Perfect square for every n

by MM » Wed Sep 17, 2008 9:27 am

Prove that the number $\underbrace{44\ldots4}_{n}\underbrace{88\ldots8}_{n-1}9$ is perfect square for every natural number n.

by **Rock'n'roller** » Wed Sep 17, 2008 11:59 am

Just the formula for the sum of the first n numbers of geometric progression

$10^1 + ... + 10^n = \frac{10^{n+1}-1}{9}$

The number is equal to $4(10^{2n-1} + ... + 10^n) + 8(10^{n-1} + ... + 10) + 9$

But $10^{2n-1} + ... + 10^n = (10 + ... + 10^{2n-1}) - (10 + ... + 10^{n-1}) = \frac{10^{2n} - 1}{9} - \frac{10^{n} - 1}{9}$ $= \frac{10^{2n} - 10^{n} }{9}$
So the number is

$\frac{4.(10^{2n} - 10^n) + 8.(10^n - 1) + 9}{9} = \frac{4.10^{2n} +4.10^n + 1}{9} = \frac{(2.10^n+1)^2}{9} = (\frac{2.10^n+1}{3})^2$

by **mrbrenner** » Wed Oct 01, 2008 10:51 am

Rock'n'roller - The approach was great, and the result was correct ... but you either started with the wrong formula, or applied it wrongly ... then had to fudge the intermediate result (+9 magically got to the numerator of a fraction with the denominator 9, also as 9, instead of 81) ... just to make the result correct.

However, all it is needed is really to start with the correct formula and apply it correctly ... and you will indeed obtain the same result. In the geometric progression, for this particular case, the correct formula is with the common ratio r = 10, and the initial factor a = 1, and applied would have been with the first term being "the initial factor times the power 0 of the ratio r (that is 1)" and the last term as the "the initial factor times the (n-1) power of 10". Then follow exactly your reasoning (excellent) and nothing needs to be fudged anymore to get to your (very correct!) result.

Perfect square for every n

Perfect square for every n

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