Any two members out of (a1, a2, a3, a4, a5)
add up to a square number.
List the ten squares.
Rem: Existence of similar sixnumber set is not resolved yet.
A trivial solution has all the members = 2x^2 but I guess that's not what's wanted.
I use s1s10 to represent the 10 squares and a=first member.
Then the desired set = (a, s1a, s2a, s3a, s4a). The next three squares are given by the equations:
s1+s22a = s5
s1+s32a = s6
s1+s42a = s7 from which
s12a = s5s2 = s6s3 = s7s4
So we're looking for integers that are the sum of two squares in two ways. There are plenty of such squares, eg, 50,65,200,221. The trick is to find such with factors that can be used in the other formulations. Ideally all the squares would be distinct.

Posted by xdog
on 20171217 13:12:22 