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## Re: [PrimeNumbers] Re: primes and squares

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• ... Fermat has proven that all primes p == 1 mod 4 are uniquely represented as a sum of two squares, while primes p == 3 mod 4 have no such representation. p
Message 1 of 6 , Apr 3, 2005
On Sunday 03 April 2005 23:48, you wrote:
> The prime 100999 is of the mod8=7 variety. So
> x^2 + y^2 = 100999 should (and does) have no solution.

Fermat has proven that all primes p == 1 mod 4 are uniquely represented as a
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.
>
> x^2 + y^2 = 100521 has 2 solutions.

I believe this is related to the fact that 100521 can be factorized in two
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]
• ... 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,
Message 2 of 6 , Apr 4, 2005
--- 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:

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
Message 3 of 6 , Apr 4, 2005
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:
>