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How to find all naturals $n$ such that $\sqrt{1\smash{\underbrace{4\cdots4}_{n\text{ times}}}}$ is an integer?

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  • $\begingroup$ Do you mean $\sqrt{1\underbrace{4\cdots4}_{n-times}}$ or perhaps$\sqrt{\underbrace{1414\cdots14}_{n-times}}$? $\endgroup$
    – Quixotic
    Commented Jul 14, 2012 at 1:24
  • $\begingroup$ I mean The first one . $\endgroup$
    – Frank
    Commented Jul 14, 2012 at 1:27
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    $\begingroup$ There will be only two integer cases for $n = 2$ and $n = 3$ and for everything else $$ \sqrt{1\underbrace{4\cdots4}_{n-times}}=2\sqrt{36\underbrace{11\cdots1}_{(n-2)-times}}$$ $\endgroup$
    – Quixotic
    Commented Jul 14, 2012 at 1:37
  • $\begingroup$ I changed the TeX code from {}_{n-times} to {}_{n\text{ times}, so that instead of $1\underbrace{4\cdots4}_{n-times}$ we see $1\underbrace{4\cdots4}_{n\text{ times}}$. The hyphen looked like a minus sign (longer than a hyphen) since it was in math mode, and the "times" got italicized and needed something to artificially separate it from the $n$, since it was in math mode. $\endgroup$ Commented Jul 14, 2012 at 1:47

1 Answer 1

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For $n \geq 4$ your number is equal to $4444$ modulo $10000$, and in particular modulo $16$. If it were a square, then $4444$ would be a square modulo $16$, implying $1111$ is a square modulo $4$. But $1111=3$ mod $4$, contradiction.

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