S = (T\cap U) \cup (T\setminus U) \cup (U\setminus T) \cup(S\setminus(T\cup U)). \tag{1}$$ We are given that the involution $\,f\,$ has no fixed points, and maps $\,T\,$ to $\,S\setminus T\,$ and $\,U\,$ to $\,S\setminus U.\,$
Then it must map their intersection $\,T\cap U\,$ to $$ (S\setminus T)\cap(S\setminus U) = S\setminus(T\cup U). \tag{2}$$ The exact same reasoning applies to their complements. This means that $\,f\,$ maps $\,S\setminus T\,$ to $\,T\,$ and $\,S\setminus U\,$ to $\,U.\,$ Thus their intersection as in equation $(2)$ gets mapped to $\,T\cap U.\,$ The two remaining parts of $\,S\,$ are $\,T\setminus U\,$ and $\,U\setminus T.\,$ Each gets mapped to the other part since the first and fourth parts are mapped by $\,f\,$ to each other and since both $\,T\,$ and $\,U\,$ get mapped to their complements.
Equation $(1)$ is true for any set $\,S\,$ and with any given $\,T\subseteq S\,$ and $\,U\subseteq S.\,$ The subsets $\,T\,$ and $\,U\,$ may or may not be disjoint. Their union $\,T\cup U\,$ may or may not be equal to $\,S.\,$ One of the subsets may or may not be a subset of the other. In other words, any of the four parts of the partition in equation $(1)$ may or may not be the empty set depending of the specific subsets $\,T\,$ and $\,U.\,$
An example is for $\,p=13\,$ where $\,(1,1,-3)\in S\,$ and $\,(1,3,-1)\in S\,$ which are not in $\,T\cup U. $