## September 2003

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Solution:

A rational root of *P _{n}*(

*x*) is an integer, because

*n*!

*P*(

_{n}*x*) =

*n*! +

*n*!

*x*+ ... +

*x*is a monic polynomial with integer coefficients (to see this let

^{n}*x*=

*a*/

*b*in lowest terms, substitute into

*n*!

*P*(

_{n}*x*) = 0 and multiply through by

*b*; then

^{n}*b*divides

*a*, a contradiction).

^{n}Hence suppose P

_{n}(

*x*) = 0 with

*x*an integer. Now choose a prime

*p*dividing

*n*; then

*p*is odd since

*n*is odd, and since

*n*! +

*n*!

*x*+ ... +

*nx*

^{n-1}= 0, it follows that

*p*also divides

*x*.

Next, by definition of

*P*,

_{n}*e*=

^{x}*P*(

_{n}*x*) +

*R*(

_{n}*x*) and

*e*

^{-x}=

*P*(-

_{n}*x*) +

*R*(-

_{n}*x*)

where the remainder

*R*(

_{n}*x*) =

*x*

^{n + 1}/ (

*n*+ 1)! + ... It follows on multiplying these together that

*P*(

_{n}*x*) +

*P*(-

_{n}*x*) = 1 +

*x*

^{n + 1}

*g*(

_{n}*x*)

where

*g*(

_{n}*x*) is a polynomial. Multiplying this through by (

*n*)

^{2}gives

(

*n*!)^{2}*P*(_{n}*x*) +*P*(-_{n}*x*) = (*n*!)^{2}+*x*^{n + 1}*G*(_{n}*x*)where

*G*(

_{n}*x*) = (

*n*!)

^{2}

*g*(

_{n}*x*) must have integer coefficients because the coefficients on the left hand side are integers. Since

*P*(

_{n}*x*) = 0, we find that (

*n*!)

^{2}is divisible by

*x*

^{n + 1}, hence by

*p*

^{n + 1}where

*p*is the odd prime chosen above.

But by Legendre's formula the greatest power of

*p*that divides

*n*! is

[

*n*/*p*+*n*/*p*^{2}+ ... ≤*n*/*p*+*n*/*p*^{2}+ ... =*n*/ (*p*- 1)where the [ ] means "greatest integer less than"(strict inequality holds but we do not need this).

Therefore

*n*+ 1 ≤ 2

*n*/ (

*p*- 1). But then

*p*≤ 1 + 2

*n*/ (

*n*+ 1) < 3, a contradiction since

*p*is an odd prime.]

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