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

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The discussion centers on proving 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 for any integer $n>1$. The participants reference a proof from BGR (1989) to support this claim. The equation's structure suggests that the coefficients derived from factorials contribute to the absence of rational solutions. The conversation emphasizes the mathematical reasoning behind the proof and the implications of the findings. Ultimately, the conclusion reinforces the assertion that rational roots do not exist for this polynomial equation.
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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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