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Prove that every number of the sequence $49,4489,444889,\ldots$ is a perfect square.

We can observe that $49=7^2, 4489=67^2, 444889=667^2, \ldots$

I have tried expanding terms of the sequence, and to express it as a whole square. But it was too tough for me.

Any help will be appreciated.

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4 Answers 4

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We can prove it as follows: \begin{align*} \underbrace{666\ldots6}_{n}\,7^2 &= \left[7 + 6(10^1 + 10^2 + \cdots + 10^n) \right]^2 \\ &= \left[7 + 6 \frac{10^{n+1} - 10}{9}\right]^2 \\ &= \left[ \frac{2 \cdot 10^{n+1} + 1}{3}\right]^2 \\ &= \frac{4 \cdot 10^{2n+2} + 4 \cdot 10^{n+1} + 1}{9} \\ &= 4 \cdot \frac{10^{2n+2} - 1}{9} + 4 \cdot \frac{10^{n+1} - 1}{9} + 1 \\ &= 4 \cdot \underbrace{111 \ldots 1}_{2n+2} + 4 \cdot \underbrace{111 \ldots 1}_{n+1} + 1 \\ &= \underbrace{444 \ldots 4}_{2n+2} + \underbrace{444 \ldots 4}_{n+1} + 1 \\ &= \underbrace{444 \ldots 4}_{n+1}\,\underbrace{888 \ldots 8}_{n}\,9. \end{align*}

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  • $\begingroup$ @user333900 You're welcome. $\endgroup$ Sep 3, 2016 at 8:45
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The $n$th term is $T_{n}=9+8.10+8.10^2+...8.10^{n-1}+4.10^n+...+4.10^{2n-1}$:

$$ \begin{align} T_n &= 9+8.10(1+10+...10^{n-2})+4.10^n(1+...+10^{n-1}) \\ &= 9+8.10(\frac{10^{n-1}-1}{9})+4.10^n (\frac{10^{n}-1}{9}) \\ &=9+8.\frac{10^{n}-10}{9}+4.\frac{10^{2n}-10^n}{9} \\ &=\frac{81+8.10^n-80+4.10^{2n}-4.10^n}{9} \\ &=\frac{1}{9}.(1+4.10^n+4.10^{2n}) \\ &={(\frac{{1+2.10^n}}{3}})^2 \end{align} $$.

Now $10\equiv 1\pmod{3}\Rightarrow 10^n\equiv 1\pmod{3}\Rightarrow 2.10^n+1\equiv 0\pmod{3}$, thus $3$ divides the numerator and we are done.

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  • $\begingroup$ @gowrath thanks for the edit. $\endgroup$
    – user345851
    Sep 3, 2016 at 10:25
  • $\begingroup$ @JonathanRichardLombardy No worries. $\endgroup$
    – gowrath
    Sep 3, 2016 at 10:26
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Very similar to this question, which asks for a proof that $1111\ldots5556$ is a square. The proof is just as simple:

Multiply $4444\ldots8889$ by $9$ and you get $4000\ldots0004000\ldots0001$, which is $2000\ldots0001^2$.

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  • $\begingroup$ Thanks for such a simple solution. $\endgroup$
    – user333900
    Sep 3, 2016 at 11:37
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You can see that $$(1+\sum_{i=0}^{n+1}6\times10^i)^2=(6\times10^{n+1}+(1+\sum_{i=0}^{n}6\times10^i))^2.$$ Expand it to get the expression $$(6\times10^{n+1})^2+2\times6\times10^{n+1}\times(1+\sum_{i=0}^{n}6\times10^i)+(1+\sum_{i=0}^{n}6\times10^i)^2.$$ Now make an induction hypothesis that the squares of $66...67$ makes the sequence in the question. Induction basis is already proven. Then the last term in the expanded expression becomes $$\color{red}{\sum_{i=n+1}^{2n+1}4\times10^i+\sum_{j=1}^{n}8\times10^j+9}.$$ The second term in the expanded expression can be simplified by taking the factor $3$ of $6$ and multiply it with $(1+\sum_{i=0}^{n}6\times10^i)$, giving a simple expression $2\times10^{n+1}+1.$ With this expression, the second term becomes $$2\times2\times10^{n+1}\times(2\times10^{n+1}+1)=\color{red}{8\times10^{2n+2}+4\times10^{n+1}}.$$ The first term is just $\color{red}{36\times10^{2n+2}}$. You can then easily verify that the sum of the three terms (in red) is the expected expression.

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