Let the four consecutive integers be n – 1, n, n + 1, and n + 2.

Then the sum of their squares is (n - 1)² + n²+ (n + 1)² + (n + 2)² = 4n² + 4n + 6 = (2n + 1)² + 5

Therefore we have (2n + 1)² + 5 = k² – 2 for some integer k.

Hence k² - (2n + 1)² = 7, or using difference of squares factorization, (k + 2n + 1)(k – 2n – 1) = 7.

As k and n are integers this means that

either

k + 2n + 1 = 7 and k – 2n – 1 = 1, which solved simultaneously gives n = 1 (and k = 4)

or

k + 2n + 1 = 1 and k – 2n – 1 = 7, which solved simultaneously gives n = - 2 (and k = 4)

The mean of the four consecutive integers is (4n + 2)/4, so the values for the mean are:

1.5 (when n = 1) and -1.5 (when n = -2)

Challenge 305: Four Square Sum