The solution found by Diophantus (A.D. 250) is as follows: take 4 squares, say a²,b²,c²,d², and let n be their sum. Then the 4 rationals n/3-a²,n/3-b²,n/3-c²,n/3-d² are such that any 3 sum to a square; for example, the first 3 sum to n-(a²+b²+c²), which is d². To solve the problem we need 4 squares such that the numbers in this solution are positive; that is, we want n/3 to be larger than each of the squares. Using square integers, this happens when the four squares are 64,81,100,121, since their sum is 366 and 366/3=122>121. This yields the solution: 1,22,41,58.
It turns out that the method of Diophantus generates all such 4-tuples. |
