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$6, 96, 996, 9996,\dots, 99999\cdots996,\dots$

Consider the above number sequence. Here in the $n$ $^{th}$ term $n-1$ digits are $9$. How can we tell about the existence of a perfect square number in this sequence?

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    $\begingroup$ If the perfect square were $10^{2n}-4$, then its factors would be $10^n+2,10^n-2$. So the perfect square must be of the form $10^{2n-1}-4$. $\endgroup$
    – abiessu
    Mar 4, 2016 at 17:57

3 Answers 3

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The sum of the digits of the $n^{th}$ term of this sequence is $$\text{Sum}=9(n-1)+6=9n-3=3(3n-1)$$ Thus, every number in the sequence is divisible by $3$ but not by $9$. Hence, no number in the sequence can be a perfect square.

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HINT: consider divisibility by a suitable small number $n$ and also by $n^2$

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Another method to approach this problem.

Using $\pmod{11}$:

  • if the number of $9’s$ in $\underbrace{99\cdots9}_{n\text{ times}}$ is even, then the number will be divisible by $11$, so $\underbrace{99\cdots9}_{n\text{ times}}6\equiv-3\equiv8\pmod{11}$, if $n$ is odd.
  • If $n$ is even, then $11\mid\underbrace{99\cdots9}_{n\text{ times}}0$, so $\underbrace{99\cdots9}_{n\text{ times}}6\equiv6\pmod{11}$ if $n$ is even.

But the quadratic residues $\pmod{11} \in ({1, 3, 4, 5, 9})$, and since $\underbrace{99\cdots9}_{n\text{ times}}6$ is always either equal to $2$ or $6$ $\pmod{11}$, then $\underbrace{99\cdots9}_{n\text{ times}}6$ will never be a perfect square.

Bonus:

  • For any $n\geq4$, then $\underbrace{99\cdots9}_{n\text{ times}}6$ is equal to $12\pmod{16}$, as after dividing it by four, we will obtain $24\underbrace{99\cdots9}_{n\text{ times}}$, which is equal to $3\pmod{4}$ whenever $n\geq2$. This is a contradiction.
  • In addition, if $n$ is even, then $\underbrace{99\cdots9}_{n\text{ times}}6+4=10^{n+1}$ is a perfect square, another contradiction to Catalan’s Conjecture. As an exercise, prove this.
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