Power of $2$ that divides $\lceil(3+\sqrt5)^{2n}\rceil$

Question

$\lceil(3+\sqrt5)^{2n}\rceil$ is divisible by

A. $2^{n+1}$

B. $2^n$

C. $2^{n-1}$

D. not divisible by $2$

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Answer 1

If $f(n)=a^n+b^n,a>b$ where $a,b$ are the roots of $t^2-28t+16=0$

So,$f(m+2)-28f(m+1)+16f(m)=0,m\ge0$

As $0<b<1,$ we need the highest power of $2$ that divides $f(n)$ be $h(n)$

$h(0)=1$

$h(1)=2$

So, by strong induction, $h(n)\ge n+1$