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I've come across an odd way of estimating the square root of a number, going like this:

  1. Given a number n,
  2. Subtract the odd numbers from n in a rising order (1, 3, 5...), until $n \leq 0$
  3. Count how many numbers you subtracted from n. This is the approximation of the square root.


  1. $n = 16$
  2. $n = (16 - 1) = 15$
  3. $n = (15 - 3) = 12$
  4. $n = (12 - 5) = 7$
  5. $n = (7 - 7) = 0$
  6. We have subtracted 4 odd numbers. Since $\sqrt16 = 4$, the approximation works.

(This script is an implementation in Python.)

However, try as I might, I can't understand why this works. Is there a good explanation available?

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You can extend this slightly: If you were looking for $\sqrt{19}$ you could subtract the first four odd numbers leaving a remainder of $3$. The next odd number is $9$, which is too big, so your approximation would be $4\frac{3}{9}$ or about $4.33$. In fact $\sqrt{19} \approx 4.36$ so it is not a bad method. –  Henry Feb 2 '12 at 13:35
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2 Answers 2

up vote 8 down vote accepted

The differences between the successive perfect squares are 1, 3, 5, 7, 9, 11 ...

Your algorithm is equivalent to summing odd numbers until the sum exceeds the input -- and it works because the sum of the first $n$ odd numbers is exactly $n^2$.

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Thanks, that makes sense! –  semi-anonymous Feb 2 '12 at 13:27
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I myself have created a method much like this I'll tell you why it works

A squares area can be given using this method For example the are of a 3by 3 is equal to 1+3+5 And for a 4 by 4 1+3+5+7 Draw a square And divide it with little squares and this pattern will show

The difference of 1^2 and 0^2= 1 The difference of 2^2 and 1^2=3 The difference of 3^2 and 2^2=5 .......... And so these odd numbers form

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