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Suppose $n>1$ is a nonsquare integer and $[a_0; a_1, a_2, a_3, \ldots]$ is its simple continued fraction expansion, with $a_0=\lfloor\sqrt n\rfloor.$ Of course the continued fraction is periodic, but does the period always start at $a_1$?

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The answer is yes. More precisely, if $n$ is a square-free integer, then the expansion of $\sqrt{n}$ in continued fraction is $$\sqrt{n} = [a_0; \, \overline{a_1, \, a_2, \, a_3, \ldots, a_2, \, a_1, \, 2a_0}],$$ where the repeating part (excluding the last term) is symmetric upon reversal.

For more details, see here.

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