Introducing The Fastest Way to Find all the Roots.

[Olvind]

Introducing The Most Fastest Way to Find Perfect Square Roots.

To find perfect square root of √3249

1st Step - Using the below formula by substituting 3249 at X to get m.

3249 / 72 = 45.125 …..
By ignoring the decimals, we get m = 45
We will take m value as m1 to input it in the below repeated subtraction series.
m1 = 45

2nd Step – Finding the Key Integer ‘c’ .
To find key integer c use the below series Successive Series.
Successive Series -

0 -- 1 has zero difference.
1 --- 3 has only a difference of one.
3 ---- 6 has a difference of two.
6 ---- 10 has a difference of three.
.
.
45 ---- 55 has a difference of nine.
e.g.
From 1st step, we have m = 45
As per the above successive series, any m that appear from 45 to 55, then we can take c = 9
So, 45 appears at this particular set of series and series has a difference of nine. Therefore, we get c = 9
( Note - This series is a form of successive series where sequence of numbers on the above right hand side, that is 1, 3, 6, 10….is the result of addition of sequence of positive integers such as 1, 2, 3, 4, 5, 6, 7, 8….
E.g., 1 + 2 = 3, 3 + 3 = 6, 6 + 4 = 10 ….

As show above, one can continue create series by adding 10 + 5 = 15, 15 + 6 = 21…..record it and use it whenever required to calculate key integer ‘c' at any other problems of finding square roots).

3rd Step – Multiplying c by constant 6.
From 2nd step, we have c = 9
Therefore, 9 × 6 = 54
Now we get c = 54
Checking whether c is final answer by dividing 3249 by 54 . We found it is not divisible.
Therefore, we will proceed to below 3rd step of checking Rules of Finding ‘c.

4th Step – Checking the rules to find perfect square root.
√X = √3249 is divisible integer divisible by 3. Therefore it follows the fourth rule of finding c.
Fourth Rule - If the √ X is any odd integer and is divisible by 3, then the final answer c will be adding s by 3 i.e. ‘c + 3'.
54 + 3 = 57
Since 3249 is divisible by 57
Therefore, √3249 = 57

Note - The rules are available at paper, check the below attachment. (The paper has NOT been peer-reviewed or published)

Moderator Note: The version attached below is obsolete. Olvind has attached a corrected version in post #5.

 

[Deleted member 4993]

Introducing The Most Fastest Way to Find Perfect Square Roots.

To find perfect square root of √3249

1st Step - Using the below formula by substituting 3249 at X to get m.

3249 / 72 = 45.125 …..
By ignoring the decimals, we get m = 45
We will take m value as m1 to input it in the below repeated subtraction series.
m1 = 45

2nd Step – Finding the Key Integer ‘c’ .
To find key integer c use the below series Successive Series.
Successive Series -

0 -- 1 has zero difference.
1 --- 3 has only a difference of one.
3 ---- 6 has a difference of two.
6 ---- 10 has a difference of three.
.
.
45 ---- 55 has a difference of nine.
e.g.
From 1st step, we have m = 45
As per the above successive series, any m that appear from 45 to 55, then we can take c = 9
So, 45 appears at this particular set of series and series has a difference of nine. Therefore, we get c = 9
( Note - This series is a form of successive series where sequence of numbers on the above right hand side, that is 1, 3, 6, 10….is the result of addition of sequence of positive integers such as 1, 2, 3, 4, 5, 6, 7, 8….
E.g., 1 + 2 = 3, 3 + 3 = 6, 6 + 4 = 10 ….

As show above, one can continue create series by adding 10 + 5 = 15, 15 + 6 = 21…..record it and use it whenever required to calculate key integer ‘c' at any other problems of finding square roots).

3rd Step – Multiplying c by constant 6.
From 2nd step, we have c = 9
Therefore, 9 × 6 = 54
Now we get c = 54
Checking whether c is final answer by dividing 3249 by 54 . We found it is not divisible.
Therefore, we will proceed to below 3rd step of checking Rules of Finding ‘c.

