Problem: Let $p$ be an odd prime.
An integer $x$ is called a \emph{quadratic non-residue}
if $p$ does not divide $x-t^2$ for any integer $t$.

Denote by $A$ the set of all integers $a$
such that $1 \le a < p$,
and both $a$ and $4-a$ are quadratic non-residues.
Calculate the remainder
when the product of the elements of $A$ is divided by $p$.

Solution: The answer is that $\prod_{a \in A} a \equiv 2 \pmod p$
regardless of the value of $p$.
In the following solution,
we work in $\FF_p$ always
and abbreviate ``quadratic residue'' and ``non-quadratic residue''
to ``QR'' and ``non-QR'', respectively.

We define
\begin{align*}
  A &= \left\{ a \in \FF_p \mid a, 4-a \text{ non-QR} \right\} \\
  B &= \left\{ b \in \FF_p \mid b, 4-b \text{ QR}, b \neq 0, b \neq 4 \right\}.
\end{align*}
Notice that
\[ A \cup B = \left\{ n \in \FF_p \mid
    n(4-n) \text{ is QR} \;
    n \neq 0, 4 \right\}. \]
We now present two approaches both based on the set $B$.

\paragraph{First approach (based on Holden Mui's presentation in Mathematics Magazine).}
The idea behind this approach is that $n(4-n)$
is itself an element of $B$ for $n \in A \cup B$,
because $4 - n(4-n) = (n-2)^2$.
This motivates the following claim.
\begin{claim*}
  The map
  \[ A \cup B \setminus \{2\} \to B
    \qquad \text{by}\quad n \mapsto n(4-n) \]
  is a well-defined two-to-one map,
  i.e.\ every $b \in B$ has exactly two pre-images.
\end{claim*}
\begin{proof}
  Since $n \notin \{0,2,4\}$,
  we have $n(4-n) \notin \{0,4\}$,
  so as discussed previously, $n(4-n) \in B$.
  Thus this map is well-defined.

  Choose $b \in B$.
  The quadratic equation $n(4-n) = b$ in $n$
  rewrites as $n^2-4n+b=0$,
  and has discriminant $4(4-b)$,
  which is a nonzero QR.
  Hence there are exactly two values of $n$, as desired.
\end{proof}
Therefore, it follows that
\[ \prod_{n \in A \cup B \setminus \{2\}} n
  = \prod_{b \in B} b \]
by pairing $n$ with $4-n$ on the left-hand side.
So, $\prod_{a \in A} a = 2$.

\paragraph{Second calculation approach (along the lines of official solution).}
We now do the following magical calculation in $\FF_p$:
\begin{align*}
  \prod_{b \in B} b
  = \prod_{b \in B} (4-b)
  &= \prod_{\substack{1 \le y \le (p-1)/2 \\ y \neq 2 \\ 4-y^2 \text{ is QR}}} (4-y^2) \\
  &= \prod_{\substack{1 \le y \le (p-1)/2 \\ y \neq 2 \\ 4-y^2 \text{ is QR}}} (2+y)
  \prod_{\substack{1 \le y \le (p-1)/2 \\ y \neq 2 \\ 4-y^2 \text{ is QR}}} (2-y) \\
  &= \prod_{\substack{1 \le y \le (p-1)/2 \\ y \neq 2 \\ 4-y^2 \text{ is QR}}} (2+y)
  \prod_{\substack{(p+1)/2 \le y \le p-1 \\ y \neq p-2 \\ 4-y^2 \text{ is QR}}} (2+y) \\
  &= \prod_{\substack{1 \le y \le p-1 \\ y \neq 2, p-2 \\ 4-y^2 \text{ is QR}}} (2+y) \\
  &= \prod_{\substack{3 \le z \le p+1 \\ z \neq 4,p \\ z(4-z) \text{ is QR}}} z \\
  &= \prod_{\substack{0 \le z \le p-1 \\ z \neq 0,4,2 \\ z(4-z) \text{ is QR}}} z.
\end{align*}
Note $z(4-z)$ is a nonzero QR if and only if $z \in A \cup B$.
So the right-hand side is almost the product over $z \in A \cup B$,
except it is missing the $z=2$ term.
Adding it in gives
\[ \prod_{b \in B} b
  = \half \prod_{\substack{0 \le z \le p-1 \\ z \neq 0,4 \\ z(4-z) \text{ is QR}}} z
  = \half \prod_{a \in A} a \prod_{b \in B} b.  \]
This gives $\prod_{a \in A} a = 2$ as desired.