250. The Court Mathematician's Salary. Once, the court mathematician received his entire annual salary in silver talers, from which he formed nine piles, together constituting a magic square. The king, upon seeing this, was impressed but regretfully noted that in none of the piles was the number of coins a prime number.

- If I had nine more coins, - said the mathematician, - by adding one coin to each pile, I would get a magic square consisting only of prime numbers.

The king calculated and was convinced that this was indeed the case. He was about to give the mathematician nine more talers when the court jester, unexpectedly exclaiming, "Wait!", removed one coin from each pile, and everyone saw that a new magic square, containing only prime numbers, had been formed. Thus, the jester saved the king nine talers. What is the court mathematician's annual salary?
250. As established by L. S. Fryer, the elements of an arbitrary magic square of the third order can actually be represented using three parameters:

$$
\begin{array}{ccc}
(e+x), & (e-x-y) & (e+y) \\
(e-x+y), & e, & (e+x-y) \\
(e-y), & (e+x+y) & (e-x)
\end{array}
$$

It is clear that these numbers can be rearranged so that each row forms an arithmetic progression with the same difference, and each column also forms an arithmetic progression with a common difference:

$$
\begin{array}{ccc}
(e-x-y), & (e-y), & (e+x-y) \\
(e-x), & e, & (e+x), \\
(e-x+y), & (e+y), & (e+x+y)
\end{array}
$$

We need to construct a magic square of the third order from a set of prime numbers \( p_i \) that have the property that all \( p_i + 2 \) are also prime. Further, note that all \( p_i \) (not equal to 3 and 5) must end in 9, 7, or 1. Therefore, the parameter \( e \) represents the arithmetic mean of two prime numbers that end in either 9 and 9, 7 and 1, 7 and 7, or 1 and 1. Consequently, the elements of our square should be sought among the sequence of the smaller of consecutive prime twins:

$$
\begin{aligned}
& 3, \quad 5, \quad 11, \quad 17, \quad 29, \quad 41, \quad 59, \quad 71, \quad 101, \quad 107, \\
& 137, \quad 149, \quad 179, \quad 191, \quad 197, \quad 227, \quad 239, \quad 269, \quad 281, \ldots
\end{aligned}
$$

The smallest value of \( e \) ending in 9, which represents the arithmetic mean of at least four pairs of prime numbers, is determined by the equations

$$
\begin{aligned}
2(149) & =107+191=101+197=71+227=59+239= \\
& =29+269=17+281
\end{aligned}
$$

(For any prime number \( < 149 \) and ending in 1 and 7, there are no more than three members of the given sequence with the same final digit.) Substituting into the middle row of the above square, sequentially replacing \( e \) with 149, and \( (e-x) \) and \( (e+x) \) with the obtained pairs of numbers, we find that only one arrangement allows three of the remaining pairs to be placed in the correct positions, namely:

From this, we immediately find three magic squares as mentioned in the problem:

| 191 | 17 | 239 |  | 192 | 18 | 240 |  | 193 | 19 | 241 |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 197 | 149 | 101 |  | 198 | 150 | 102 |  | 199 | 151 | 103 |
| 59 | 281 | 107 |  | 60 | 282 | 108 | 61 | 283 | 109 |  |

The court mathematician was paid annually

\( 9 e + 9(149 + 1) = 1350 \) thalers.

This result was also obtained by another method \( -A . M . M ., \mathbf{5 5}\), 429 (September 1948).

The next solution was found by W. Benson:

| 2088 | 1302 | 4800 |
| ---: | ---: | ---: |
| 5442 | 2730 | 18 |
| 660 | 4158 | 3372 |

In this case, the annual salary of the court mathematician is 24570 thalers.