On the graph of a polynomial with integer coefficients, two points are chosen with integer coordinates. Prove that if the distance between them is an integer, then the segment that connects them is parallel to the horizontal axis.
Problem
Source: Spanish Mathematical Olympiad. 2015. Problem 1
Tags: algebra, polynomial
30.04.2015 17:54
tchebytchev wrote: On the graph of a polynomial with integer coefficients, two points are chosen with integer coordinates. Prove that if the distance between them is an integer, then the segment that connects them is parallel to the horizontal axis. Just write (P(a)−P(b))2+(a−b)2=c2 and notice that a−b|P(a)−P(b) So (P(a)−P(b)a−b)2+1=d2 And it's easy to conclude from n2+1=m2 that n=0 Q.E.D.
08.07.2015 17:43
This problem was on Moscow's Math Olympiad many years ago.
29.04.2024 16:39
The distance between two points with x coordinates p and q respectively, is given by D=√(q−p)2+(f(q)−f(p))2 As f is a polynomial with integer coefficients, we have q−p|f(q)−f(p). We rewrite this divisibility as f(q)−f(p)=k(f(q)−f(p)), where k is an integer. This forces (q−p)√(k2+1)∈Z+, which means k=0. Thus we have f(p)=f(q).
02.08.2024 15:11
Let |a−b|=x,|P(a)−P(b)|=kx, where x,k∈N0. From the Pythagorean Theorem, we require k2+1=m2 for some m∈Z, but from here we can easily conclude k=0. ◻