Prove by strong induction that $2^n$ divides $p_n$ for all integers n ≥ 1 
Let $p_1 = 4$, $p_2 = 8$, and $p_n = 6p_{n−1} − 4p_{n−2}$ for each integer $n ≥ 3$.
  Prove by strong induction that $2^n$ divides $p_n$ for all integers $n ≥ 1$

I got up to the base step where by you prove for $p_3$ but unsure about the strong induction step

If $2^n$ divides $p_n$ then we have that $p_n = \lambda_n 2^n$ for some integer $\lambda_n$.  Therefore, the inductive hypothesis states that:
$$
p_n = \lambda_n2^n = 6p_{n - 1} - 4p_{n - 2}
$$
If we assume that it holds for both $n - 1$ and $n - 2$ (this is the strong induction) then we get that: $p_{n - 1} = \lambda_{n - 1}2^{n - 1}$ and $p_{n - 2} = \lambda_{n - 2}2^{n - 2}$.  Putting this together we find that:
\begin{align}
6p_{n - 1} - 4p_{n - 2} =&\ 6\lambda_{n - 1}2^{n - 1} - 4\lambda_{n - 2}2^{n - 2}\\
=&\ 2^n \left(\frac{6\lambda_{n - 1}}{2^1} - \frac{4\lambda_{n - 2}}{2^2}\right) \\
=& 2^n \left(\frac{6}{2}\lambda_{n - 1} - \frac{4}{4}\lambda_{n - 2}\right) \\
=&\ 2^n \left(3\lambda_{n - 1} - \lambda_{n - 2}\right)
\end{align}
Therefore $2^n$ divides $p_n$ by a factor of $3\lambda_{n - 1} - \lambda_{n - 2}$ (which are both assumed to be integers through the inductive hypothesis).
