No Rational Roots of $x^n+\cdots+1=0$

[melese]

(BGR,1989) Prove that for any integer $n>1$ the equation $\displaystyle \frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=0$ has no rational roots.

 

[Albert1]

(BGR,1989) Prove that for any integer $n>1$ the equation $\displaystyle \frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=0$ has no rational roots.

Spoiler

$\displaystyle \frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=0----(*)$
if n=2 then we have :$x^2+2x+2=0$ no real solution
if m is a solution of original equation then m<0
multiply both sides with n ! we obtain :
$x^n+nx^{n-1}+-----+n! x+ n!=0$
using "the rational zero theorem"
if the original equation has a rational solution m<0 then n! must be a multiple of it(m is a negative integer)
replacing x with any negative factor of n! to (*)will not be zero ,so no rational root exists
in fact :
---+$\displaystyle \frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=e^x$
[FONT=&#32048](Maclaurin expasion of $e^x$)[/FONT]

 

[melese]

I agreed with you up to

Spoiler

replacing x with any negative factor of n! to (*)will not be zero

Spoiler

This part is a little vauge to me. To see what I mean, what if my orginal question was to show that there are no integer solutions? - Then your step appears hasty.

መለሰ

 

[Albert1]

$\displaystyle \frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=0----(*)$

Spoiler

if n is even and m (a negative integer ) is a root then (*) becomes
$\displaystyle \frac{m^n}{n!}+\frac{m^{n-2}}{(n-2)!}\cdots+\frac{m^2}{2!}+1$
$- \dfrac{k^{n-1}}{(n-1)!}- \dfrac{k^{n-3}}{(n-3)!}\cdots-\dfrac{k}{1!}$
will not be zero,(the calculation is very tedious)
here k=-m is a positive integer
likewise if n is odd then (*) becomes
$\displaystyle -\frac{k^n}{n!}-\frac{k^{n-2}}{(n-2)!}\cdots-\frac{k^2}{2!}+1$
$+ \dfrac{m^{n-1}}{(n-1)!}+ \dfrac{m^{n-3}}{(n-3)!}\cdots+\dfrac{m}{1!}$
also will not be zero
so there is no integer root for original equation

 

[CellCoree]

I would first clarify that the statement provided is not entirely accurate. The actual statement should be "no rational roots of $x^n+\cdots+1=0$ exist for any integer $n>1$." This is because the statement implies that there are no rational roots at all, which is not true. For example, the equation $x+1=0$ has a rational root of $-1$.

Moving on to the actual content, the claim that the equation $\frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=0$ has no rational roots is also not entirely accurate. It should be stated that for any integer $n>1$, the equation has no rational roots other than the possible root of $-1$. This is because when $x=-1$, all the terms in the equation become $0$, making the equation true.

To prove this statement, we can use the rational root theorem. According to this theorem, if a polynomial equation has a rational root, it must be of the form $\frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. In this case, the constant term is $1$ and the leading coefficient is $\frac{1}{n!}$.

For the equation to have a rational root other than $-1$, $p$ must be a factor of $1$ and $q$ must be a factor of $\frac{1}{n!}$. Since $n>1$, $n!$ is greater than $1$ and therefore, the only possible factor of $\frac{1}{n!}$ is $1$. This means that $q$ can only be $1$, making the root $\frac{p}{q}=p$, where $p$ is any integer.

Substituting $x=p$ in the equation, we get $\frac{p^n}{n!}+\frac{p^{n-1}}{(n-1)!}\cdots+\frac{p^2}{2!}+\frac{p^1}{1!}+1=0$. This can be simplified to $\frac{p^n}{n!}+\frac{p^{n-1}}