# How fast/slow in comparison with other methods for proving that certain integer (also finding square root) is odd perfect square in next method?

Example 1: $49 - (49-1)/4 = 49 - 12 = 37$ where $12 = 3 * 4$ and $37=4^3-3^3$ -centered hexagonal number and its square root is $3+4=7$

Example 2: $121 -(121-1)/4 = 121 - 30 = 91$ where $30 = 5 * 6$ and $91 = 6^3-5^3$ and square root is $5+6=11$

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Why do you choose the factorisation $30 = 5 * 6$ and not $30 = 3 * 10$, for example? –  Ben Millwood Aug 31 '12 at 17:37
Oh, hmm. I think I can see something in here (you're looking at $\frac{1}{4}(n^2 - 1) = (n/2-1/2)(n/2+1/2)$) but it's completely unclear. Please specify the algorithm more carefully. –  Ben Millwood Aug 31 '12 at 17:39
I did not choose that, I have just observed that relation. How difficult is to factorize number if you know that difference of two factors is 1?(12=3*4 or 30=5*6) –  Bojan Vasiljević Aug 31 '12 at 17:42
About as difficult as finding the square root of the number, I'd say... –  Ben Millwood Aug 31 '12 at 17:45
(2n+1)^2 - ((2n+1)^2-1)/4 = 3n^2+3n+1 –  Bojan Vasiljević Aug 31 '12 at 17:45