4th Step – Checking the rules to find perfect square root.
√X = √3249 is divisible integer divisible by 3. Therefore it follows the fourth rule of finding c.
Fourth Rule - If the √ X is any odd integer and is divisible by 3, then the final answer c will be adding s by 3 i.e. ‘c + 3'.
54 + 3 = 57
Since 3249 is divisible by 57
Therefore, √3249 = 57

Note - The rules are available at paper, check the below attachment.

Was the attached paper accepted & published in a peer-reviewed Journal?

 

[Olvind]

Was the attached paper accepted & published in a peer-reviwed Journal?

No, but you find it at academia online publishing journal.

 

[Dr.Peterson]

Introducing The Most Fastest Way to Find Perfect Square Roots.

To find perfect square root of √3249

1st Step - Using the below formula by substituting 3249 at X to get m.

3249 / 72 = 45.125 …..
By ignoring the decimals, we get m = 45
We will take m value as m1 to input it in the below repeated subtraction series.
m1 = 45

2nd Step – Finding the Key Integer ‘c’ .
To find key integer c use the below series Successive Series.
Successive Series -

0 -- 1 has zero difference.
1 --- 3 has only a difference of one.
3 ---- 6 has a difference of two.
6 ---- 10 has a difference of three.
.
.
45 ---- 55 has a difference of nine.
e.g.
From 1st step, we have m = 45
As per the above successive series, any m that appear from 45 to 55, then we can take c = 9
So, 45 appears at this particular set of series and series has a difference of nine. Therefore, we get c = 9
( Note - This series is a form of successive series where sequence of numbers on the above right hand side, that is 1, 3, 6, 10….is the result of addition of sequence of positive integers such as 1, 2, 3, 4, 5, 6, 7, 8….
E.g., 1 + 2 = 3, 3 + 3 = 6, 6 + 4 = 10 ….

As show above, one can continue create series by adding 10 + 5 = 15, 15 + 6 = 21…..record it and use it whenever required to calculate key integer ‘c' at any other problems of finding square roots).

3rd Step – Multiplying c by constant 6.
From 2nd step, we have c = 9
Therefore, 9 × 6 = 54
Now we get c = 54
Checking whether c is final answer by dividing 3249 by 54 . We found it is not divisible.
Therefore, we will proceed to below 3rd step of checking Rules of Finding ‘c.

4th Step – Checking the rules to find perfect square root.
√X = √3249 is divisible integer divisible by 3. Therefore it follows the fourth rule of finding c.
Fourth Rule - If the √ X is any odd integer and is divisible by 3, then the final answer c will be adding s by 3 i.e. ‘c + 3'.
54 + 3 = 57
Since 3249 is divisible by 57
Therefore, √3249 = 57

Note - The rules are available at paper, check the below attachment. (The paper has NOT been peer-reviewed nor published)

I don't know whether your method is correct, as it isn't clear enough to check. (But the fact that you compare a number to triangular numbers, which are related to squares, suggests there may be a nugget of truth in there.)

What I know is that it isn't the "most fastest way" (which would be hard to prove even if it were). I can find √3249 in just three steps of the "divide and average" method (here), with no guessing, and no long series of subtractions:

Starting with 100, the greatest square power of ten less than 3249, we get

3249/100=32, rounded​

(100+32)/2=66​

3249/66=49, rounded​

(66+49)/2=57.5, round to 58 (deliberately not making a choice that would speed the work)​

3249/58=56, rounded​

(58+56)/2=57​

3249/57=57, so we have the root​

How is your work above faster?

But it could be interesting to discuss how you came up with your method, how you proved it is valid, and how it might be stated more clearly or improved. My guess is that you could do a lot better once you have more knowledge of algebra, series, proofs, and so on.

 

[Olvind]

Introducing The Most Fastest Way to Find Perfect Square Roots.

To find perfect square root of √3249

1st Step - Using the below formula by substituting 3249 at X to get m.

3249 / 72 = 45.125 …..
By ignoring the decimals, we get m = 45
We will take m value as m1 to input it in the below repeated subtraction series.
m1 = 45

2nd Step – Finding the Key Integer ‘c’ .
To find key integer c use the below series Successive Series.
Successive Series -

0 -- 1 has zero difference.
1 --- 3 has only a difference of one.
3 ---- 6 has a difference of two.
6 ---- 10 has a difference of three.
.
.
45 ---- 55 has a difference of nine.
e.g.
From 1st step, we have m = 45
As per the above successive series, any m that appear from 45 to 55, then we can take c = 9
So, 45 appears at this particular set of series and series has a difference of nine. Therefore, we get c = 9
( Note - This series is a form of successive series where sequence of numbers on the above right hand side, that is 1, 3, 6, 10….is the result of addition of sequence of positive integers such as 1, 2, 3, 4, 5, 6, 7, 8….
E.g., 1 + 2 = 3, 3 + 3 = 6, 6 + 4 = 10 ….

