35 This entry had its birth in my fooling around with my 1966 attempt to determine the length of 1/4 of a circle's circumference without Pi. tpo = term position sqrt = square root sq = square nnss = natural number summation sequence tpo0 0/4 = 0 (0/4 = 0) + 1 = 1 - tpo1 4/4= 1 8/4 = 2 8 + 1 = 9 2 - 1 = 1 2 x 2 3 x 3 - tpo2 16/4 = 4 24/4 = 6 24 + 1 = 25 6 - 4 = 2 4 x 4 5 x 5 - tpo3 36/4 = 9 48/4 = 12 48 + 1 = 49 12 - 9 = 3 6 x 6 7 x 7 - tpo4 64/4 = 16 80/4 = 20 80 + 1 = 81 20 - 16 = 4 8 x 8 9 x 9 - tpo5 100/4 = 25 120/4 = 30 120 + 1 = 121 30 - 25 = 5 10 x 10 11 x 11 - tpo6 144/4 = 36 168/4 = 42 168 + 1 = 169 42 - 36 = 6 12 x 12 13 x 13 - tpo7 196/4 = 49 224/4 = 56 224 + 1 = 225 56 - 49 = 7 14 x 14 15 x 15 - tpo8 256/4 = 64 288/4 = 72 288 + 1 = 289 72 - 64 = 8 16 x 16 17 x 17 - tpo9 324/4 = 81 360/4 = 90 360 + 1 = 361 90 - 81 = 9 18 x 18 19 x 19 Am letting tpo0 be compressed. Thereafter, each tpo visibly consists of two terms displaying division by "4". As can be seen, an sq results in the first term of each tpo. In an odd tpo, an odd sq; in an even tpo, an even sq. Am defining each set by the results in it: odd-even set and even-even set. General: - 1) Each set's tpo = the difference between the division-by-4 results for that set. 2) The sum of each set's results divided by its tpo = the sqrt of the square for that set's 2nd term. 3) The square for a set's 2nd term is its intial integer + 1. 4) Excluding 0, the sum of each set's results = every 2nd term in a paired natural number summation sequence. 5) Including 0, summing the division-by-4 result of the 2nd term of a set with the division-by-4 result of the following set's 1st term = every 1st term in a paired natural number summation sequence. 6) tpo times (tpo + 1) = division-by-4 result for the 2nd term in that tpo's set. 7) In each set/ the first term's division-by-4 result = the square of that set's tpo. 8) Deriving from its intial integer, the 1st sqrt in each set is always an even integer. Therefore, if dividing its square by "4" equals an even integer, it is the first square root/square term in an even- even tpo set. Therefore the difference between the (2nd sq - 1) and the first square/ will be an even integer, and

thatinteger will be that set's tpo. 9) "2" times a tpo = the square root of that tpo's initial integer. Still, how can one know when an integer is a square. Using the above, one knows immediately that when the initial integer in the first term of a tpo is divided by "4"/ the result is also a square. One knows too that the sqrt of that sq = that set's tpo. Further, one knows that an odd sq - 1 is evenly divisible by eight, and that the resulting integer is a term in the natural number summation sequence. Here is an example I think might be helpful. I got it from the 3.14159265 expansion of Pi; so I already know the answer. 8464/4 = 2116 2116/4 = 529 528/8 = 66 Depending on how the natural number summation sequence is treated, "66" is either the 11th or 12th term in that sequence. Admittedly, it is easier to ignore 0 and see "66" as the 11th term, but what one is trying to do might not make that the best choice. 1 3 6 10 15 21 28 36 45 55 66 78 So, the squares are proved, but what is the square root of 529 and what is the square root of 2116 and what is the square root of 8464? The square root of 2116 equals the tpo where the initial integer is 8464. 2 times the sqrt of 529 = the sqrt of 2116, and 2 times the sqrt of 2116 = the sqrt of 8464. I'm taking a sabbatical. Okay, PM 8 is drawing near, and here is the setup: 1 2 3 wherein each integer is a square root in the first group. Am approaching the problem as if I do not know anything. The even square root rules, but first this: the infinite set of odd integers runs between the squares, connecting one to the next. And, yes, I am leaving 0 behind. About those odd integers: 1) the 1st one is "1" less than 2 x the even square root - 2) the 2nd one is "1" more than 2 x the even square root - 3) if an odd integer is the starting point, then that integer + 1 = a sum which when divided by "2" = the square root of the square it is connecting to - - - Back to 1 2 3: new look: 1 x 1 = 1 2 x 2 = 4 3 x 3 = 9 because 1 + 3 = 4 and 4 + 5 = 9. Note also that 3 + 5 = 8 as does 4 + 4; and 8 into 8 = 1. [nnss] Next group: 3 4 5 - the connecting odd integers here are 7 and 9 wherein (7 + 1)/2 = 4 and (9 + 1)/2 = 5. 3 x 3 = 9 4 x 4 = 16 5 x 5 = 25 because 9 + 7 = 16 and 16 + 9 = 25. Remember also (because I forgot) that (2 x the even sqrt) - 1 = the 1st connecting odd integer and (2 x the even sqrt) + 1 = the 2nd connecting odd integer. That's the method, but it still does not reveal the answer I am seeking. The odd integer connecting 9 x 9 to 100 is 19. (19 + 1)/2 = 10 and 2 x 10 = 20. Another way to look at 10 x 10 is 5 x 20, and 20 is important here in that (9 x 9) - 1 = 80 and (11 x 11) - 1 = 120. 20 + 20 = 40 = 19 + 21. [ A further point of interest which applies generally is: (2 x the lesser odd sqrt) + 1 = the connecting integer related to it, and (2 x the greater odd sqrt) - 1 = the integer related to it. ] This might be useful if only a square root is known. Back to 8464 and the paired nnss terms and 66 (term 11). That 66/6 = 11 is interesting, but that 11 x 2 = 22 may be more than just interesting: 22 + 22 = 44 = 21 + 23 44 x 2 = 88 529 - 88 = 441 440/8 = 55 (term 10 in the nnss). This says that 23 x 23 = 529, and that 46 x 46 = 2116, and that 46 is the tpo, and that 92 x 92 = 8464. By the way, 22 x 22 = 484, and 45 is the integer that connects 484 to 529, and 45 + 1 = 46, and 46/2 = 23. May 26, 2008 addenda: 10) Regarding 2nd term in each two-term set: (odd sq - 1)/tpo = 4 x (tpo + 1) and (odd sq - 1)/4 = tpo x (tpo + 1) and (odd sq - 1)/4 divided by (tpo + 1) = tpo. 11) Regarding 1st term in each two-term set: sq/4tpo = tpo. #Brian A. J. Salchert

