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For $9, 99, 999, 9999, 99999,\dots$, except $9$, are the rest of the numbers $9$'s perfect squares? Are there other perfect squares with all $9$'s.

This problem is given for K-12 students, which I have no idea how to do it.

Thank you.

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  • $\begingroup$ Can you do any special cases? $\endgroup$
    – hardmath
    Oct 5, 2015 at 1:35
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    $\begingroup$ 9k is a perfect square if and only if k is a perfect square, so this question is equivalent to asking whether a number that's all 1's is a perfect square, other than 1 itself. $\endgroup$ Oct 5, 2015 at 1:35
  • $\begingroup$ What are special cases? To me, 9 is the only perfect square among 9, 99, 999,... I'm not sure if it's true. $\endgroup$
    – dh16
    Oct 5, 2015 at 1:36
  • $\begingroup$ How about when the count of 9's is even? Do you see why these cases are not squares? $\endgroup$
    – hardmath
    Oct 5, 2015 at 1:37
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    $\begingroup$ To be honest, this is not my field. I can only think I can write those even count numbers like: $10^{2k} -1$, which fails to be a perfect square because suppose yes. Then $10^{2k}$ is a perfect square, which implies we have 2 perfect squares different at most 1 (a contradiction). $\endgroup$
    – dh16
    Oct 5, 2015 at 1:44

3 Answers 3

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Consider these numbers modulo 4.

Theorem:

A number $n^2$ with $n$ an integer satisfies either $n^2\equiv 0(\text{mod}4)$ or $n^2\equiv 1(\text{mod}4)$

Proof of theorem:

Case 1: $n\equiv 0$ or $2(\text{mod}4)$. Then $n^2\equiv 0(\text{mod}4)$

Case 2: $n\equiv 1(\text{mod}4)$. Then $n^2\equiv 1(\text{mod}4)$

Case 3: $n\equiv 3(\text{mod}4)$. Then $n^2\equiv 1(\text{mod}4)$

Corollary:

A number which is $2(\text{mod}4)$ or $3(\text{mod}4)$ cannot be a perfect square.

Now, look at $9\dots 99(\text{mod}4)$

A number modulo 4 is congruent to the final two digits modulo 4, and $99(\text{mod}4)$ is congruent to $3$, therefore is not a perfect square.

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There are not a lot of tools at K-12 for this problem, so use the simplest one: take the last several digits. The last digit is a square, so look at the final two digits.

There is no perfect square ending in 99

because

any perfect square leaves a remainder of 0 or 1 when divided by 4

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All integers can be written as $10n\pm k$, where $0\le k\le5$. Squaring, we have $100n^2\pm20nk+k^2$, which for $0\le k^2<10\iff0\le k\le3$ has an even tens digit, and the remaining two options do not produce a $9$ at the last position. $($In general, no “repdigit” in base $10$ can ever be a square$)$.

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