Complete solution for 4x4
semi-bimagic squares with one magic or bimagic diagonal
by Lee Morgenstern, August-September 2011.
Complete solution for rational 4x4 semi-bimagic squares with one magic diagonal.
Choose rational t and x. Compute w from either equation [1] or [2].
t^2
- 1
[1] w = ----------------------
2(t^2
- 2tx + x^2 - x)
(t^2
- 1)(x^2 + x - 1)
[2] w = -------------------------------------------------
2((t^2)(x^2)
+ 2tx^3 + x^4 - xt^2 - 2x^2 - x + 1)
- t^2 - 2t + 2x^2
+ 2x + 1
Z = ---------------------------
t^2
- 1
wZ^2 + Zwx + 3Zw
- Z - wx + 2w + 3x - 2
D = ---------------------------------------
Z(Zw
+ 3w - 1)
J = (D - 1)(Z + 1)
M = 1 + x/(Zw)
G = x + x/(Zw)
H
= M - 1 - (M - 1) / (G - M + 1)
C = J + (D - 1) / (D - J - 1)
L = C +
(M - 1)(G - J) / [(C - J)(G - M + 1) + M - 1]
Q = G - D + 1 + [(D - 1)(D
- G) - J(J - M)] / (D - G - J + M - 1)
B = - C + H + L + Q
E = D - G +
L + Q - 1
I = G + H - M + 1
K = D - G - J + M + Q - 1
N = C + D - J
- 1
P = - C + H + J + L - M + 1
F = 1
A = 0
A B C D
E F G H
I J K L
M N P Q
D,G,J,M will be a magic diagonal.
Example using equation [1] for w.
t = 6, x = -5 --> w = 5/36, Z = -1/5
0 351 -99 -129
-269 1 175 216
211 -104 -209 225
181 -125 256 -189
Complete formulation for all rational 4x4 semi-bimagic
squares with one bimagic diagonal,
which can then be translated to a 4x4
nearly bimagic square with 19/20 correct sums.
Finding a rational solution to either of
equations [3] or [4] below generates a 19/20 bimagic square solution.
Finding
all rational solutions to both equations [3] and [4] generates all 19/20 bimagic
square solutions.
Solve equation [3] for rational u and x.
[3] (x^4)(u-1)(u-1)(u-2)(u-2)
+
(x^3)(u)(u-1)(u-2)(u-2)
+ (x^2)(u^4 - 5u^3 + 12u^2 - 14u
+ 7)
+ (x)(u)(-2u^2 + 5u - 5)
+ u^2
- u + 1 = 0
Set t
= ux.
Compute w using equation [1]
above.
Compute Z, D, etc as above.
--------------
Two rational solutions
to equation [3] are
(x,u) = (1,1)
(x,u)
= (1,2)
but they both lead to magic
squares with duplicate entries.
--------------
Solve equation [4] for rational t and x.
[4] (t^4)(x^2)(x^4 - 3x^3 + 4x^2 - 3x + 1)
+
(t^3)(x)(6x^6 - 9x^5 + 3x^4 - x^3 + x^2 + x - 1)
+ (t^2)(13x^8
- 2x^7 - 9x^6 - 5x^5 + 2x^4 + 5x^3 + x^2 - 3x + 1)
+ (t)(12x^9
+ 12x^8 - 6x^7 - 17x^6 - 9x^5 + 9x^4 + 13x^3 - 5x^2 - 5x + 2)
+
4x^10 + 8x^9 + 3x^8 - 6x^7 - 10x^6 - 6x^5 + 6x^4 + 8x^3 - 2x^2 - 3x + 1 = 0
Compute w using equation [2] above.
Compute
Z, D, etc as above.
--------------
Two rational solutions to equation
[4] are
(x,t) = (0,-1)
(x,t) = (1,-1)
but
they both lead to magic squares with duplicate entries.
Are there any other rational solutions
to equations [3] and [4]?
I can't find any. I graphed both equations [3]
and [4] using a surface plot, height = f(t,x).
Places where height = 0 would
be solutions.
Places where height = 0 and t and x are both rational would
be rational solutions.
But for both equations, the entire surface (except
for two points) appear to be above the height = 0 plane.
So, visually, there
aren't any other rational (or real) solutions.
An idea for a rigorous proof is to try
and exploit the height >= 0 feature.
If equations [3] and [4] could be
expressed as the sum of two or more squared expressions, that would do it.
The sum of squares must be non-negative.
That would explain the height >= 0 feature. Also, to be 0, each
expression must be 0. So you would have at least two simultaneous equations
in two unknowns.
That would mean there are a finite number of solutions and
the rational ones could be easily enumerated.
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