Example 1. Find all pairs of non-negative integers x and y such that y2(x + 1) = 1576 + x2.
Example 2. Sixteen numbers are put into the boxes of a four-by-four array, so as to form a magic square. This means that the four row sums, the four column sums and the two diagonal sums are each equal to the same number s. Show that s is also the sum of the four numbers in the corners of the array (do not assume that the sixteen numbers are the integers 1, 2, 3, . . . , 16).
Solution 1. Let x and y are non-negative integers such that y2(x + 1) = 1576 + x2. Then
y2(x + 1) = 1576 + x2
or, y2(x + 1) = 1577 + (x2 – 1)
or, y2(x + 1) = 1577 + (x – 1) (x + 1)
or, y2(x + 1) – (x – 1) (x + 1) = 1577
or, (x + 1) (y2 – x + 1) = 19 × 83.
In particular, (x + 1) is a positive divisor of 19 × 83 with quotient (y2 – x + 1).
- If x + 1 = 1, then x = 0 and y2 – x + 1 = 1577 so that y2 = 1576.
- If x + 1 = 19, then x = 18 and y2 – x + 1 = 83 so that y2 = 83 + 17 = 100.
- If x + 1 = 83, then x = 82 and y2 – x + 1 = 19 so that y2 = 100.
- If x + 1 = 1577, then x = 1576 and y2 – x + 1 = 1 so that y2 = 1576.
Since √1576 ≈ 39.7 is not an integer, the two possibilities are (x, y) = (18, 10) and (82, 10).
Solution 2. First observe that the sum of all 16 numbers is the sum of the four rows, which is 4s.
The sum of the middle two rows, the middle two columns and the two diagonals is 6s and this uses all of the numbers once except for the middle four, each of which is used three times.
If we write m to denote the sum of the middle four numbers, we see that 6s is the sum of all the numbers plus an extra 2m.
This gives the equation 6s = 4s + 2m, from which we deduce that m = s.
Finally note that the sum of the two diagonals is 2s, and this is the sum of the four middle boxes plus the sum of the four corners. Since the middle boxes sum to m = s, it follows that the corner boxes also sum to s, as wanted.