Copy and complete the magic square so that the sum of numbers in each row or column or diagonal is 18. V2 − U2 = 2as, where v=3, u=2, a=1; find the value of s

$$M_{31} + M_{32} + M_{33} = 18$$ $$7 + 8 + M_{33} = 18$$ $$15 + M_{33} = 18$$ $$M_{33} = 18 - 15$$ $$M_{33} = 3$$

The square becomes:

$$\begin{bmatrix} \phantom{x} & \phantom{x} & 4 \\ \phantom{x} & \phantom{x} & \phantom{x} \\ 7 & 8 & 3 \end{bmatrix}$$

$$M_{13} + M_{22} + M_{31} = 18$$ $$4 + M_{22} + 7 = 18$$ $$11 + M_{22} = 18$$ $$M_{22} = 18 - 11$$ $$M_{22} = 7$$

The square becomes:

$$\begin{bmatrix} \phantom{x} & \phantom{x} & 4 \\ \phantom{x} & 7 & \phantom{x} \\ 7 & 8 & 3 \end{bmatrix}$$

$$M_{11} + M_{22} + M_{33} = 18$$ $$M_{11} + 7 + 3 = 18$$ $$M_{11} + 10 = 18$$ $$M_{11} = 18 - 10$$ $$M_{11} = 8$$

The square becomes:

$$\begin{bmatrix} 8 & \phantom{x} & 4 \\ \phantom{x} & 7 & \phantom{x} \\ 7 & 8 & 3 \end{bmatrix}$$

$$M_{11} + M_{12} + M_{13} = 18$$ $$8 + M_{12} + 4 = 18$$ $$12 + M_{12} = 18$$ $$M_{12} = 18 - 12$$ $$M_{12} = 6$$

The square becomes:

$$\begin{bmatrix} 8 & 6 & 4 \\ \phantom{x} & 7 & \phantom{x} \\ 7 & 8 & 3 \end{bmatrix}$$

$$M_{11} + M_{21} + M_{31} = 18$$ $$8 + M_{21} + 7 = 18$$ $$15 + M_{21} = 18$$ $$M_{21} = 18 - 15$$ $$M_{21} = 3$$

The square becomes:

$$\begin{bmatrix} 8 & 6 & 4 \\ 3 & 7 & \phantom{x} \\ 7 & 8 & 3 \end{bmatrix}$$

$$M_{21} + M_{22} + M_{23} = 18$$ $$3 + 7 + M_{23} = 18$$ $$10 + M_{23} = 18$$ $$M_{23} = 18 - 10$$ $$M_{23} = 8$$

The completed square is:

$$\begin{bmatrix} 8 & 6 & 4 \\ 3 & 7 & 8 \\ 7 & 8 & 3 \end{bmatrix}$$

Let's verify all sums:
Rows:
$8+6+4 = 18$ (Correct)
$3+7+8 = 18$ (Correct)
$7+8+3 = 18$ (Correct)

Columns:
$8+3+7 = 18$ (Correct)
$6+7+8 = 21$ (Incorrect, should be 18)
$4+8+3 = 15$ (Incorrect, should be 18)

The given numbers (4, 7, 8) are inconsistent with the properties of a 3x3 magic square where all rows, columns, and diagonals must sum to 18. Specifically, the center element of a 3x3 magic square must be one-third of the magic sum ($18/3 = 6$). However, based on the given numbers, the anti-diagonal sum forces the center element to be 7. This contradiction means a complete magic square cannot be formed with the given numbers and sum.

Since a complete magic square cannot be formed, we provide the square filled as much as possible following the initial steps, and note the inconsistency.

The completed square based on the initial steps is:

$$\boxed{ \begin{bmatrix} 8 & 6 & 4 \\ 3 & 7 & 8 \\ 7 & 8 & 3 \end{bmatrix} }$$

Note: This square satisfies the sum of all rows and both diagonals, but not all columns.

$$s = \frac{5}{2}$$ $$s = 2.5$$

The value of $s$ is $\boxed{\mathbf{2.5}}$.