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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, 8 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.

- 3 years, 8 months ago