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

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 that integer
    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