Find a (all) rational solutions, 0<(x,y,z)<1 to the equation:
$$\frac{x-x^3}{(x^2+1)^2} + \frac{y-y^3}{(y^2+1)^2} = \frac{z-z^3}{(z^2+1)^2}$$
or else show there are no rational solutions.
This equation arises in the following context:
A Magic square is an 3x3 positive integer matrix where all the rows, columns and diagonals have the same sum. An open question is whether a Magic Square of Squares exists (all the entries are themselves squares). A hourglass configuration examines the top row, middle entry and bottom row of the Magic Square. A valid hourglass configuration has the top row, bottom row, center column and both diagonals having the same sum.
Showing the above equation has no solutions implies there is no valid hourglass configuration of Squares and thus no Magic Square of Squares exists.
- With $m_i,n_i$ positive integers, $m_i>n_i$ and one of $m_i,n_i$ even.
For i = 1,2,3, let
$$a_i=(m_i^2+n_i^2)^2$$ $$b_i=4m_i n_i (m_i^2-n_i^2)$$ $$c_i=(m_i^2+n_i^2)^2$$
So that $a_i$ and $a_i±b_i$ are squares as $b_i$ are congruum.
With $a=a_1 a_2 a_3$
$a±b_1 a_2 a_3= a_2 a_3(a_1±b_1) $
$a±a_1 b_2 a_3=a_1 a_3 (a_2±b_2) $
$a±a_1 a_2 b_3=a_1 a_2 (a_3±b_3) $
We have 3 arithmetic sequences of 3 squares, all with center a.
- If $a_2 a_3 b_1+a_1 a_3 b_2-a_1 a_2 b_3=0$ , we can form a valid hourglass configuration as:
$$[ a_1 a_2 a_3+b_1 a_2 a_3 , a_1 a_2 a_3+a_1 b_2 a_3 , a_1 a_2 a_3-a_1 a_2 b_3$ ]$$
$$ [ - ,a_1 a_2 a_3, - ] $$
$$[ a_1 a_2 a_3+a_1 a_2 b_3 , a_1 a_2 a_3-a_1 b_2 a_3 , a_1 a_2 a_3 - b_1 a_2 a_3 ]$$
- Rearranging
$a_2 a_3 b_1+a_1 a_3 b_2-a_1 a_2 b_3=0$
$\frac{b_1}{a_1} +\frac{b_2}{a_2}-\frac{b_3}{a_3}=0$
$\frac{4m_1 n_1 (m_1^2-n_1^2 )}{(m_1^2+n_1^2)^2} +\frac{4m_2 n_2 (m_2^2-n_2^2 )}{(m_2^2+n_2^2)^2}-\frac{4m_3 n_3 (m_3^2-n_3^2 )}{(m_3^2+n_3^2)^2}=0$
Divding by 4 and multiplying each term by $\frac{1}{m_i^4}$ and let $x=\frac{n_1}{m_1},y=\frac{n_2}{m_2},z=\frac{n_3}{m_3}$
You obtain the original equation:
$$\frac{x-x^3}{(x^2+1)^2} + \frac{y-y^3}{(y^2+1)^2} = \frac{z-z^3}{(z^2+1)^2}$$
So showing this equation has no rational solutions -> there is no valid Magic Square Square hour glass configuration -> there is no valid Magic Square of Squares
Note:
You can obtain a closely related equation that has rational solutions, but I don't know how to use a solution from the related equation to find a rational solution main equation. The related equation is obtained by noting that in $$\frac{x-x^3}{(x^2+1)^2}$$ Multiplying top and bottom by $$\frac{1}{x^2}$$ Yields $$\frac {\frac{1}{x}-x} {(\frac{1}{x}+x)^2}$$ Which equals $$\frac {\frac{1}{x}-x} {(\frac{1}{x}-x)^2+4}$$ substituting in the main equation $$a=\frac{1}{x}-x , b=\frac{1}{y}-y, c=\frac{1}{z}-z$$ Gives $$\frac{a}{a^2+4} + \frac{b}{b^2+4} = \frac{c}{c^2+4}$$ With further substituting $$a =2a, b=2b, c=2c$$ has rational solutions. See: rational number solutions to $\frac{a}{a^2+1} + \frac{b}{b^2+1} = \frac{c}{c^2+1}$ with $abc\ne 0$
In this equation, setting c to a fixed value yields a series of elliptic curves. You can generate a parameterized solution via the following MAGMA code:
R<c>:=PolynomialRing(RationalField());
K<c>:=FieldOfFractions(R);
A<a,b>:=AffineSpace(K,2);
C := Curve(A,a*(b^2+4)*(c^2+4)+b*(a^2+4)*(c^2+4) -c*(a^2+4)*(b^2+4));
Genus(C);
Cp := ProjectiveClosure(C);
E, f:=EllipticCurve(Cp,Cp![c,0]);
boo,g:=IsInvertible(f);
T,h:=TorsionSubgroup(E);
P:=Cp![0,c,1];
Q:=f(P);
g(2*Q);
((-c^7 - 4c^5 - 48c^3 + 64c)/(c^6 - 12c^4 - 16c^2 - 64) : (8c^5 + 128c)/(c^6 - 4c^4 - 16*c^2 + 64) : 1)
so for any c:
$$a=\frac{((-c^7 - 4c^5 - 48*c^3 + 64c)}{(c^6 - 12c^4 - 16c^2 - 64)}$$
$$b=\frac{(8c^5 + 128c)}{(c^6 - 4c^4 - 16c^2 + 64)}$$
is a solution to:
$$\frac{a}{a^2+4} + \frac{b}{b^2+4} = \frac{c}{c^2+4}$$
as $a=\frac{1}{x}-x , b=\frac{1}{y}-y, c=\frac{1}{z}-z$
$$a=\frac{-(z^{14} - 3 z^{12} + 49 z^{10} - 203 z^8 + 203 z^6 - 49 z^4 + 3 z^2 - 1)}{(z^{13} - 18 z^{11} + 47 z^9 - 124 z^7 + 47 z^5 - 18 z^3 + z)}$$ $$b=\frac{(8 z (-1 + 5 z^2 - 26 z^4 + 26 z^6 - 5 z^8 + z^{10}))}{(1 - 5 z^2 - 5 z^4 + z^6)^2}$$ and: $$x=\frac{-a\pm\sqrt{a^2+4}}{2}$$ $$y=\frac{-b\pm\sqrt{b^2+4}}{2}$$
so if the discriminants:
$$a^2+4$$ $$b^2+4$$
are squares, we have a solution to the original equation.
$$\frac{x-x^3}{(x^2+1)^2} + \frac{y-y^3}{(y^2+1)^2} = \frac{z-z^3}{(z^2+1)^2}\tag{1}$$
$\bullet\ x=1$
Clearly, if $y=z$ then equation $(1)$ has infinitely many positive solutions as follows.
$(x,y,z)=(1,y,y).$
$\bullet\ y=1$
Similarly, equation $(1)$ has infinitely many positive solutions as follows.
$(x,y,z)=(x,1,x).$
$\bullet\ z=1$
Substitute $z=1$ to equation $(1)$ then, we get
$(x-x^3)(y^2+1)^2+(y-y^3)(x^2+1)^2=-(xy-x+1+y)(xy-1)(xy+x+1-y)(x+y).$
Hence if $xy-x+1+y=0$ or $xy=1$ then equation $(1)$ has infinitely many positive rational solutions as follows.
$(x,y,z)=(x,\frac{1}{x},1),(x, \frac{x-1}{x+1}, 1).$
Assuming all $(x,y,z) \ne 1$, no positive rational solution was found with height$\lt100.$