The problem is to find the missing numbers in two magic squares. In a magic square, the sum of the numbers in each row, each column, and each diagonal is the same.

1. Problem Description

The problem is to find the missing numbers in two magic squares. In a magic square, the sum of the numbers in each row, each column, and each diagonal is the same.

2. Solution Steps

Problem 1:

The magic square is:

$$

\begin{bmatrix}

7 & a & 5 \\

2 & 4 & b \\

c & 8 & 1

\end{bmatrix}

$$

The sum of the diagonal from top left to bottom right is $7 + 4 + 1 = 12$.

Therefore, the sum of each row, column, and diagonal must be

1

2.

From the first row, $7 + a + 5 = 12$, so $a = 12 - 7 - 5 = 0$.

From the second row, $2 + 4 + b = 12$, so $b = 12 - 2 - 4 = 6$.

From the third column, $5 + b + 1 = 12$, and we already found $b = 6$. So, $5 + 6 + 1 = 12$ is consistent.

From the third row, $c + 8 + 1 = 12$, so $c = 12 - 8 - 1 = 3$.

From the first column, $7 + 2 + c = 12$, and we found $c=3$. So, $7 + 2 + 3 = 12$ is consistent.

The square becomes:

$$

\begin{bmatrix}

7 & 0 & 5 \\

2 & 4 & 6 \\

3 & 8 & 1

\end{bmatrix}

$$

Problem 2:

The magic square is:

$$

\begin{bmatrix}

8 & 1 & c \\

a & 5 & b \\

4 & c & d

\end{bmatrix}

$$

The sum of the diagonal from top left to bottom right is $8 + 5 + d$.

The sum of the numbers in each row, each column, and each diagonal is the same.

The sum of the first column is $8 + a + 4 = 12 + a$.

The sum of the second column is $1 + 5 + c = 6 + c$.

The sum of the first row is $8 + 1 + c = 9 + c$.

Let's assume that all the numbers are different integers from 1 to

9. We know that $9+c = 12 + a = 6+c = 8+5+d$.

From $9 + c = 6 + c$, we can see that these two can't be same unless 9 = 6, which is not true.

$8 + 5 + d = 12 + a$, or $13 + d = 12 + a$, then $a = d + 1$.

$8 + 5 + d = 9 + c$, or $13 + d = 9 + c$, then $c = d + 4$.

Considering the diagonal $4 + 5 + 8 = 17$. It looks to be sum of all the columns/rows should be the same.

$8 + 1 + c = 17$, so $c = 8$.

$4 + c + d = 17$, so $4 + 8 + d = 17$, $12 + d = 17$, $d = 5$.

$a + 5 + b = 17$.

$8 + a + 4 = 17$, so $a = 5$. but a should not be

5.

However, $4 + c + d= 4 + 8 +5 = 17$.

$a + 5 + b = 17$

$8 + a + 4 = 17$ gives $a = 5$. But this cannot be the case, since $5$ is already present.

So, we have to look for a trick.

Try $8+1+6=15$.

So, if 15 is the magic number, $c = 6$.

$a + 5 + b = 15$

$8 + a + 4 = 15$, so $a = 3$.

$4 + c + d = 15$, so $4 + 6 + d = 15$, so $d = 5$.

$8 + 5 + d = 15$ implies that $d = 2$. We already know $d=5$.

The question does not provide that numbers from 1 to 9 are used, therefore some other values can be used.

I do not see an easy solution to this problem given the information available.

3. Final Answer

For the first magic square: $a = 0$, $b = 6$, $c = 3$.

For the second magic square, there is not enough information to determine the values of $a, b, c, d$. A possible solution would be $c = 6, a=3, d=5$, $b=7$ which gives:

$$

\begin{bmatrix}

8 & 1 & 6 \\

3 & 5 & 7 \\

4 & 6 & 5

\end{bmatrix}

$$

But that is incorrect since all rows, columns, and diagonals are not equal to

1

5.

The first magic square is:

$a = 0$

$b = 6$

$c = 3$

```

7 0 5

2 4 6

3 8 1

```