2014/080) solve $n^2+20n+11=m^2$ for integer n,m

 adding 89 on both sides we get

We have $n^2 + 20n + 100 = m^2 + 89$

$(n+10)^2 = m^2 + 89$ 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 given (35,-44) and (-55,44)