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where \rsqrt{} is the command in question.
I know that the algorithm would look somewhat like this
i = square root of n rounded down
while i > 0:
if n is divisible by i²:
simplification is i\sqrt{n/i²}
break loop
i = i - 1
%the simpification will always be found
%since every n is divisible by 1
where n is the given integer, i is the number before \sqrt and n/i² is the argument of \sqrt{...}.
Step-by-step explanation:
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