mathematical induction

Let P(n) be the proposition that for any positive integer n
(3 + √5)^n + (3 - √5)^n is divisible by 2^n
When n = 1, LHS = (3 + √5) + (3 - √5) = 6
RHS = 2
LHS is divisible by RHS => P(1) is true
When n = 2, LHS = (3 + √5)^2 + (3 - √5)^2
= 9 + 6√5 + 5 + 9 - 6√5 + 5
= 28
RHS = 2^2 = 4
LHS is divisible by RHS => P(2) is true
Suppose P(k) is true, i.e. (3 + √5)^k + (3 - √5)^k = A(2^k) where A is an integer;
and P(k+1) is true, i.e. (3 + √5)^(k+1) + (3 - √5)^(k+1) = B[2^(k+1)] where B is an integer; then
(3 + √5)^(k+1) + (3 - √5)^(k+1) = B[2^(k+1)]
[(3 + √5)^(k+1) + (3 - √5)^(k+1)][(3 + √5) + (3 - √5)] = B[2^(k+1)][(3 + √5) + (3 - √5)]
(3 + √5)^(k+2) + (3 + √5)(3 - √5)^(k+1) + (3 - √5)(3 + √5)^(k+1) + (3 - √5)^(k+2) = B[2^(k+1)](6)
(3 + √5)^(k+2) + (3 - √5)^(k+2) + (3 + √5)(3 - √5)[(3 - √5)^k + (3 + √5)^k] = 3B[2^(k+2)]
(3 + √5)^(k+2) + (3 - √5)^(k+2) + (9 - 5)[(3 - √5)^k + (3 + √5)^k] = 3B[2^(k+2)]
(3 + √5)^(k+2) + (3 - √5)^(k+2) + 4[A(2^k)] = 3B[2^(k+2)]
(3 + √5)^(k+2) + (3 - √5)^(k+2) + A[2^(k+2)] = 3B[2^(k+2)]
(3 + √5)^(k+2) + (3 - √5)^(k+2) = (3B - A)[2^(k+2)] => P(k+2) is true
Hence by the principle of MI, P(n) is true for all positive integer n.