\newcount\x \newcount\nx
\def\mysqrt#1{\x=#1 \nx=\x \isqrt{#1}
\big\lfloor\sqrt{#1}\big\rfloor=\number\x}
\def\isqrt#1{\nx=#1
\divide\nx by\x \advance\nx by\x \divide\nx by2
\ifnum\x=\nx \else\x=\nx \isqrt{#1}\fi}
$\mysqrt{361}$
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