is a tiny wandering imaginary dinosaur which migrated from AOL in October of 2008.

## About Me

- brian (baj) salchert
- Rhodingeedaddee is my node blog. See
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## Guide

[6-16-2009 Update Insert: Most of what is in this space is now moot. I found out what I was doing wrong and have reinstated Archives and Labels searches. They do work. However, in certain cases you may prefer Labels to Archives. Example: 1976 Today begins in November of 2006 and concludes in December of 2006, but there are other related posts in other months. Note: Labels only shows 20 posts at a time. There are 21 hubs, making 21 (which is for 1976 Today) an older hub.] ********************************* to my online poems and song lyrics using Archives. Use hubs for finding archival locations but do not link through them. Originally an AOL Journal, where the archive system was nothing like the system here, this blog was migrated from there to here in October of 2008. Today (Memorial/Veteran's Day, May 25, 2009) I discovered a glitch when trying to use a Blogger archive. Now, it may be template-related, but I am unable to return to S M or to the dashboard once I am in the Archives. Therefore, I've decided on this approach: a month-by-month post guide. The sw you see in the codes here stood for Salchert's Weblog when I began it in November of 2006. It later became Sprintedon Hollow. AOL provided what were called entry numbers, but they weren't consistent, and they didn't begin at the first cardinal number. That is why the numbers after "sw" came to be part of a post's code. ************** Here then is the month-by-month post guide: *2006* November: 00001 through 00046 - December: 00047 through 00056 -- *2007* January: 00057 through 00137 - February: 00138 through 00241 - March: 00242 through 00295 - April: 00296 through 00356 - May: 00357 through 00437 - June: 00438 through 00527 - July: 00528 though 00550 - August: 00551 through 00610 - September: 00611 through 00625 - October: 00626 through 00657 - November: 00658 through 00729 - December: 00730 through 00762 -- *2008* January: 00763 through 00791 - February: 00792 through 00826 - March: 00827 through 00849 - April: 00850 through 00872 - May: 00873 through 00907 - June: 00908 through 00931 - July: 00932 through 00955 - August: 00956 through 00993 - September 00994 through 01005 - October: 01006 through 01007 - November: 01008 through 01011 - December: 01012 through 01014 -- *2009* January: 01015 through 01021 - February: 01022 through 01028 - March: 01029 through 01033 - April: 01034 through 01036 - May: 01037 through 01044 - ******************************************************* 1976 Today: 2006/11 and 2006/12 -- Rooted Sky 2007: 2007/01/00063rsc -- Postures 2007: 2007/01/sw00137pc -- Sets: 2007/02/sw00215sgc -- Venturings: 2007/03/00216vc -- The Undulant Trees: 2007/03/00266utc -- This Day's Poem: 2007/03/00267tdpc -- Autobio: 2007/04/sw00316ac -- Fond du Lac: 2007/04/00339fdl -- Justan Tamarind: 2007/05/sw00366jtc -- Prayers in December: 2007/05/sw00393pindc -- June 2007: 2007/06/sw00440junec -- Seminary: 2007/07/sw00533semc -- Scatterings: 2008/08/00958sc ** Song Lyrics: 2008/02/sw00797slc ********** 2009-06-02: Have set S M to show 200 posts per page. Unfortunately, you will need to scroll to nearly the bottom of a page to get to the next older/newer page.

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## Saturday, May 24, 2008

### sw00903math--squares-sets-from-4-divisor

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