(<var>n</var>+1)2-<var>n</var>2 = 2<var>n</var>+1. </blockquoteSo your algorithm works for any square integer. But it's not very fast, as it takes time proportional to O(sqrt(<var>n</var>)), where <var>n</var> is the number you're rooting, not its size!
For comparison, Newton-Raphson / the Babylonian algorithm double the number of correct digits at each iteration.
In reply to Re: Re: Re: Re: Square Root algorithm by ariels
in thread Square Root algorithm by nysus
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