IBM Ponder This #056 · December 2002

This month's puzzle comes from S V Nagaraj.

Find four distinct integers a,b,c,d such that
100 <= a,b,c,d <= 12000 and the sum of "all" the four of them
and the sum of "every" two of them are perfect squares.

(That is, a+b, a+c, a+d, b+c, b+d, c+d, and a+b+c+d are all
squares of integers.)

Solution

Best opened after a real attempt

To be added.