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Proof That Equation Has No Integer Roots


Date: 05/09/2000 at 05:23:33
From: Kai Xi Liu
Subject: Polynomials

Dear Dr Maths,

I am wondering if you can help me with this question regarding 
polynomials:

Prove that if p is a prime number, then the equation:

     x^5 - px^4 + (p^2-p)x^3 + px^2 - (p^3+p^2)x - p^2 = 0 

has no integer roots.

How do I start on such a daunting problem? Can you please help me?

Awaiting your reply,
Kai Xi Liu


Date: 05/09/2000 at 14:42:56
From: Doctor Rob
Subject: Re: Polynomials

Thanks for writing to Ask Dr. Math.

The Rational Root Theorem says that if a polynomial has a rational 
root, then the numerator of the root must divide the constant term, 
and the denominator must divide the leading coefficient.

Applying that to your polynomial, the numerator of the root must 
divide p^2, and the denominator must divide 1. Putting these together, 
you can see that there are only six possible rational roots, all of 
which are integers:

     x = 1, -1, p, -p, p^2, or -p^2.

Try each and see that none of them works.

An alternative approach is this: according to the equation, if there 
is an integer solution x, then:

     x^5 = p*(x^4 - [p-1]*x^3 - x^2 + [p^2+p]*x + p),

so p divides x^5, so p divides x. Write x = p*y, substitute, and 
simplify.

   p^2*(p^3*y*5 - p^3*y^4 + [p-1]*p^2*y^3 + p*y^2 - [p+1]*p*y - 1) = 0

         p^3*y*5 - p^3*y^4 + [p-1]*p^2*y^3 + p*y^2 - [p+1]*p*y - 1 = 0

               p*(p^2*y^5 - p^2*y^4 + [p-1]*p*y^3 + y^2 - [p+1]*y) = 1

which implies that p is a divisor of 1. This is a contradiction, so 
there is no value of y that works, so there is no value of x that 
works.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Number Theory
High School Basic Algebra
High School Number Theory

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