As show above, one can continue create series by adding 10 + 5 = 15, 15 + 6 = 21…..record it and use it whenever required to calculate key integer ‘c' at any other problems of finding square roots).

3rd Step – Multiplying c by constant 6.
From 2nd step, we have c = 9
Therefore, 9 × 6 = 54
Now we get c = 54
Checking whether c is final answer by dividing 3249 by 54 . We found it is not divisible.
Therefore, we will proceed to below 3rd step of checking Rules of Finding ‘c.

4th Step – Checking the rules to find perfect square root.
√X = √3249 is divisible integer divisible by 3. Therefore it follows the fourth rule of finding c.
Fourth Rule - If the √ X is any odd integer and is divisible by 3, then the final answer c will be adding s by 3 i.e. ‘c + 3'.
54 + 3 = 57
Since 3249 is divisible by 57
Therefore, √3249 = 57

Note - The rules are available at paper, check the below attachment. (The paper has NOT been peer-reviewed nor published)

Ignore the above Attached PDF. Get this new corrected PDF on Square roots.

 

[Olvind]

I don't know whether your method is correct, as it isn't clear enough to check. (But the fact that you compare a number to triangular numbers, which are related to squares, suggests there may be a nugget of truth in there.)

What I know is that it isn't the "most fastest way" (which would be hard to prove even if it were). I can find √3249 in just three steps of the "divide and average" method (here), with no guessing, and no long series of subtractions:

Starting with 100, the greatest square power of ten less than 3249, we get

3249/100=32, rounded​

(100+32)/2=66​

3249/66=49, rounded​

(66+49)/2=57.5, round to 58 (deliberately not making a choice that would speed the work)​

3249/58=56, rounded​

(58+56)/2=57​

3249/57=57, so we have the root​

How is your work above faster?

But it could be interesting to discuss how you came up with your method, how you proved it is valid, and how it might be stated more clearly or improved. My guess is that you could do a lot better once you have more knowledge of algebra, series, proofs, and so on.

Divide the x by 72 , 3249 / 72 = 45.125 get the 45 ( ignore decimal).
Subtract the 45 until you get negative result. (Finding of Key Integer c)
45 - 1 = 44
44 -2 = 42
42 - 3 = 39
39 - 4 = 35
35 - 5 = 30
30 - 6 = 24
24- 7 = 17
17 - 8 = 9
9 - 9 = 0 .... (zero cannot be further subtracted by 10 therefore we stop at this step ( if we continue further, we will get negative difference as -10 ) This condition of stopping further calculation is explained at paper).

Take above 9 (bold)
9 x 6 = 54 ..... as explained at paper to multiply by six.

The rule to use .....(explained at paper).
√X = √3249 integer divisible by 3. Therefore it follows the fourth rule of finding c.
Fourth Rule - If the √ X is any odd integer and is divisible by 3, then the final answer c will be adding s by 3 i.e. ‘c + 3'.

54 + 3 = 57

Therefore, √3249 = 57

This is the first easy method of getting answer 57.

Second method is where we can create below successive series, keep record of it and use it any time while finding answers, in order to reduce calculation time. So no need to calculate subtraction series as explained above.

Successive Series -

0 -- 1 has zero difference
1 --- 3 has only as difference of one i.e. 2
3 ---- 6 has a difference of two i.e. 4 and 5
6 ---- 10 has a difference of three i.e. 7 to 9
10 ---- 15 has a difference of four i.e. 11 to 14
15 ---- 21 has a difference of five i.e. 16 to 20
21 ---- 28 has a difference of six i.e. 22 to 27
28 ---- 36 has a difference of seven i.e. 28 to 35
36 ---- 45 has a difference of eight i.e. 37 to 44
45 ---- 55 has a difference of nine i.e. 46 to 54
.
.
.


