sw00817math-page2.subtracting.squares

entry 18

tpo = termposition      pown = positive odd whole number
                                 pewn = positive even whole number

Today am beginning subtraction sequence with 1² - 0² = 1.
Am interested in how it is not like the sequence presented in entry 17.

1² - 0² = 1

2² - 1² = 3      3² - 2² = 5      4² - 3² = 7

5² - 4² = 9 = 3²

[ note: this puts 9 at tpo 3 and tpo 5 instead of at tpo 2 and tpo 4 --
          is this better, worse, or just different? ]
[ note: a pown whose only multiples are itself and 1/ never appears
          above level one in a descension-by-squares sequence/
          because it is not a square of any whole number square root ]
[ note: the "itself" must be a pown greater than "1" ]

6² - 5² = 11      7² - 6² = 13

[ note: 25 (5²) first appears in a d-b-s sequence as the
          minuend in the squares' subtraction whose remainder is 9 ]
[ note: my personal view is that today's sequence is easier to work with ]
[ note: if you are wondering why I am talking about levels greater than
          level one but am not showing any such, it is because for clarity
          I have chosen a collapsed form -- explanation: ordinarily a d-b-s
          declines oppositionally from its major minuend/ so that at its
          highest level 5² would show 2² as its subtrahend/ and then 4²
          and so reveal the remainders "21" and "9" and a d-b-s (DF10)
          consisting of two levels -- not so bad, but at higher levels can 
          quickly get unwieldy ]          (revised 02/21/08)
[ note: DF = descension family      and the 10 = sum of factors in DF
          for 5 x 5:      5 + 5  or  3 + 7  or  1 + 9 ]          (added 02/21/08) 

Okay/// so in this sequence the tpo will always equal (the remainder
plus 1) divided by 2.  Therefore, if the minuend square root is 128, 
the remainder will be (128 times 2) minus 1.  I prefer letting the minuend
square root be the tpo.  Still, there seems to be no easy way to know if
a given pown is only divisible by itself and 1.  Mathematicians, using
a specialized formula, have found such numbers that are figuratively
beyond our solar system; but I am interested in something more basic.

8² - 7² = 15 = 3 x 5

9² - 8² = 17      10² - 9² = 19

11² - 10² = 21 = 3 x 7

12² - 11² = 23

13² - 12² = 25 = 5²

[ note: each pown tpo (square root) equals the square root of the 
          remainder of that term in this sequence which relates to it,
  thus 5² finds its equal at tpo 13 or (5 x 3) - 2 ]

Here are some examples:
-
(3 x 2) - 1      tpo 3 and tpo 5          (5 x 3) - 2      tpo 5 and tpo 13
(7 x 4) - 3      tpo 7 and tpo 25          (9 x 5) - 4      tpo 9 and tpo 41
(11 x 6) - 5      tpo 11 and tpo 61          (13 x 7) - 6      tpo 13 and tpo 85

[ note: the matching tpo locations are separated from each other in just the
          same manner as they are in the sequence shown on entry 17, but are
          one tpo higher here -- what is 2 and 4 there is 3 and 5 here ]

Though I wanted to avoid mentioning facts I have long known, the 
usefulness of them suggests otherwise.  Only odd numbers matter,
and among those only those numbers with final digits that are not
"5".  Any pown whose digits add to 3 or 6 or 9 is divisible by "3".
"861" is such a number in that 8 + 6 = 14 and 14 + 1 = 15 and
1 + 5 = 6.  Now, there is what I have come to call an Elimination
Table (E T) which effectively bypasses all numbers divisible by "3".
Am not going to explain that here.  However--to continue with
more of the obvious, a greater-than-1 pown does not acquire its
power to invalidate the primality of other powns until it is squared.
"2" is the first prime number, and it is the reason 3, 5, and 7 are
prime numbers.  "3" is the reason 9, 15, and 21 are not prime #s,
but 11, 13, 17, 19, and 23 are.

Another way to ferret out the truth about a given pown is by adding
squares to it, beginning with 1².  This works with "143" since 144
is 12 x 12.  Here observation tells one "143" in not divisible by 3, or
by 5, or by 7; but must be divisible by two prime numbers.  Which
two?  1² indicates that whichever two they are they must be 2
apart from each other.  So, (12 - 1) = 11 and (12 + 1) = 13.  Let's
see.  Yes.  So where are the squares?  The squares run along a
line separated by the powns in ascending order.  From 0 to 1 is 1;
from 1 to 4 is 3; from 4 to 9 is 5; from 9 to 16 is 7; but is there a
trick of sorts one can use?  Am not sure; but do know every pown
square is 1 greater than a number divisible by 8, and that every
pewn square is divisible by 4.  However, something different is
teasing my brain.  1/2 = .5; 4/2 = 2; 9/2 = 4.5; 16/2 = 8; 25/2 =
12.5; 36/2 = 18; 49/2 = 24.5  |  2 - .5 = 1.5; 4.5 - 2 = 2.5; 8 -
4.5 = 3.5; 12.5 - 8 = 4.5; 18 - 12.5 = 5.5; 24.5 - 18 = 6.5  |
I know: I left my flyswatter on the moon.      .5, 1.5, 2.5, 3.5,
4.5, 5.5, 6.5 is a run of numbers wherein each greater number
is 1greater than the one before it.  Have also noted another fact
about these numbers, but am not sure about its value.  If .5 is
allowed to represent 1, 1.5 to represent 4, 2.5 to represent 9,
3.5 to represent 16, 4.5 to represent 25, 5.5 to represent 36,
and 6.5 to represent 49, how does one use them?  The other
fact I noted is that if .5 is added to one of these numbers, the
resulting number is the square root of the number represented.
That's fine, but is not what I was hoping for.  Similarly, 2 times
one of these .5 numbers gives us the difference between that
number it represents and next lesser represented number; but
that too is not what I was hoping for.  Hmm: 1 to 5 = 4; 5 to 9 =
4; 9 to 13 = 4 -- this is for 1, 9, 25, and 49.  Likewise: 3 to 7 =
4, and 7 to 11 = 4 -- this is for 4, 16, and 36.  On the moon?  How
about on an asteroid.  For what it's worth: If the result of dividing 4
into a pewn is a square, then that pewn is a square.       



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Brian A. J. Salchert