{3,3,3,3,3,3} is a set of six integers such that the sum of the squares of the reciprocals totals 2/3.
(1/3)^2 + (1/3)^2 + (1/3)^2 + (1/3)^2 + (1/3)^2 + (1/3)^2 = 2/3

Does there exist a set of integers with fewer than 6 members whose sum of the squares of the reciprocals totals 2/3?

Problem inspired by Find the triplets

(In reply to Here is another... No computer used this time. by Steven Lord)

Actually, there are two solutions with exactly 5 reciprocals of integer squares. There is also one with 4 reciprocals of squares and of course (1/3)^2 = 1/9, but that's a dubious case, since it's not a sum.

Posted by Charlie on 2020-08-26 07:13:04