Let $p$ be an odd prime.
Prove that the integer part of
(\sqrt{5}+2)^p - 2^{p+1}
is divisible by $20p$.
Solution. Let
$(\sqrt{5}+2)^p - 2^{p+1} = I + F$, where $I$ is an integer and $0 <
F <1$.
We need to show that $20p$ divides $I$.
Let $f=(\sqrt{5}-2)^p$. Since $0<\sqrt{5}-2<1$, we have $0<f
< 1$. Therefore, $-1 < -f < 0$ and $-1 < F-f < 1$.
But
\begin{eqnarray}
N+F-f & = & (\sqrt{5}+2)^p - 2^{p+1} - (\sqrt{5}-2)^p \\
& = & 2(p\cdot 1\cdot 2\cdot 5^{(p-1)/2} + p\cdot 3\cdot
2^3\cdot 5^{(p-3)/2} \\
& \qquad & {}+\cdots + p\cdot (p-2)\cdot 2^{p-2}\cdot 5 + 2^p)
- 2^{p+1},
\end{eqnarray}
since $p$ is odd. Therefore $F-f$ is an integer, which must be 0 since
$-1 < F-f < 1$.
Thus, by factoring the greatest common factor, we get
N=4\cdot 5(p\cdot 1\cdot 5^{(p-3)/2} + p\cdot 3\cdot
2^2\cdot 5^{(p-5)/2} + \cdots + p(p-2)\cdot 2^{p-1}),
But $p$ is a multiple of every term, thus $20p$ divides $N$.