Let $n >1$ be a natural number and $T_{n}(x)=x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ... + a_1 x^1 + a_0$. Assume that for each nonzero real number $t $ we have $T_{n}(t+\frac {1}{t})=t^n+\frac {1}{t^n} $. Prove that for each $0\le i \le n-1 $ $gcd (a_i,n) >1$. Proposed by Morteza Saghafian
Problem
Source: Iran RMM TST 2019,day1 p2
Tags: polynomial, algebra, chebyshev polynomial
ThE-dArK-lOrD
30.07.2019 21:30
The polynomial $T_n$ in the problem must equal to $2\mathcal{T}_n(x/2)$ where $\mathcal{T}_n$ is the Chebyshev polynomial. We then get
$$T_n(x)=n\sum_{k=0}^{\lfloor n/2\rfloor }\frac{(-1)^k(n-k-1)!}{k!(n-2k)!}x^{n-2k}.$$The $\nu_p$ of the coefficient is equal to $$\nu_p(n)+\sum_{t=1}^{\infty}{\left( \left\lfloor \frac{n-k-1}{p^t}\right\rfloor -\left\lfloor \frac{k}{p^t}\right\rfloor -\left\lfloor \frac{n-2k}{p^t}\right\rfloor \right)} =\sum_{t=1}^{\infty}{\left( \left\lfloor \frac{n-k-1}{p^t}\right\rfloor -\left\lfloor \frac{k}{p^t}\right\rfloor -\left\lfloor \frac{n-2k}{p^t}\right\rfloor +1_{p^t\mid n}\right)}.$$Not hard to prove the term in the last summation is non-negative so the coefficient is indeed an integer.
To prove the required conclusion, if the term is zero for all prime $p\mid n$ we get that $p^t\mid k$ for all $t$ that $p^t\mid n$.
When $p$ runs through all prime divisor of $n$ we get $n\mid k$ or $k=0\implies n-2k=n$.
jeusson
05.04.2020 15:43
I solved it by induction (p->pm) First step, Tn(x) is integer cofficient polynomial. Second step, n=p (so easy, express (x+1/x)^p) Third step, Tpm(x)=Tp(Tm(x)) and induction
AFSA
21.05.2020 16:41
I have completely different and short solution. note that $\mathcal{T'}_n(x)=n\mathcal{U}_{n-1}(x)$ where $\mathcal{T}_n(x)$ is first chebyshev polynomial and $\mathcal{U}_n(x)$ is second chebyshev polynomial The polynomial $T_n$ in the problem must equal to$\mathcal{C}_n(x)=2\mathcal{T}_n(x/2)$ from the first equation we know $\mathcal{C'}_n(x)=2n\mathcal{U}_{n-1}(x/2)$ notice that $\mathcal{U}_n(x/2)\in\mathbb{Z}[x]$ and if n is odd $a_0=0$ and if n is even $a_0=2$ so we are done.