Maths Olympiad Prep

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, 2012

Algebra Difficulty 5.6 AIME, harder Prove it Saudi Arabia

Sequence (x1,x2,… )(x_1, x_2, \dots) is defined as x1=20x_1 = 20, x2=12x_2 = 12,
xn+2=xn+xn+1+2xnxn+1+121, x_{n+2} = x_n + x_{n+1} + 2\sqrt{x_n x_{n+1} + 121},
for n≥1n \ge 1.
1) Compute x10x_{10}.
2) Determine with justification if every term in the sequence is an integer?

Solution

It is clear that x3=20+12+2⋅19=70x_3 = 20 + 12 + 2 \cdot 19 = 70. We note that
xn+3=xn+1+xn+2+2xn+1xn+2+121=xn+1+xn+2+2xn+1(xn+xn+1+2xnxn+1+121)+121=xn+1+xn+2+2xn+12+2xn+1xnxn+1+121+xnxn+1+121=xn+1+xn+2+2(xn+1+xnxn+1+121)=3xn+1+xn+2+2xnxn+1+121=3xn+1+xn+2+xn+2−xn−xn+1=2xn+2+2xn+1−xn. \begin{aligned} x_{n+3} &= x_{n+1} + x_{n+2} + 2\sqrt{x_{n+1}x_{n+2} + 121} \\ &= x_{n+1} + x_{n+2} + 2\sqrt{x_{n+1}(x_n + x_{n+1} + 2\sqrt{x_n x_{n+1} + 121}) + 121} \\ &= x_{n+1} + x_{n+2} + 2\sqrt{x_{n+1}^2 + 2x_{n+1}\sqrt{x_n x_{n+1} + 121} + x_n x_{n+1} + 121} \\ &= x_{n+1} + x_{n+2} + 2(x_{n+1} + \sqrt{x_n x_{n+1} + 121}) \\ &= 3x_{n+1} + x_{n+2} + 2\sqrt{x_n x_{n+1} + 121} \\ &= 3x_{n+1} + x_{n+2} + x_{n+2} - x_n - x_{n+1} \\ &= 2x_{n+2} + 2x_{n+1} - x_n. \end{aligned}
Therefore, (x1,x2,… )(x_1, x_2, \dots) is an integer sequence and it is straightforward to find x4=144x_4 = 144, x5=416x_5 = 416, x6=1050x_6 = 1050, x7=2788x_7 = 2788, x8=7260x_8 = 7260, x9=19046x_9 = 19046, and x10=49824x_{10} = 49824.

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