Multiple solutions to Diophantine Equation $X = \tfrac{1}{2} xy(x^2 - y^2) $ I am trying to figure out if the following is true or false:
$X = \tfrac{1}{2} xy(x^2 - y^2) $ has three (or more) solutions of $x, y$ if and only if $X = 6561555*n^4 = 3*5*7*11*13*19*23*n^4 $ where n is an odd number. (Where X, x, y are all Natural Numbers and x, y have opposite parity).
I can prove the "if" part rather easily as there are solutions of
$x1 = n*2*3*23$, $y1 = n*5 $
$x2 = n*7*11$, $y2 = n*2*19 $
$x3 = n*2*3*13$, $y3 = n*5*11 $
Proving if that there are no other values of X with three solutions is difficult.  I would be happy with either a proof or a counter example.  Thanks.
 A: The statement that the only solution is $X =6561555n^4$ is false. The problem is to find three Pythagorean triples such that the products of their legs is equal,
$$\tfrac{1}{2}x_1y_1(x_1^2-y_1^2) = \tfrac{1}{2}x_2y_2(x_2^2-y_2^2) = \tfrac{1}{2}x_3y_3(x_3^2-y_3^2) = X\tag1$$
However, a second primitive solution is,
$$X = 1285021492755 = 3\cdot5\cdot7\cdot11\cdot13\cdot19\cdot23\cdot37\cdot67\cdot79$$
where,
$$x_1,y_1 =1610,\, 869\\ 
x_2,y_2 =2002,\, 1817\\
x_3,y_3 =2622,\, 143$$
P.S. Note that there is an infinite number of integer solutions to,
$$x_1y_1(x_1^2-y_1^2) = x_2y_2(x_2^2-y_2^2)\tag2$$
so there may a large subset (presumably infinite) such that $(1)$ is true as well.
A: (This is an addendum to my answer.)
Incidentally, another interesting thing about, 
$$a=2^4X_1 = 2^4\cdot6561555=104984880\tag1$$ 
is that, given four different cuboids with dimensions $a,b,c,$
$\hskip2.0in$
$$C_1(a, b, c) = 104984880, \,18503891, \, 602547660 \\
C_2(a, b, c) = 104984880, \,69423660 , \, 263812549\\
C_3(a, b, c) = 104984880, \,123913972, \, 172044675\\
C_4(a, b, c) =104984880, \, 208276965, \, 365714812 $$
then all four cuboids obey,
$$a^2+b^2 =x_1^2\\
a^2+c^2 =x_2^2\\
a^2+b^2+c^2 =x_3^2$$
where the $x_i$ are integers. This was found by Randall Rathbun. I suspect one can do the same for,
$$a=2^4X_2 = 2^4\cdot 1285021492755 = 20560343884080\tag2$$
but it would take my old computer too long to find $b,c$.
