Three consecutive positive odd integers a,b and c satisfy b^{2}-a^{2}=\(48\) and c^{2}-b^{2}>0. What is the value of c^{2}-b{^2}?
Let a = 2n+ 1 and b = 2n + 3
So
(2n + 3)^2 - (2n + 1)^2 = 48
4n^2 + 12n + 9 - 4n^2 - 4n - 1 = 48
8n + 8 = 48 divide through by 8
n + 1 = 6
n = 5
a = 11 b = 13 c = 15
And
c^2 - b^2 =
15^2 - 13^2 = 56
