2014/016)solve n^2 + 20n + 11 = m^2 for integer m, n.

We have n^2 + 20n + 11 = m^2 

adding 89 on both sides we get
(n+10)^2 = m^2 + 89 or taking n+ 10 = p we get
P^2 – m^2 = 89 or (p+m)(p-m) = 89
As 89 is prime there are 4 cases
1) p+m = 89,p-m = 1 => p = 45, m = 44 or n = 35, m = 44
2) p+m = – 89,p-m = – 1 => p =- 45, m = – 44 or n = – 55 , m = -44
3) p +m = -1, p-m = – 89 =>
4) p+m = – 89, p-m = 1
one can solve (3) and (4) to give (35,-44) and (-55,44)