Remember
a^2-b^2=(a-b)(a+b)
16=4^2
x^4=(x^2)^2
so
16x^4-1=(4x^2)^2-(1)^2=(4x^2-1)(4x^2+1)
and the 4x^2-1 can be further factored
(2x)^2-(1)^2=(2x-1)(2x+1)
factoref rom is
(4x^2+1)(2x+1)(2x-1)
D is answer
24= 3*2^3
88= 11*2^3
664= 83*2^3 -> 83=11+72 = 11 + 2^3*3^2
664=2^3 (11+2^3*3^2) = 88 +(2^3*2^3*3^2) = 88 +(24^2)
8408= 1051 * 2^3 -> 1051= 83+968 -> 968 = 2^3 * 11^2
8408= 2^3 (83+2^3*11^2) = 664 +(2^3*2^3*11^2) = 664 +(88^2)
So:
a(n) = a(n-1) + a(n-2)^2
Lets check: 88+24^2= 664
664+88^2= 8408