Supposing that there is a set (a,b,c,d,e,f) and a+b, a+c, a+d, b+c... are all squares. Then each member of the set can be paired with 5 others but a+b=b+a, so there are 5*6/2=15 combinations. 5(a+b+c+d+e+f)=sum of all 15 squares.
If a,b and c are odd, a+b, b+c and a+c are even, and must be multiples of 4 since they are square. But a+b+c is odd, and a+b+csad(a+b)+(b+c)+(a+c))/2 which is even. Therefore at most 2 of the integers can be odd.
If a is odd, and the rest are even, a+b,b+c≡1(mod4), b+c≡2-a(mod4), a≡2(mod4).
If a and b are odd, and the rest are even, a+b≡0(mod4), a+c,b+c≡1(mod4), a+2c+b≡2(mod4), c≡1(mod4) but c is meant to be even.
The only remaining possibility is that all 6 are even. Every pair a,b is either both 2(mod4) or both 0(mod4) since (2x)^2==0(mod4). If they are all 0(mod4), they can all be divided by 4 until a set is obtained which is all 2(mod4).
2(mod4)≡2 or 6 (mod8). If a and b are 2(mod8), a+b≡(4(mod8) and cannot be square. If a and b are 6(mod(8), a+b are 12(mod8) and cannot be square. In any set of 6 intergers ≡2(mod8), there will always be a pair both 2(mod8) or both 6(mod8).

Question: What is the largest set possible such that every 3 sum to a square? Are there any sets of 4 such that every 3 sum to a cube?

Last edited by namealreadychosen (2011-07-27 17:29:10)