Bound of Chebyshev polynomials I'm learning about Chebyshev polynomial (First kind) $T_n$ and I'm confusing with this problem:
" Prove that with all $n>5$ we have this inequality:
$\vert(\sqrt 3)^{n}\cdot T_n(\dfrac{1}{\sqrt 3})\vert >1$ "
 A: The combinatorial facts needed are that if $n=2k$, ${n \choose 2}=2$ iff $n=2$. 
This means that $\Re (1+i\sqrt 2)^{2N}=-1$ iff $N=2$ (all other terms in the expansion are divisible by $4$)
For $n=4m+1$, if $\nu_2(n-1)=k \ge 2$ ($\nu_2$ being the $2$ valuation so highest power of $2$ dividing the number) then $\nu_2{n-1 \choose 2^q}=k-q, q \le k$ so $2^{2^{q-1}}{n-1 \choose 2^q}$ divides by $2^{k+1}$ for $q \ge 3$ and then $2^r{n-1 \choose 2r}$ divides by $2^{k+1}$ for any $r \ge 3$ since${n-1 \choose 2^q}$ are local minima as is easily seen by writing the combinations out. 
This means that if $n=4m+1$ the only way  $\Re (1+i\sqrt 2)^{4N+1}=1$ iff $N=1,5$ (as all terms except $2{4N+1 \choose 2}, 4{4N+1\choose 4}$ divide by a higher power of $2$ hence to get $1$ either $4N+1=1$ or $2{4N+1 \choose 2}=4{4N+1\choose 4}$ so $4N+1=5$
Let $a_n=T_n(\dfrac{1}{\sqrt 3})$. The recurrence for $T_n(x)=\cos n\theta, x=\cos \theta, -1 \le x \le 1$ is $T_{n+1}(x)-2xT_n(x)+T_{n-1}(x)=0, T_0(x)=1, T_1(x)=x$ so,
$a_{n+1}-\frac{2}{\sqrt 3}a_n+a_{n-1}=0, a_0=1, a_1=\frac{1}{\sqrt 3}$
Solving the recurrence we get $a_n=\frac{1}{2}(c^n+\bar c^n), c=\frac{\sqrt 3 + i\sqrt 6}{3}$, so 
$|\vert(\sqrt 3)^{n}\cdot T_n(\dfrac{1}{\sqrt 3})\vert =|\frac{1}{2}(b^n+\bar b^n)|=|\Re b^n|, b=1+i \sqrt 2$. 
Now clearly $|\Re b^n|$ is an odd integer so we need to show that it is not $\pm 1$ when $n >5$ (note that it is $1$ for $n=5$)
If $\Re b^n=-1$ then $b^{2n}+2b^n+3^n=0$ so $x^2-2x+3/x^{2n}+2x^n+3^n$ and with $x=-1$ we get $n$ even hence the first paragraph shows $n=2$
If $\Re b^n=1$ then $b^{2n}-2b^n+3^n=0$ so $x^2-2x+3/x^{2n}-2x^n+3^n$ and with $x=-1$ we get $n$ odd hence the first paragraph shows $n=1,5$ (as ${n \choose 2}$ must be even since the next term in the expansion of $\Re (1+i\sqrt 2)^{n}$ is divisible by $4$, $n=4N+1$)
Done!
Notes: I would love an easier solution than the above - finding a solution is equivalent (by taking norms) to showing that the Diophantine equation $3^n=2q^2+1$ has only $(1,1),(2,2), (5,11)$ as roots in non zero positive integers, but I do not see a direct approach to that. Also as $T_n$ has $n$ roots at the appropriate rational multiples of $\pi$, the problem essentially requires showing that $\cos^{-1} \frac{1}{\sqrt 3}$ stays away far enough from those numbers, so the problem is definitely non-trivial.
