# Is this method new for finding square roots of specific perfect squares

I added some explanation.

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I really want to, but I can't make heads or tails of this. I think the numbers-to-words ratio is way off. Perhaps some more exposition and explanation would help. –  Antonio Vargas May 2 '12 at 15:19
I am an amateur. Thats my best. :) –  Bojan Vasiljević May 2 '12 at 15:26
If you want help then I suggest you take my advice. I am certainly not going to spend my time sifting through your confusion for you on the chance that something of value is hidden there, and I would bet others feel the same. –  Antonio Vargas May 2 '12 at 15:28
Fair enough Antonio. I know now your answer. thanks –  Bojan Vasiljević May 2 '12 at 15:32
@Bojan: To communicate all but the most routine mathematics, we need to use words. Plenty of words, as in whole sentences and paragraphs (if not pages) of them. Please do "use your words". If you are not comfortable writing in English, then feel free to give the explanations in your native language: for this worldwide audience, and with contemporary software, it will not be so hard for us to do the translation. –  Pete L. Clark May 4 '12 at 13:02
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## 1 Answer

This is not an answer but may provide a starting point for someone else.

The first sequence ($1, 7, 19, 37, 61, \cdots$) are the hexagonal centered (hex) numbers, A003215. The general form of the $n^{th}$ term is $3n(n+1)+1.$

In the second sequence, $6n-1$ occurs at position $2n-1$, and $(6n-1)^2$ at position $4n(3n-1).$ So the number of terms (jumps) between the two is \begin{align} 4n(3n-1) - (2n-1) &= 12n^2 - 6n + 1. \\ & = 3(2n-1)(2n)+1 \\ & = \text{the }(2n-1)^{th}\text{ hex number.} \end{align} Thus there is a connection between certain odd squares and the hex numbers.

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