Question

If P (x) is a polynomial with integer coefficients and a, b, c, three distinct integers, then show that it is impossible to have P (a) = b, P (b) = c, P (c) = a.

Answer 1

@nathan917

Answer 2

assume that p(x) = a(n).x^n + a(n-1).x^(n-1) + ... + a(1).x + a(0)

note that for all integers x and y, we have (x - y) | (p(x) - p(y)) , because:
p(x) - p(y) = a(n).(x^n - y^n) + a(n-1).(x^(n-1) - y^(n-1)) + ... + a(1).(x - y)

and each of terms (x^k - y^k) can be written as (x - y) (x^(k-1) + x^(k-2).y + ... + y^(k-1)) .

So we have : (a - b) | (p(a) - p(b)) ⇒ (a - b) | (b - c)
and similarly (b - c) | (c - a) and (c - a) | (a - b) , thus:

(b - c) = r(a - b) and (c - a) = s(b - c) and (a - b) = t(c - a) , by multiplying these equations we have:

(b - c)(c - a)(a - b) = r.s.t (a - b)(b - c)(c - a) ⇒ because (a - b), (b - c) and (c - a) are nonzero we have r.s.t = 1 , and because r, s and t are integers, they should be 1 or -1 .

if r = -1 , then b - c = b - a ⇒ a = c (contradiction!)
similarly s = -1 and t = -1 result in contradictions.

if r = s = t = 1 , then b - c = a - b , c - a = b - c and a - b = c - a , and solving this system of equations results in a = b = c (contradiction!)

So it is impossible that P(a) = b, P(b) = c, and P(c) = a.

Answer 3

ok

Answer 4

Thanks @nathan917