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

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The equation $\frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}+\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=0$ has been proven to have no rational roots for any integer $n>1$. This conclusion is established in the work of BGR (1989), which provides a comprehensive proof of the absence of rational solutions for this polynomial equation. The discussion emphasizes the significance of factorial terms in the polynomial and their impact on the rational root theorem.

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(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.
 
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melese said:
(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.
$\displaystyle \frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=0----(*)$
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 $e^x$)
 
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I agreed with you up to
Albert said:
replacing x with any negative factor of n! to (*)will not be zero
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.
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Albert said:
$\displaystyle \frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=0----(*)$
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
 
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