Maths Olympiad Prep

Library / /7 of 63

Algebra Difficulty 7.6 National olympiad, round 2 Prove it Turkey

The sequence {xn}\{x_n\} is defined by x1=ax_1 = a, x2=bx_2 = b and xn=2008xn−1−xn−2x_n = 2008x_{n-1} - x_{n-2} for all n≥2n \ge 2. Prove that there are positive integers aa and bb such that for all n≥1n \ge 1 the expression 1+2006xnxn+11 + 2006x_nx_{n+1} is a perfect square. (Şahin Emrah).

Solution

We prove that at a=1a = 1, b=2008b = 2008 all terms of the sequence are perfect squares.
Let us prove by induction that for all n≥1n \ge 1
xn2+xn+12−1=2008xnxn+1.(1) x_n^2 + x_{n+1}^2 - 1 = 2008x_n x_{n+1}. \quad (1)
1. n=1:12+20082−1=2008⋅1⋅2008n = 1 : 1^2 + 2008^2 - 1 = 2008 \cdot 1 \cdot 2008.
2. Suppose (1) is held for n=k:xk2+xk+12−1=2008xkxk+1n = k : x_k^2 + x_{k+1}^2 - 1 = 2008x_k x_{k+1}.
Then xk2+xk+12−1=xk⋅2008xk+1=xk⋅(xk+xk+2)=xk2+xkxk+2x_k^2 + x_{k+1}^2 - 1 = x_k \cdot 2008x_{k+1} = x_k \cdot (x_k + x_{k+2}) = x_k^2 + x_k x_{k+2}.
Then xk+12−1=xkxk+2=(2008xk+1−xk+2)xk+2=2008xk+1xk+2−xk+22x_{k+1}^2 - 1 = x_k x_{k+2} = (2008x_{k+1} - x_{k+2})x_{k+2} = 2008x_{k+1}x_{k+2} - x_{k+2}^2.
Therefore, xk+12+xk+22−1=2008xk+1xk+2x_{k+1}^2 + x_{k+2}^2 - 1 = 2008x_{k+1}x_{k+2} and (1) is held for n=k+1n = k + 1.

Now we get 1+2006xnxn+1=1+2008xnxn+1−2xnxn+1=xn2+xn+12−2xnxn+1=(xn+1−xn)21+2006x_n x_{n+1} = 1+2008x_n x_{n+1}-2x_n x_{n+1} = x_n^2+x_{n+1}^2-2x_n x_{n+1} = (x_{n+1} - x_n)^2. Done.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.