Consider the grid of points $$X = \{(m, n) : 0 < m, n \le 4\}$$. We say a pair of points $$(a, b), (c, d)$$ in $$X$$ is a knight-move pair if $$(c = a \pm 2 \text{ and } d = b \pm 1)$$ or $$(c = a \pm 1 \text{ and } d = b \pm 2)$$. The number of knight-move pairs in $$X$$ is:

All points lie on the $$4 \times 4$$ integer grid
$$X=\{(m,n)\mid 1\le m\le 4,\;1\le n\le 4\}$$
so there are $$16$$ points in total.

A knight step changes one co-ordinate by $$\pm 2$$ and the other by $$\pm 1$$.
For every point in the grid, count how many such steps keep the knight inside the grid. The number depends only on the “type’’ of square, i.e. on its distance from the edges.

Case 1: corner squares
Corners are $$(1,1),(1,4),(4,1),(4,4)$$. From a corner the legal moves are
$$\bigl(+2,+1\bigr)\quad\text{and}\quad\bigl(+1,+2\bigr)$$ (or their symmetric versions),
so each corner has $$2$$ moves.
Total contributions: $$4\times 2 = 8$$ ordered pairs.

Case 2: edge but not corner
Eight such squares exist (for example $$(1,2),(1,3),(2,1),(3,1)$$ and the symmetric ones). From any of these squares the knight can move in $$3$$ legal ways. Total contributions: $$8\times 3 = 24$$ ordered pairs.

Case 3: interior squares
The four central squares are $$(2,2),(2,3),(3,2),(3,3)$$. A knight located here has $$4$$ legal moves. Total contributions: $$4\times 4 = 16$$ ordered pairs.

Add all the contributions:
$$8+24+16 = 48$$

In this count every ordered pair $$\bigl((a,b),(c,d)\bigr)$$ that satisfies the knight-move condition appears exactly once, which is what the problem asks for. Hence,

Number of knight-move pairs $$= 48$$.