Proof.
In defining
![$ \nu$](img58.png)
, we expressed each number in
as congruent to a number in the set
No number
![$ 1, 2, \ldots \frac{p-1}{2}$](img62.png)
appears more than once, with either choice of sign, because if it
did then either two elements of
![$ S$](img63.png)
are congruent modulo
![$ p$](img3.png)
or
0 is the sum of two elements of
![$ S$](img63.png)
, and both events are impossible.
Thus the resulting set must be of the form
where each
![$ \varepsilon _i$](img65.png)
is either
![$ +1$](img19.png)
or
![$ -1$](img20.png)
. Multiplying together
the elements of
![$ S$](img63.png)
and of
![$ T$](img66.png)
, we see that
so
The lemma then follows from Proposition
1.1.