A Problem Seminar - D.J. Newman

paul_mathematics · Hodge Conjecture Offline Joined: 17 Mar 2004 Posts: 58

A and B are positive integers. xi = (A + 1/2)^i + (B + 1/2)^i. Prove that only finitely many of x1, x2, x3, ... are integers.

**Posted:** Tue Feb 15, 2005 1:34 pm

pbornsztein

Let $x=2A+1$ and $y=2B+1$ , which both are odd integers.
The number $x_n$ is an integer if and only if $2^n$ divides $x^n+y^n$ .
But, if $n>0$ is even $x^n+y^n = 2 \mod [4]$ thus $x_n$ is not an integer.
If $n$ is odd then $x^n+y^n = (x+y)(x^{n-1} - x^{n-2}y + \cdots + y^{n-1})$ and the second factor is clearly odd. Thus $x_n$ is an integer if and only if $2^n$ divides $x+y$ which can only occur a finite number of times.

Pierre.

**Posted:** Tue Feb 15, 2005 1:45 pm