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## Re: [PrimeNumbers] Re: A Property

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• - I have designed a new (i think it is new) and quite simple compression algorithm which can compress a string of ANY length into a string of 666 bytes (using
Message 1 of 64 , Oct 21, 2005
- I have designed a new (i think it is new) and quite simple compression
algorithm which can compress a string of ANY length into a string
of 666 bytes (using the well-known fact that 36 n^2 - 666 n + 1277 is prime
for all n between 0 and 42,inclusive )(1)

- I'm currently working on the corresponding decompression algorithm ...

(1)
-> for(n=0;n<=42;n++) {p = 36*n**2 - 666*n +1277; println(n,p,isprime(p));}

0 1277 2
1 647 2
2 89 2
3 -397 2
4 -811 2
5 -1153 2
6 -1423 2
7 -1621 2
8 -1747 2
9 -1801 2
10 -1783 2
11 -1693 2
12 -1531 2
13 -1297 2
14 -991 2
15 -613 2
16 -163 2
17 359 2
18 953 2
19 1619 2
20 2357 2
21 3167 2
22 4049 2
23 5003 2
24 6029 2
25 7127 2
26 8297 2
27 9539 2
28 10853 2
29 12239 2
30 13697 2
31 15227 2
32 16829 2
33 18503 2
34 20249 2
35 22067 2
36 23957 2
37 25919 2
38 27953 2
39 30059 2
40 32237 2
41 34487 2
42 36809 2
-------------------------------------
http://www.echolalie.com
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• ... Thank you David. I had been confused by consideration of the following: if x y = z and y x, then if we set t = (y+x)/2 and s = (y-x)/2 then z = t^2 - s^2
Message 64 of 64 , Oct 20, 2012
On 10/20/2012 10:07 AM, primenumbers@yahoogroups.com wrote:
> 2.2. Re: Question
> Date: Fri Oct 19, 2012 9:02 am ((PDT))
>
>
>
> Kermit Rose<kermit@...> wrote:
>
>> >Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
>> >
>> >for what values of x would
>> >x^2 + 5 x + 6 be a perfect square.
> Set A = 1, B = 0, C = -1, D = 5, E = 0, F = 6.
>
> and you will get the obvious answer from Dario:
>
> x = -2
> y = 0
> and also:
> x = -3
> y = 0
> Calculation time: 0h 0m 0s
>
> David

Thank you David.

I had been confused by consideration of the following:

if x y = z and y > x,
then if we set t = (y+x)/2 and s = (y-x)/2

then z = t^2 - s^2
and x = y - 2 t.

So z = x y = (y - 2t) y = y^2 - 2 t y

y^2 - 2 t y - z = 0

Transforming w = y - k yields

(w + k)^2 - 2 t (w + k) - z = 0

w^2 + (2 k - 2 t) w + (k^2 - 2 t k - z) = 0

which has integral solution if and only if

(2k - 2 t)^2 + (z + 2 t k - k^2) is an integral square

In all that derivation, I forgot that I still did not know what value t was.

I confused myself too easily.

Kermit

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