- Hi Michael

Yes I wasn't very clear. Here's an example.

The prime 17 is in the mod8==1 group. (Actually the smallest member.)

According to the model there should be

exactly one way in which x^2 + y^2 = 17,

exactly one way in which x^2 + 2*y^2 = 17

exactly one way in which x^2 - 2*y^2 = 17.

where y < 17/2.

Solutions are:

1^2 + 4^2 = 17 (x=1,y=4)

3^2 + 2*2^2 = 17 (x=3,y=2)

5^2 - 2*2^2 = 17 (x=5,y=2)

Of course it would be more interesting to try higher primes like

100913, which is also of the mod8=1 variety. So like 17 it should

(and does) have exactly one solution for each of the three equations.

The prime 100999 is of the mod8=7 variety. So

x^2 + y^2 = 100999 should (and does) have no solution.

x^2 + 2*y^2 = 100999 should (and does) have no solution.

x^2 - 2*y^2 = 100999 should (and does) have exactly one solution.

Compare this to the composite number 100521.

x^2 + y^2 = 100521 has 2 solutions.

x^2 + 2*y^2 = 100521 has 10 solutions.

x^2 - 2*y^2 = 100521 has 2 solutions.

I'm embarrassed to say that I have no idea how mundane such

observations are.

Mark

--- In primenumbers@yahoogroups.com, "Michael Gian" <work.gian@s...>

wrote:>

from

> Mark,

>

> Please humor a newbie; I'm not sure I understand what "has a

> solution" means in context of this message.

>

> --- In primenumbers@yahoogroups.com, "Mark Underwood"

> <mark.underwood@s...> wrote:

> >

> > Hi all,

> >

> > Some observations follow. These will be old news to number

> theorists

> > (I assume), but I've found it quite interesting. I make no

> guarantee

> > that the (rather hasty) observations below are entirely

> correct :)

> >

> >

> > ********

> >

> >

> > One way to consider all the (odd) primes is by grouping them

> > according to the remainder when divided by 8. Four (about equal)

> > groups result, each with their particular modulus eight value:

> 1,3,5

> > or 7. Four sample primes from each group respectively are

> 17,19,13

> > and 23.

> >

> > Now it turns out that each of these groups has a distinctive

> feature

> > as they relate to 3 quadratic expressions:

> >

> > x^2 + y^2

> > x^2 + 2*y^2

> > x^2 - 2*y^2

> >

> > where y is less than half the expression.

> >

> > Each prime from the mod8 == 1 group has exactly one solution for

> each

> > of the above three quadratic relations.

>

> When you say "has exactly one solution for each of the above three

> quadratic relations" do you mean that there exists a unique pair,

> (x,y) such that:

> x^2 + y^2 == 1 mod 8 and

> x^2 + 2*y^2 == 1 mod 8 and

> x^2 - 2*y^2 == 1 mod 8 ?

>

> I count at least four (x,y)that solve all three:

> {(1,0), (3,0), (5,0), (7,0)},

> and many more that solve one, if not two of them.

> I am also finding many pairs which solve {3, 5, 7} mod 8, per what

> is said below.

>

> Perhaps you can explain where I am missing the meaning of "solves".

>

> Michael

>

> >

> > Each prime from the mod8 == 3 group has exactly one solution for

> the

> > x^2 + 2y^2 relation and none for the other two.

> >

> > Each prime from the mod8 == 5 group has exactly one solution for

> the

> > x^2 + y^2 relation and none for the other two.

> >

> > Each prime from the mod8 == 7 group has exactly one solution for

> the

> > x^2 - 2y^2 relation and none for the other two.

> >

> > It appears that a composite number n can and will look like a

> prime

> > in this regard if and only if

> >

> > n = p * 2^m

> >

> > n = p^2 * 2^m

> >

> > n = p1^2 * p2 * 2^m The two primes involved must be from

> different

> > groups, and not from the mod8 ==1 group.

> >

> > n = p1^2 * p2^2 * p3 * 2^m The three primes involved must be

> > three different groups and not from the mod8 == 1 group.

> >

> >

> > ****

> >

> >

> > More observations, but I'll spare you :)

> >

> >

> > Mark - On Sunday 03 April 2005 23:48, you wrote:
> The prime 100999 is of the mod8=7 variety. So

Fermat has proven that all primes p == 1 mod 4 are uniquely represented as a

> x^2 + y^2 = 100999 should (and does) have no solution.

sum of two squares, while primes p == 3 mod 4 have no such representation. p

== 7 mod 8 is also == 3 mod 4.

