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Polynomial Proof


From: Karan
Subject: Theory of equations

What do I do? p(x) is a polynomial of nth degree with integer 
coefficients in x.

Prove that:

     p(a) = b
     p(b) = c
     p(c) = a

It can't simultaneously take place if a, b, and c are integers.

Thanks,
Karan


Date: 03/27/2001 at 13:56:40
From: Doctor Rob
Subject: Re: Theory of equations

Thanks for writing to Ask Dr. Math, Karan.

You must assume that no two of a, b, and c are equal. If any two are 
equal, all three are equal, and there exist such polynomials:

     p(x) = f(x)*(x-a) + a

for any polynomial f(x) with integer coefficients.

Assume a, b, and c do satisfy the three equations. Suppose that 
a = min(a,b,c). (The argument is similar when the smallest one is 
b or  c.) Then define:

     q(x) = p(x+a) - a

Then

     q(0) = b - a = d > 0
     q(d) = c - a = e > 0
     q(e) = 0

Then, from the first equation, you get:

     q(x) = x*r(x) + d

Applying this to the second and third equations,

     e = q(d) = d*r(d) + d = d*(r[d]+1)

     0 = q(e) = e*r(e) + d

The former of these two implies that d|e. The latter implies that e|d. 
Since both are positive, that implies that d = e, so b-a = c-a, so 
b = c, a contradiction.


Here's a slightly different, short proof:

Recall that for all integers n >= 0

     (x^n-y^n)/(x-y) = x^(n-1) + x^(n-2)*y + ... + x*y^(n-2) + y^(n-1)

This implies that for any polynomial p(x)

     (p(x)-p(y))/(x-y)

is also a polynomial with integer coefficients. This means that if x 
and y are integers, so is this quotient.

Now consider:

     (p(a)-p(b))/(a-b) = (b-c)/(a-b) = B
     (p(b)-p(c))/(b-c) = (c-a)/(b-c) = C
     (p(c)-p(a))/(c-a) = (a-b)/(c-a) = A

A, B, and C are all integers. Multiply them together, and you can see 
that A*B*C = 1, so each of A, B and C is either 1 or -1. If 
a = min(a,b,c), then A < 0; if b = min(a,b,c), then B < 0; and if 
c = min(a,b,c), then C < 0. In each case you get the two larger 
numbers equal, a contradiction.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Modern Algebra

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