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 $2$

https://math.stackexchange.com/questions/127893/showing-that-lceil-sqrt3-12n-rceil-is-divisible-by-2n1 — lab bhattacharjee, Apr 30 '19 at 09:54
See also https://math.stackexchange.com/questions/2202564/product-of-integral-and-fractional-part-of-binomial-expansion — lab bhattacharjee, Apr 30 '19 at 09:55

score 1 · Answer 1 · answered Apr 30 '19 at 08:36

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, $h(n)\ge n+1$

answered Apr 30 '19 at 08:36

lab bhattacharjee

1 Answers1