Using the identity (A^{2} + C^{2})(B^{2} + D^{2}) = (AB + CD)^{2} + (AD  BC)^{2},
(A^{2} + C^{2})(B^{2} + D^{2}) = 2016^{2} + 1^{2}
= 4064257
= 317 x 12821 (both primes) *
= (11^{2} + 14^{2})(70^{2} + 89^{2}) *
Since 14 x 70 > 11 x 89, of the four possible allocations of these
numbers to A, B, C, D, only two satisfy the original equations: viz
{A, B, C, D} = {14, 89, 11, 70} or {A, B, C, D} = {70, 11, 89, 14}
Done without a computer? Well not quite (*), but Srinivasa would
have no trouble.
Edited on May 26, 2016, 7:44 pm
Edited on May 26, 2016, 7:45 pm

Posted by Harry
on 20160526 19:44:07 