Ponder This Challenge - September 2003 - Irrational root of polynomial

Ponder This Challenge:

This month's puzzle was submitted by Peter Beech.

For n>1 let P(x)= 1 + x1/1! + x2/2! + ... + xn/n! be the polynomial of degree n formed from the first n+1 terms of the power series for ex. When n is odd, this polynomial has one real root. Prove that this root is irrational for all odd n>1.

We will post the names of those who submit a correct, original solution! If you don't want your name posted then please include such a statement in your submission!

We invite visitors to our website to submit an elegant solution. Send your submission to the ponder@il.ibm.com.

If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com

Solution

• Click here to view the solution

  Solution:

  A rational root of Pn(x) is an integer, because n!Pn(x) = n! + n!x + ... + xn is a monic polynomial with integer coefficients (to see this let x = a/b in lowest terms, substitute into n!Pn(x) = 0 and multiply through by bn; then b divides an, a contradiction).

  Hence suppose Pn(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 Pn,

  ex = Pn(x) + Rn(x) and e-x = Pn(-x) + Rn(-x)

  where the remainder Rn(x) = x**n + 1 / (n + 1)! + ... It follows on multiplying these together that

  Pn(x) + Pn(-x) = 1 + x**n + 1gn(x)

  where gn(x) is a polynomial. Multiplying this through by (n)2 gives

  (n!)2Pn(x) + Pn(-x) = (n!)2 + x**n + 1Gn(x)

  where Gn(x) = (n!)2gn(x) must have integer coefficients because the coefficients on the left hand side are integers. Since Pn(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 / p2 + ... ≤ n / p + n / p2 + ... = n / (p - 1)

  where the [ ] means "greatest integer less than"(strict inequality holds but we do not need this).

  Therefore n + 1 ≤ 2n / (p - 1). But then p ≤ 1 + 2n / (n + 1) < 3, a contradiction since p is an odd prime.]

  ---

  If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com

Solvers

• Hagen von Eitzen (9.3.2003 @04:52:39 PM EDT)
• Coserea Gheorghe (9.4.2003 @01:17:52 PM EDT)
• Iqstar (9.4.2003 @05:00:00 PM EDT)
• Dave Blackston (9.5.2003 @03:06:17 PM EDT)
• David Friedman (9.5.2003 @03:48:21 PM EDT)
• Joseph DeVincentis (9.5.2003 @10:05:42 PM EDT)
• Patrick J. LoPresti (9.6.2003 @01:30:34 PM EDT)
• Henry Shin (9.6.2003 @09:27:03 PM EDT)
• Dmitry Litvinenko (9.6.2003 @01:32:27 AM EDT)
• Dan Dima (9.8.2003 @10:45:50 AM EDT)
• Mihai Cioc (9.9.2003 @12:03:32 PM EDT)
• John G. Fletcher (9.9.2003 @09:14:17 PM EDT)
• Ilan Algor (9.10.2003 @02:45:36 AM EDT)
• Surendar Nanchary (9.10.2003 @11:59:00 AM EDT)
• Francis Golding (9.10.2003 @12:10:47 PM EDT)
• Clive Tong (9.10.2003 @04:34:00 PM EDT)
• Andreas Knappe (9.15.2003 @04:19:00 AM EDT)
• Karl D'Sousa (9.15.2003 @11:44:00 AM EDT)
• R. Nandakumar (9.16.2003 @09:54:03 AM EDT)
• Vineet Gupta (9.16.2003 @06:50:37 PM EDT)
• Peter Huggins (9.17.2003 @05:43:53 PM EDT)
• David Woodruff (9.19.2003 @08:56:23 PM EDT)
• Sarang Aravamuthan (9.20.2003 @02:18:00 PM EDT)
• Frank Mullin (9.24.2003 @09:20:32 AM EDT)