> x^2 + 2*y^2 = 100999 should (and does) have no solution.

I believe this is related to the theory of imaginary quadratic fields (and in

particular elliptic curves with complex multiplication, which I'm a little

bit more familiar with, despite what David might rightly say after my

blunders on the primeform list). Now, someone correct me if I'm wrong, but

you can show that there's indeed only one solution, related to the

decomposition of 100999 in the field Q(sqrt(-2)). You can use Cornacchia's

algorithm to efficiently search for solutions or show that there are none.

> x^2 - 2*y^2 = 100999 should (and does) have exactly one solution.

Now this is out of my territory, but by symmetry I would risk the very

uneducated guess that it is related to the theory of real quadratic fields.

> Compare this to the composite number 100521.

I believe this is related to the fact that 100521 can be factorized in two

>

> x^2 + y^2 = 100521 has 2 solutions.

different ways (up to units, of course) in the ring of Gaussian integers

Z[i]: 3^4*(4 - i)*(3 + 8i) and 3^4*(4 + i)*(3 + 8*i).

> x^2 + 2*y^2 = 100521 has 10 solutions.

Similarly, I believe this should be factorable in Q(sqrt(-2)) in 10 different

ways up to units.

> x^2 - 2*y^2 = 100521 has 2 solutions.

Won't comment here, but I'm hoping the idea about real quadratic fields and

its extension here is correct.

Décio

[Non-text portions of this message have been removed] - --- In primenumbers@yahoogroups.com, "Mark Underwood"

<mark.underwood@s...> wrote:>

...

> x^2 - 2*y^2 = 100999 should (and does) have exactly one solution.

...

>

> Mark

Solutions of x^2 - 2*y^2 = 100999:

>

x0=357, y0=115

Xn+1 = 3 Xn + 4 Yn

Yn+1 = 2 Xn + 3 Yn

so we can deduce:

x1=1531, y1=1059

x2=8829, y2=6239

x3=51443, y3=36375

x4=299829, y4=212011

x5=1747531, y5=1235691

x6=10185357, y6=7202135

x7=59364611, y7=41977119

x8=346002309, y8=244660579

x9=2016649243, y9=1425986355

....

We can also negate x_n and/or y_n.

You can find the solution using my Quadratic Diophantine Equation

Solver at:

http://www.alpertron.com.ar/QUAD.HTM

Best regards,

Dario Alpern

Buenos Aires - Argentina - Thank you Decio and Dario

Decio I looked up Fermat's "4n+1" Theorem and it is as you say, and

proved by Euler. (That is, every prime of the form 4n+1 can be

uniquely expressed as x^2 + y^2.) Thanks for the input on the other

forms. I have a lot to learn.

Dario, your data was great, thanks. Because of it I am going to

investigate something further. I had only done solutions counts, not

investigated the properties of the actual solutions before.

Your data showed many solutions, and I said their was only one

solution for p = x^2 -2*y^2 , p = 8m+7. What gives?

Brings out another one of my mistakes! I had said that y<p/2, but

what I meant to say was that y < p^2 /2. In such a case there is only

one solution.

I've book marked your very cool site BTW!

Mark

--- In primenumbers@yahoogroups.com, "Dario Alpern" <alpertron@h...>

wrote:>

> --- In primenumbers@yahoogroups.com, "Mark Underwood"

> <mark.underwood@s...> wrote:

> >

> ...

> > x^2 - 2*y^2 = 100999 should (and does) have exactly one solution.

> >

> ...

> > Mark

> >

>

> Solutions of x^2 - 2*y^2 = 100999:

>

> x0=357, y0=115

>

> Xn+1 = 3 Xn + 4 Yn

> Yn+1 = 2 Xn + 3 Yn

>

> so we can deduce:

>

> x1=1531, y1=1059

> x2=8829, y2=6239

> x3=51443, y3=36375

> x4=299829, y4=212011

> x5=1747531, y5=1235691

> x6=10185357, y6=7202135

> x7=59364611, y7=41977119

> x8=346002309, y8=244660579

> x9=2016649243, y9=1425986355

> ....

>

> We can also negate x_n and/or y_n.

>

> You can find the solution using my Quadratic Diophantine Equation

> Solver at:

>

> http://www.alpertron.com.ar/QUAD.HTM

>

> Best regards,

>

> Dario Alpern

> Buenos Aires - Argentina