8.4. Can the numbers from 1 to 9 be placed in the cells of a $3 \times 3$ square so that the sum of any two numbers in adjacent cells (cells sharing a side) is a prime number?
# 8.4. Answer. No.

Solution. Let $a, b, c, d, e, f, g, h, k$ be the desired arrangement.

| $a$ | $b$ | $c$ |
| :--- | :--- | :--- |
| $d$ | $e$ | $f$ |
| $g$ | $h$ | $k$ |

Then, from the simplicity of the sums $a+b, b+c, \ldots, a+d, \ldots, f+k$, it follows that they are odd (each sum is greater than 2). Therefore, in any pair of adjacent cells, one number is even and the other is odd. But from 1 to 9, there are only 4 even numbers, so the even numbers are $b, d, f, h$, i.e., together with the central cell, they form the sums $e+2, e+4, e+6, e+8$. But among four consecutive odd numbers, one is necessarily divisible by 3, and the only prime number divisible by 3 is 3. Therefore, $e=1$. But then the sum $e+8=9$ is not a prime number.