<s>[INST] Check my solution of no.2 please 1.Show that the closest integer to $(2 + \sqrt{5})^{2017} -2^{2018}$  is divisible by $2017$. (Solved)
2.Suppose that $a,b,c$ are positive real numbers and that for some integer $2 \leq n$, $a^n + b^n = c^n$    
Let $k$ be an integer with $1 \leq k <n$. Prove that there is triangle with side length $a^k,b^k,c^k$
My $Sol$ of 2) $f(n)=a^n +b^n ,g(n)=c^n, 
f(n)-g(n) =h(n)=0, h'(n)=a^n log(a) + b^n log(b) - c^n log(c)$. $h'(n)$ is decreasing function, so by $k<n,h(k)=a^k+b^k-c^k>0 =h(n) $.
($a,b<c$ is trivial)
This proof is work?
 [/INST] Hint for 1. $(2+\sqrt{5})^{2017}+(2-\sqrt{5})^{2017}$ is an integer and $(2-\sqrt{5})^{2017}$ is very small. 
By setting $a_n=(2+\sqrt{5})^n+(2-\sqrt{5})^n$ we have $a_0=2, a_1=4$ and $a_{n+2}=4 a_{n+1}+a_n$.
$2017$ is a prime and $5$ is not a quadratic residue $\!\!\pmod{2017}$. By considering $\mathbb{F}_{2017}[x]/(x^2-4x-1)\simeq\mathbb{F}_{2017^2}$ and by taking $\alpha$ as a root of $x^2-4x-1$, by Vieta's Theorem and Frobenius map $x\to x^{2017}$ we get $\alpha+\alpha^p = \alpha^p+\alpha^{p^2}=4$.
It follows that
$$(2+\sqrt{5})^{2017}+(2-\sqrt{5})^{2017}-2^{2018}\equiv 4-4\equiv\color{red}{0}\pmod{2017}.$$
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