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, 2012

Algebra Difficulty 4.6 AIME Prove it Saudi Arabia

Prove that for every positive integer $n$ the number
$N = \underbrace{44\dots4}_{n} \underbrace{88\dots8}_{n} - 1 \underbrace{33\dots3}_{n-1} 2,$
is a perfect square.

Solution

We have
$\begin{aligned} N &= \underbrace{44\dots4}_{n-1} \underbrace{355\dots56}_{n-1} = \underbrace{44\dots4}_{2n} - \underbrace{88\dots8}_{n} \\ &= 4(10^{2n-1} + \dots + 10 + 1) - 8(10^{n-1} + \dots + 10 + 1) \\ &= 4 \cdot \frac{10^{2n}-1}{9} - 8 \cdot \frac{10^n-1}{9} = \frac{1}{9}(4 \cdot 10^{2n} - 8 \cdot 10^n + 4) \\ &= \left(2 \cdot \frac{10^n-1}{3}\right)^2, \end{aligned}$
and we are done, since $\frac{10^n - 1}{3}$ is an integer. Notice that, in fact,
$N = \underbrace{66\dots6}_{n}^2.$

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