This series goes infinitely. Its is explained in the paper on how to create it and use it for finding Key Integer c.
Solution e.g.,
Divide the x by 72 , 3249 / 72 = 45.125 get the 45 ( ignore decimal).
45 is at ninth step of this series.
therefore, take 9
Next,
9 x 6 = 54 ..... as explained at paper - to multiply by six.

The rule to use .....(explained at paper).
√X = √3249 integer divisible by 3. Therefore it follows the fourth rule of finding c.
Fourth Rule - If the √ X is any odd integer and is divisible by 3, then the final answer c will be adding s by 3 i.e. ‘c + 3'.


54 + 3 = 57

Since 3249 is divisible by 57
Therefore, √3249 = 57

 

[Dr.Peterson]

You didn't answer my questions:

Why do you say this is fast? It seems incredibly slow and cumbersome. It shouldn't take 64 pages to explain a useful method!

Why do you think it is correct? I don't see anything that claims to be a proof, which is necessary for anything to be called mathematical. Nor is there even a full and concise description of the method.

 

[mmm4444bot]

academia online publishing journal

Please provide an URL for that site. Thank you!

?
[imath]\;[/imath]

 

[Cubist]

The "Fourth rule" seems correct to me (see below). Sorry but I don't have the time to find, or verify, any of the other rules that might be in the document.

Proof
The part of OP's method where the triangular sequence is traversed boils down to finding an n that satisfies the following...

[math]\frac{n(n+1)}{2} \le \frac{x^2}{72} < \frac{(n+1)(n+2)}{2}[/math]
...since the n^th triangular number is given by n*(n+1)/2

The example half way down post#6 states, "Fourth Rule - If the √ x is any odd integer and is divisible by 3, then the final answer c will be adding s by 3 i.e. ‘c + 3'." and I think this is equivalent to saying x=6n+3 because c=6n (and the red √ in the quote was a mistake, I think). Therefore...

[math]\frac{n(n+1)}{2} \le \frac{(6n+3)^2}{72} < \frac{(n+1)(n+2)}{2}[/math]
Multiplying by 72, expanding and simplification yields...

[math]0 \le 9 < 72(n + 1)[/math]
This is obviously true, therefore the "Fourth rule" seems correct.

--

@Olvind it's great that you are interested in this topic and have started documenting it. The thing that concerns me:- is this actually faster? Which method of finding integer roots have you tested this against? What were the actual results, and on what computer? Or are you thinking of a human using a calculator or pen+paper? The "lookup" of triangular numbers will be the slow part, and I think you'd probably end up having to use Newton's method (or similar) to do this for bigger integers which seems to defeat the point of this method?

 

[Cubist]

Proof

Sorry, I missed the "ignoring the decimals" which results in a floor operation in the proof...

The part of OP's method where the triangular sequence is traversed boils down to finding an n that satisfies the following...

[math]\frac{n(n+1)}{2} \le \left\lfloor \frac{x^2}{72} \right\rfloor< \frac{(n+1)(n+2)}{2}[/math]
...since the n^th triangular number is given by n*(n+1)/2

The example half way down post#6 states, "Fourth Rule - If the √ x is any odd integer and is divisible by 3, then the final answer c will be adding s by 3 i.e. ‘c + 3'." and I think this is equivalent to saying x=6n+3 because c=6n (and the red √ in the quote was a mistake, I think). Therefore...

[math]\frac{n(n+1)}{2} \le \left\lfloor\frac{(6n+3)^2}{72} \right\rfloor < \frac{(n+1)(n+2)}{2}[/math]
Multiply by 72, expand and simplify
[math]36\left(n^2+n\right) \le 72\left\lfloor\frac{(6n+3)^2}{72} \right\rfloor < 36\left(n^2+n\right) + 72(n + 1)[/math]Consider the middle expression...
[math] 72\left\lfloor{\frac{4\left( n^2+n\right)+\red{1}}{8}}\right\rfloor[/math]The red "1" doesn't matter. "n" is an integer and the (n^2+n) will always be even. Therefore...
[math]=36\left(n^2+n\right)[/math]And we have
[math]0 \le 0 < 72(n + 1)[/math]which is true, therefore the "Fourth rule" seems correct.