Because you so nasty!

I want to share this amazing way of making spirals from integers. The
equation is:

repeat
B:= Trunc(Sqrt(A));
C:= 2*Sqr(B) - A;
if A = C then Exit_This_Block;
A:= C;
until A <= 3;

Functional progaram is available at:
www.geocities.com/dedanoe/delphiart.zip I was interested if you can *****
up with some solid theory for this. My tendency was to get this type of
matrix equation:

+- -+ +- -+ +- -+
| Sqrt(A) B | | x1 x2 | | B Sqrt(C) |
| | = | | X | |
| B Sqrt(A) | | y1 y2 | | Sqrt(C) B |
+- -+ +- -+ +- -+

The funny thing is that this is how my dynamic lever functions. Go on,
check my usenet posts. Then I intend to turn the matrix {{x1, x2}, {y1,
y2}} into {{0, -1}, {1, B}}. I hope that will help me write A (if A is
not prime) as product of two integers. For that purpose I can extend
one row (column) and add it to the other row (column). Yes, indeed the
matrix {{x1, x2}, {y1, y2}} has norm =1 because:

|Sqrt(A) B| |B Sqrt(C)|
A-B^2=| |=| |=B^2-C
|B Sqrt(A)| |Sqrt(C) B|

Some examples to get you interested:
16 - 9 = 9 - 2; BigestCommonDivider(16, 2) = 2 <> 1;
15 - 9 = 9 - 3; BigestCommonDivider(15, 3) = 3 <> 1;
14 - 9 = 9 - 4; BigestCommonDivider(14, 4) = 2 <> 1;
13 - 9 = 9 - 5; 13 and 5 are both primes;
12 - 9 = 9 - 6; BigestCommonDivider(12, 6) = 6 <> 1;
11 - 9 = 9 - 7; 11 and 7 are both primes;
10 - 9 = 9 - 8; BigestCommonDivider(10, 8) = 2 <> 1;

Note this:
Step 1: 117 - 100 = 100 - 83;
Step 2: 83 - 81 = 81 - 79;
Step 3: 79 - 64 = 64 - 49;
End: 49 = Sqr(7);
117 = 3 * 3 * 13

***** on give me some theory, right?

Dobri Karagorgov -- Dedanoe
schizo-paranoid-lever-sex-pert
http://dedanoe.tripod.com

Appendix

+- -+ +- -+ +- -+
| Sqrt(A) B | | x1 x2 | | B Sqrt(C) |
| | = | | X | |
| B Sqrt(A) | | y1 y2 | | Sqrt(C) B |
+- -+ +- -+ +- -+

+- -+ +- -+
| Sqrt(A) B | | Sqrt(A)+kB B+kSqrt(A) |
| | ~ | |
| B Sqrt(A) | | B Sqrt(A) |
+- -+ +- -+

+- -+ +- -+
| Sqrt(A) B | | Sqrt(A)+kB B+kSqrt(A)+n(Sqrt(A)+kB) |
| | ~ | |
| B Sqrt(A) | | B Sqrt(A)+nB |
+- -+ +- -+

+-
|(Sqrt(A)+kB)(Sqrt(A)+nB)=A
|
|B(B+kSqrt(A)+n(Sqrt(A)+kB))=B^2
+-

+-
|(k+n)Sqrt(A)+nkB=0 -nkB
| => Sqrt(A) = -------
|(k+n)Sqrt(A)+nkB=0 n + k
+-

+- -+ +- -+
| Sqrt(A) B | | -nkB/(n+k)+kB B(1+nk)+(n+k)(-nkB)/(n+k) |
| | ~ | |
| B Sqrt(A) | | B -nkB/(n+k)+nB |
+- -+ +- -+

+- -+ +- -+
| Sqrt(A) B | | k^2/(n+k) 1 |
| | ~ B^2| |
| B Sqrt(A) | | 1 n^2/(n+k) |
+- -+ +- -+

+- -+ +- -+
| B Sqrt(C) | | B+kSqrt(C) Sqrt(C)+kB |
| | ~ | |
| Sqrt(C) B | | Sqrt(C) B |
+- -+ +- -+

+- -+ +- -+
| B Sqrt(C) | | B+kSqrt(C) Sqrt(C)+kB+nB+nkSqrt(C) |
| | ~ | |
| Sqrt(C) B | | Sqrt(C) B+nSqrt(C) |
+- -+ +- -+

+-
|(n+k)B+nkSqrt(C)=0 -(n+k)B
| => Sqrt(C) = ---------
|(n+k)B+nkSqrt(C)=0 nk
+-

+- -+ +- -+
| B Sqrt(C) | | k/n (n+k)/(nk) |
| | ~ B^2| |
| Sqrt(C) B | | (n+k)/(nk) n/k |
+- -+ +- -+

+- -+ +- -+ +- -+
| k^2/(n+k) 1 | | x1 x2 | | k/n (n+k)/(nk) |
| | = | | X | |
| 1 n^2/(n+k) | | y1 y2 | | (n+k)/(nk) n/k |
+- -+ +- -+ +- -+

Now we only need to pick them k and n so that k^2/(n+k) and n^2/(n+k)
are integers.

+- +-
|k^2=x(n+k) |k(k-x)=xn k x n-y
| => | => --- = ----- = -----
|n^2=y(n+k) |n(n-y)=yk n k-x y
+- +-

xy=(n-y)(k-x)=nk-nx-ky+xy

nk=nx+ky

1 = (x/k) + (y/n)

You're welcome to conclude this math!

Dobri Karagorgov -- Dedanoe
schizo-paranoid-lever-sex-pert
http://dedanoe.tripod.com

dedanoe wrote:

I want

=20
i want you to ***** yourself.
=20
we dont always get what we want.