Divisibility problem involving the $2015^{th}$ power

Question

Show that the number $$ (5+2\sqrt6)^{2015} + (5-2\sqrt6)^{2015} - 10$$ is divisible by $960$.

Answer 1

Well, the sequence $a_n=(5+2\sqrt6)^n+(5-2\sqrt6)^n-10$ must follow some linear recurrence relation like $a_n=10a_{n-1}-a_{n-2}+80$ with initial conditions $a_1=0,\;a_2=88$, and every other of these seems to be divisible by 960, which should be easy to prove by induction.

Answer 2

Your number $+10$ is equal to $$\sum_{k=0}^{2015} \binom{2015}{k} 5^{2015-k} (2 \sqrt{6})^k + \sum_{k=0}^{2015} \binom{2015}{k} 5^{2015-k} (-1)^k(2 \sqrt{6})^k$$ Now, when $k$ is odd the terms sum to $0$ so you are left to even terms, i.e. $$2 \cdot \sum_{m=0}^{1007} \binom{2015}{2m} 5^{2015-2m} (2 \sqrt{6})^{2m}$$ where $2m=k$ are the even terms of the previous sum. This is equal to $$2 \cdot \sum_{m=0}^{1007} \binom{2015}{2m} 5^{2015-2m} 8^m3^m$$ Now, for $m \neq 0,1$ the number $$5^{2015-2m} 8^m3^m$$ is divisible by $5$, by $64=8^2$ and by $3$. So we are left to the first two terms $$2 \cdot \binom{2015}{0} 5^{2015} + 2 \cdot \binom{2015}{2} 5^{2013} \cdot 24 = 5^{2013} \cdot 2 (5^2 + 2015 \cdot 1007 \cdot 24)$$ Clearly this is divisible by $5$. Modulo $3$ this is $$(-1)^{2013}(-1)(1+0) \equiv 1 \equiv 10 \mod{3}$$ while modulo $64$ we have to work harder.