Key idea. First find the possible magnitudes, then restore the signs.

1. Let \(X=|x|\), \(Y=|y|\). Since \(x^2-y^2=15>0\), we have \(X>Y\ge0\), and

   \[(X-Y)(X+Y)=15.\]

2. The positive factor pairs of 15, with the smaller first, are \((1,15)\) and \((3,5)\). If \(u=X-Y\), \(v=X+Y\), then

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

3. The pair \((1,15)\) gives \((X,Y)=(8,7)\); the pair \((3,5)\) gives \((X,Y)=(4,1)\).

4. Squares do not depend on signs. Each magnitude pair gives four solutions:

   \[\begin{gathered}(8,7),(8,-7),(-8,7),(-8,-7),\\(4,1),(4,-1),(-4,1),(-4,-1).\end{gathered}\]

5. Both magnitudes work because \(64-49=15\) and \(16-1=15\). Every factor pair was considered, so the list is complete.

Remember. For equations involving squares, separating magnitudes from signs can make completeness easier to check.