Question 567318
Let k-3, k-1, k+1, k+3 be the four numbers. Then for some integer m,


*[tex \LARGE m^2 = (k-3)(k-1)(k+1)(k+3)]


*[tex \LARGE = (k^2 - 9)(k^2 - 1)]


*[tex \LARGE = k^4 - 10k^2 + 9]


Compare with *[tex \LARGE (k^2 - 5)^2 = k^4 - 10k^2 + 25]. This is definitely a perfect square, so we need this to differ with m^2 by 16. A simple check shows that {9,25} and {0,16} are the only pairs of perfect squares that differ by 16. {0,16} is impossible because we want our perfect square to be odd. If we have {9,25} then


*[tex \LARGE m^2 = k^4 - 10k^2 + 9 = 9]
*[tex \LARGE (k^2 - 5)^2 = k^4 - 10k^2 + 25 = 25]


Hence k^4 - 10k^2 = 0, (k^2)(k^2 - 10) = 0. The only integer solution is k = 0, which yields {-3,-1,1,3}, and m^2 = 9. However, this is not in the set of natural numbers, so there are no solutions.