需要平方根的算法提供“余数”

Question

I'm writing a calculator without using decimals (supports only Rational numbers), but I'd like to be able to do a version of square root.

When a square root function is pressed for (say) the number 12, I'd like to just simplify/"reduce" the square root and return 2*sqrt(3)--by it into (2*2) * 3 and extracting the sqrt(2*2) as 2.

I'm using biginteger which has a very nice gcd() method and a pow() method that is restricted to positive parameters (which makes sense unless you are trying to do exactly what I'm trying to do.

I could come up with a few iterative ways to do this but they may take a while with numbers in the hundreds-of-digits range.

I'm hoping there is some cute, simple, non-iterative trick I haven't been exposed to.

Just to clarify: I have the intent to add imaginary numbers so I'm planning on results like this:

17 + 4i √3  
-----------  
     9

Without long streams of decimals.

Answer 1

What you're asking, in essence, is to find all repeated prime factors. Since you're dealing with numbers in the hundreds-of-digits range, I'm going to venture a guess here that there are no good ways to do this in general. Otherwise public key cryptography will all of a sudden be on somewhat shaky ground.

There are a number of methods of computing the square root . With those, you can express the result as an integer plus a remainder less than 1.

Answer 2

Maybe try finding the highest perfect square that is less than your number. That will give you part of the equation, then you would only need to handle the remainder part which is the difference between your number and the perfect square you found. This would degrade as numbers get large as well, but perhaps not as fast.