Let $n, \in \mathbb {N^*} , n\ge 3$ a) Prove that the polynomial $f (x)=\frac {X^{2^n-1}-1}{X-1}-X^n $ has a divisor of form $X^p +1$ where $p\in\mathbb {N^*} $ b) Show that for $n=7$ the polynomial $f (X) $ has three divisors with integer coefficients .
Problem
Source: Moldova TST 2018
Tags: algebra, polynomial
06.04.2018 14:41
Remark: a) If $n=2k$, the first problem is trivial, because if we denote ${{2}^{n}}-1=m>n$, we have $f(x)={{x}^{m-1}}+{{x}^{m-2}}+...+x+1-{{x}^{2k}}$, so on the basis of Bezout theoreme because we have $f(-1)=0$, result that we have $f$ divisible by $x+1$ And if $n=4k+1$, we have $f(i)=f(-i)=0$, so $f$ is divisible by $x^2+1$. It remain to see the situation $n=4k+3$.
06.04.2018 14:48
$f_n (x)=\frac {X^{2^n-1}-1}{X-1}-X^n $ a)$1+x^{v_2(n)} | f_n(x)$ b)$\frac{x^7-1}{x-1} | f_7(x)$
11.11.2021 19:57
rmtf1111 wrote: $f_n (x)=\frac {X^{2^n-1}-1}{X-1}-X^n $ a)$1+x^{v_2(n)} | f_n(x)$ It's not true when n is odd. You can't say that $2|f_n(x)$ for odd n.
11.11.2021 20:01
$(a)$ $\bullet$ If $n$ is even, choose $p=1$. $\bullet$ If $n$ is odd, choose $p=2^{v_2(n+1)}$
11.11.2021 20:08
For $(b)$ from $(a)$ we get $x^8+1$ is one of the factors. Can someone find $2$ more factors???