Using the identity (A2 + C2)(B2 + D2) = (AB + CD)2 + (AD - BC)2,
(A2 + C2)(B2 + D2) = 20162 + 12
= 4064257
= 317 x 12821 (both primes) *
= (112 + 142)(702 + 892) *
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
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Posted by Harry
on 2016-05-26 19:44:07 |