Mq = 7 (mod 120) ==> Mq = (2x)^2 + q*(3y)^2 ??

[Tony Reix]

We have the following data :D :

M7 = 127 = 120*1 + 7 = (2*4)^2 + 7*(3*1)^2
M11 = 2047 = 120*17 + 7 = (2*17)^2 + 11*(3*3)^2
M23 = ... = 120*5*11*31*41 + 7 = (2*80)^2 + 23*(3*3*67)^2
M31 = ... = 120*29*43*113*127 + 7 = (2*4*647)^2 + 31*(3*3*919)^2

Is this enough for building the conjecture :

Mq = 7 (mod 120) ==> Mq = (2x)^2 + q*(3y)^2 with (x,y) unic.

We also have the data :D :

M7 = 1^2 + 2*7*(3)^2
M11 = 43^2 + 2*11*(3)^2 = (3*15)^2 + 2*11*(1)^2
M23 = (3*797)^2 + 2*23*(241)^2 = (11*251)^2 + 2*23*(3*43)^2
M31 = 42407^2 + 2*31*(3*7*113)^2

Is this enough for building the conjectures :

Mq = 7 (mod 120) ==> Mq = (x)^2 + 2*q*(y)^2 .

(Mq = 7 (mod 120) and Mq = (x)^2 + 2*q*(y)^2 with (x,y) unic) ==> Mq is prime.

It seems more computation is required to confirm/infirm the conjectures.

[Tony Reix]

I found the previous data while looking at the theorems:

{ p = x^2 + 5*y^2 and q = 20k + 1 or 9 } ==> { q | p ==> q = x^2 + 5*y^2 }

{ p = x^2 + 5*y^2 and q = 20k + 3 or 7 } ==> { q | p ==> q = 2*x^2 + 2*xy + 3*y^2 }

If we look at small factors of M11 and M23, we can see :

Let say: f(x,y) = 2*x^2 + 2*xy + 3*y^2 .

2047 = 23 * 89 and 23 = f(-1,3) = f(-2,3) = f(1,-3) = f(2,-3)

M23 = 47 * ... and 47 = f(-5,3) = f(-2,-3) = f(2,3) = f(5,-3).

Then I saw that some Mq are of the form: 20k+7 and I tried with the form: x^2 + k*y^2 , and finaly I found the previous data.

[Tony Reix]

The following prime q such that Mq = 7 (mod 120) are:

Mq prime: 107, 127, 607, ...
Mq not prime: 43, 59, 67, ...

[Tony Reix]

I forgot M19 :

M19 = 524287 = 120*17*257 + 7 = (2*257)^2 + 19*(3*3*13)^2

M19 = (683)^2 + 2*19*(3*13)^2

Which confirms the 3 conjectures :D .

We also have the noticeable relation-ships :

M11 = 43^2 + 2*11*(3)^2 and 3*43 = 129 = 2^7 + 1
M11 = 120*17 + 7 where 17 = 2^4 + 1

M19 = (683)^2 + 2*19*(3*13)^2 and 3*683 = 2049 = 2^10 + 1
M19 = 120*17*257 + 7 where 17 = 2^4 + 1 and 257 = 2^8 + 1

But this kind of relation-ship doesn't appear with the others (M7, M11, M23, M31).

So M11 and M19 seems to belong to the same "family".
Seems there is another conjecture to propose ?

[Tony Reix]

After some more computation, it seems that for each q prime (that I've tested: 5, 7, 11, 13, 17, 19, 23), there are many k such that Mq=x^2+ky^2 has (at least) a (x,y) solution.
But it seems that only not-prime Mq may have (at least) TWO solutions for a given k.

M11 = 43^2 + 22*3^2 = 45^2 + 2*11*1^2
M23 = 196^2 + 31*519^2 = 2676^2 + 31*199^2
M23 = 2391^2 + 2*23*241^2 = 2761^2 + 2*23*129^2

So the conjecture is:

If there is some k such that Mq = x^2 + k*y^2 has at least 2 solutions, then Mq is not prime.

Well, how can it be proven

[Maybeso]

Originally Posted by Tony Reix

So the conjecture is:

Well, how can it be proven

Tony,
Start with two solutions for some k:

Mq = a^2 + kb^2
Mq = x^2 + ky^2

solve for k in both and equate them,
after solving for Mq you get:

Mq = [(bx)^2 - (ay)^2]/(b^2 - y^2) = (bx - ay)(bx + ay)/(b - y)(b + y)

Does this imply that Mq is either composite or a fraction?

Bruce

[Tony Reix]

Here are more data (Mq = 7 (mod 120)) :

Mq = (2*a)^2 + q*(3*b)^2
---------------------------
q | a b
---------------------------
7 | 4 1
11 | 17 3
19 | 257 39
23 | 80 201
31 | 2588 2757
43 | 1233175 83731
43 | 635261 136227
47 | 5130496 289491
47 | 4091984 417581
59 | 40236967 32763051
67 | 986248409 488139951
71 | 23649483968 440498497

Mq = a^2 + 2q*b^2
---------------------------
q | a b
---------------------------
3 | 1 1
7 | 1 3
11 | 45 1
11 | 43 3
19 | 683 39
23 | 2761 129
23 | 2391 241
31 | 42407 2373
43 | 2654819 142569
43 | 382699 317139
47 | 8508129 852713
47 | 8159951 888177
59 | 686086693 29935779

These data show:
- if Mq is prime, it seems there is only 1 pair (a,b) that is solution of each above form.
- If Mq is composite, it seems there is 1 or 2 pairs (a,b) that are solution of each form.
(I never found more than 2 solutions).
As an example, for M59, there is only 1 pair (a,b) for the 2 forms. That means that there exists another k such that there is 2 solutions. But finding it would take a while.

About the "family" for M11 and M19, I haven't find other q with the same exact properties.

As a conclusion, studying Mq = a^2 + k*b^2 with Mq = 7 (mod 120) has shown interesting data about Mq.
And the most interesting one is that it seems there is always 1 or 2 solution (a,b) for the form: Mq = (2*a)^2 + q*(3*b)^2 [I] for all q such that Mq = 7 (mod 120).

Notice the 2 coefficients: 2 and 3, like in: Mq = (8x)^2 - (3qy)^2 or Mq = (2a)^2 + 3*(3b)^2 !

I never found a solution for [I] for Mq != 7 (mod 120).

About Bruce's question, I have no answer ...

[Tony Reix]

We have
2^4 * 2^3 = 16 * 8 = 128 = 8 (mod 120)
Thus
2^(4k+3) = 2^(4k) * 8 = 2^(4*(k-1)) * [ 16 * 8 ] = 2^(4*(k-1)) * 8 (mod 120) = 2^(4*(k-2)) * 8 = ... = 1 * 8 = 8 (mod 120)
And finaly
M(4k+3) = 7 (mod 120)