Some more (still unuseful) maths about Mersenne numbers.

[T.Rex]

In an old MersenneForum thread, I provided the proof of the following theorem:
q prime >5, Mq = A*B ==> there is a unique pair (x,y) such that : M[q] = (8x)^2 - (3qy)^2 (I)

Here are some more information.
1)
We know: (A prime or not) A | M[q] ==> A = 1 + 2qa .
M[q] = A*B = (1+2qa)(1+2qb) = 1+2q(2qab+a+b) .
Let say: S = a+b P = a*b D = b-a .
We have: S^2 - 4P = D^2
M[q] = 1 + 2q(2Pq + S) = 1 + 4Pq^2 + 2Sq = 1 + (S^2 - D^2)q^2 + 2Sq = 1 + 2qS + (qS)^2 - (Dq)^2
M[q] = (1 + qS)^2 - (Dq)^2 (II)
(I) and (II) ==> 8x = 1 + qS and 3y = D .
That entails: D = 1 (mod 2) .
Since: M[q-1] = 2^(q-1) - 1 is odd and M[q-1] = (M[q] - 1)/2 = q(2Pq + S) , then S = 1 (mod 2) .
We know: q odd ==> q^2 = 1 (mod 8) .
Since: 2Pq^2 = 2^(q-1) - (1+Sq) and 1+Sq = 8x then 2Pq^2 = 0 (mod 8). And thus: P = 0 (mod 4) .
2)
We know: a and b are solutions of: x^2 - Sx + P = 0 .
Let consider the curve [S,P] : y = x^2 - Sx + P .
The minimum of [S,P] is the point: m ( S/2 , -(D/2)^2 ) .
For a given q, all curves [Si,Pi] cross the point: M ( -1/(2q) , M[q]/(2q)^2 ) and all points m
(from different curves [S,P] for a q) belong to the curve [Mq] : y = (M[q] - (1 + 2qx)^2) / (2q)^2 .
M is the maximum of this curve.
All points with coordinates (S,P) belong to the line: P = (M[q] - (1 + 2qS)) / (2q)^2 .
The surface between [y = 0] and [S,P] is: - D^3 / 6 .
The surface between [S,P] and [Mq] is: ((1 + qS)/(2q))^3 / 3 .
It's useful to draw the curves with q = 11 and q = 29 ( 3 prime roots).

[T.Rex]

Here are 2 examples for the paragraph 1) of previous note.

a) q = 11 , M11 = 23*89
=======================
A=23 ; B=89
a= 1 ; b= 4
S= 5 ; P= 4; D= 3
x= 7 ; y= 1

M11 = (8*x)^2 - (3*q*y)^2 = (8*7)^2 - (3*11*1)^2
M11 = (1+29*S)^2 - (D*q)^2 = (1+5*11)^2 - (3*11)^2


b) q = 29 ; M29 = 233*1103*2089
===============================

-1) A1=233 ; B1=2304167
-----------------------
a1= 4 ; b1=39727
S1= 39731; P1=158908; D1=39723
x1= 144025 ; y1=13241

2*P1*q+S1 = M[q-1]/q = 9256395

8*x1 = 8*144025 = 1 + 29*39731 = 1 + q*S1 = 1152200
3*29*y1 = 3*29*13241 = 29*39723 = q*D1 = 1151967

M29 = (8*x1)^2 - (3*q*y1)^2 = (1+29*S1)^2 - (D1*q)^2

-2) A2=1103 ; B2=486737
-----------------------
a2= 19 ; b2= 8392
S2= 8411 ; P2= 159448; D2=8373
x2= 30490 ; y2= 2791

2*P2*q+S2 = M[q-1]/q = 9256395

8*x2 = 8*30490 = 1 + 29*8411 = 1 + q*S2 = 243920
3*29*y2 = 3*29*2791 = 29*8373 = q*D2 = 242817

M29 = (8*x2)^2 - (3*q*y2)^2 = (1+29*S2)^2 - (D2*q)^2

-3) A3=2089 ; B3=256999

a3= 36 ; b3= 4431
S3= 4467 ; P3= 159516; D3=4395
x3= 16193 ; y3= 1465

2*P3*q+S3 = M[q-1]/q = 9256395

8*x3 = 8*16193 = 1 + 29*4467 = 1 + q*S3 = 129544
3*29*y3 = 3*29*1465 = 29*4395 = q*D3 = 127455

M29 = (8*x3)^2 - (3*q*y3)^2 = (1+29*S3)^2 - (D3*q)^2

[T.Rex]

On the figure are displayed the curves and points for q=11 :
M : (-1/2q , Mq/4q^2) = (-)1/22 , 2047/484)
m : (S/2 , -(D/2)^2) = (5/2 , -9/4)
(S,P) : (5,4)

[S,P] is the green curve.
[Mq] is the red curve.
The yellow line contains all points such that:
P = (M[q] - (1 + 2qS)) / (2q)^2 .

The three curves (green, yellow, red) have a common point: M .

[T.Rex]

Here are the curves for q=23 .
The line containing the possible values for (S,P) cannot be drawn due to the scale.
The green curve is the [S,P] curve for which a and b are solution of y=0 .
The red curve is the [Mq] curve.

[T.Rex]

Here are the curves for q=29 .
The red curve is the [Mq] curve with maximum M.
The green, yellow and blue curves are the [Si,Pi] curves.
The green one deals with: a1=4 b1=39727
The yellow one deals with: a2=19, b2=8392
The blue one deals with: a3=36, b3=4431

The red curve [Mq] crosses the 3 other curves at m1, m2, m3, which
are the minimum of each of these 3 curves.
m1 = (S1/2 , -(D1/2)^2) = (39731/2 , -(39723/2)^2) .

[T.Rex]

Here is a graphical (2 parts) presentation of:
Mq = (8x)^2-(3qy)^2 = (1+Sq)^2 - (Dq)^2 .

First part:

[T.Rex]

Part 2:

[T.Rex]

The same as previously plus squares q^2 .

[T.Rex]

Here is a nice visual proof of: Mq = 1 (mod 6) .