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/
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