B=2016A
A^2+2016A=C^2
A=(1+sqrt (4C^28063))/2 [1]
4C^28063 is square if 8063=4Cnn^2=n(4Cn)
As 8063=733*11 (both primes)
Or n=11 and 4Cn=733 =>C=186
Or n=733 and 4Cn=11 =>C=186
Then from [1] A=(1+361)/2 A=181 and 180
B=1835 and 2196
We are asked for positive integers.
Anyway there is a solution also con A negative
So:
181+1835=2016 181^2+1835=186^2
(180)+2196=2016 (180)^2+2196=186^2
Edited on June 22, 2016, 5:11 pm

Posted by armando
on 20160622 17:08:37 