2*b^2 + b + 1 = n
(b+2)^2 + b + 2 = n
b^2 + 4*b + 4 + b + 2 = n
b^2 + 5*b + 6 = n
Subtracting the first equation from twice this gives
9*b + 11 = n
So
2*b^2 + b + 1 = 9*b + 11
2*b^2  8*b  10 = 0
b^2  4*b  5 = 0
b = (4 + sqrt(16 + 20)) / 2
= 2 + sqrt(9)
= 5
n = 9*5 + 11 = 56

Posted by Charlie
on 20160317 10:41:11 