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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$\begingroup$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.
$\endgroup$ $\begingroup$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
$\endgroup$ $\begingroup$any perfect square leaves a remainder of 0 or 1 when divided by 4
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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