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Perfect squares

\(49, 4489, 444889, 44448889.................\) In the series of the above numbers following the same pattern, prove that, each of these are whole squares of integers..

Note by Sagnik Saha
3 years, 3 months ago

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\(T_n = 44444....(n times) 88888... (n-1 times)9\)

\(= \frac{4}{9}\times(10^n-1) \times 10^n + \frac{8}{9} \times(10^{n-1}-1) \times 10 + 9\)

\(= \frac{1}{9}({ 4\times(10^n-1)\times10^n + 8(10^{n-1}-1)\times + 81})\)

\(= \frac{1}{9}{( 4 \times 10^{2n} - 4 \times 10^n + 8 \times(10^n - 10) + 81)}\)

\(= \frac{1}{9}{( 4\times10^{2n} - 4\times10^n + 8 \times 10^n - 80 + 81)}\)

\(= \frac{1}{9}{( 4\times10^{2n} + 4 \times 10^n + 1)}\)

\(= \frac{1}{9}[(2\times10^n)^2 + 2\times (2\times10^n) + 1^2]\)

\(= \frac{1}{9}(2\times10^n + 1)^2\)

\(= (\dfrac{2 \times 10^n +1}{3})^2\)

Now note

\(2 \equiv -1 (\mod{3} )\) and

\(10 \equiv 1 (\mod{3} ) \implies 10^n \equiv 1 (\mod{3} )\)

\(\implies 2\times10^n \equiv -1 (\mod{3} )\)

\(\implies 2\times10^n + 1 \equiv 0 (\mod{3} )\) or

\(3 | 2\times10^n + 1\).

This maybe proved by induction as well. Thus \({(2*10^n + 1)/3}\) is an integer and thus \(T_n\) is a perfect square for all n. Sagnik Saha · 3 years, 3 months ago

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