Key idea. Factor the difference of squares, then check the reconstruction carefully.

1. Let \(X=|x|\), \(Y=|y|\). We have \(X>2Y\), so the two factors in

   \[(X-2Y)(X+2Y)=12\]

   are positive.

2. Set \(u=X-2Y\), \(v=X+2Y\). Then

   \[X=\frac{u+v}{2},\qquad Y=\frac{v-u}{4}.\]

   Both quantities must be integers.

3. Check every positive factor pair of 12:

   \(u\)	\(v\)	\(X\)	\(Y\)
   \(1\)	\(12\)	\(\frac{13}{2}\)	\(\frac{11}{4}\)
   \(2\)	\(6\)	\(4\)	\(1\)
   \(3\)	\(4\)	\(\frac{7}{2}\)	\(\frac{1}{4}\)

   Only the middle row gives integers.

4. Thus \(|x|=4\), \(|y|=1\). Restoring both signs gives \((4,1),(4,-1),(-4,1),(-4,-1)\). Each satisfies \(16-4=12\).

Remember. The reconstruction formula determines the divisibility conditions. Here the factor difference must be divisible by 4.