Does equation $xy(x+y)(x-y)=10z^2$ have nonzero integer solutions? Supposing that solution exists, I've found that $x$, $y$, $(x+y)$, $(x-y)$ should be pairwise coprime numbers, and hence $x$ and $y$ are coprime of different parity. And after that I can not make any step further.
 A: The original question is equivalent to asking whether $10$ is a congruent number - which it isn't.
With a general multiplier $N$, we have
\begin{equation*}
xy(x+y)(x-y)=Nz^2
\end{equation*}
Define $t=x/y$ and $w=z/y^2$ giving
\begin{equation*}
t(t+1)(t-1)=Nw^2
\end{equation*}
and then define $s=Nt$ and $r=N^2w$ leading to
\begin{equation*}
r^2=s^3-N^2s
\end{equation*}
which is the elliptic curve for the congruent number problem. 
Non-zero solutions only come from curves with rank greater than zero. This includes $N=5,6,7$ which give curves of rank $1$. $N=2730$ gives a curve of rank $2$ using Denis Simon's ellrank code.
A: The reference to congruent numbers settles the original question and similar questions with $10$ replaced with $N$. There is an even more general result relating to division polynomials and discrete analogs of Weierstrass sigma and Jacobi theta functions. This is given in my work at 
Weierstrass Elliptic Function Polynomials.
In brief, I construct four  sequences $\, (w_n,x_n,y_n,z_n) \,$ such that
$\, (x_n/x_0)^2 - (y_n/y_0)^2 = (x_1^2-y_1^2) (w_n/w_1)^2, \,$
$\, (z_n/z_0)^2 - (y_n/y_0)^2 = (z_1^2-y_1^2) (w_n/w_1)^2, \,$
$\, (z_n/z_0)^2 - (z_n/z_0)^2 = (z_1^2-x_1^2) (w_n/w_1)^2  \,$ for all integer $\, n.$ They satisfy recursions such as
$\, w_{n+1}w_{n-1}x_0^2 = w_n^2 x_1^2 - x_n^2 w_1^2, \,$ and $\, 
 x_{n+1}x_{n-1}x_0^2 = x_n^2 x_1^2 -
 (x_1^2-y_1^2)(x_1^2-z_1^2)(w_n/w_1)^2 x_0^4.  $
For an numerical example we let
 $$  x_0 \!=\! y_0 \!=\! z_0 \!=\! w_1 \!=\! y_1 \!=\! 1 \quad \text{and} \quad
 x_1 \!=\! \sqrt{2},\,  z_1 \!=\! \sqrt{3},\, x_1^2-y_1^2 \!=\! 1,\, z_1^2-y_1^2 \!=\! 2,\,
 z_1^2-x_1^2 \!=\! 1.  $$
In this particular example we have
 $$ x_n^2-y_n^2 = w_n^2, \,\, z_n^2-y_n^2 = 2w_n^2, \,\, z_n^2-x_n^2 = w_n^2,
  \,\, x_n^2-w_n^2 = y_n^2, \,\, x_n^2+w_n^2 = z_n^2. $$
This implies $\, x_n^2 w_n^2 (x_n^2+w_n^2)(x_n^2-w_n^2) = (w_nx_ny_nz_n)^2
 = w_{2n}^2 /4. \,$
Because $\, w_n \,$ is a divisibility sequence, we have $\, w_{2n} = w_2t_n \,$
where $\, w_2 = 2\sqrt{6} \,$ and $\, t_n \,$ is an integer sequence. Thus,
 $\, w_{2n}^2/4 \!=\! 6\,t_n^2. $ This gives a sequence of solutions to
 $\, x^2 w^2(x^2+w^2)(x^2-w^2) = 6\,t^2. \,$  The four sequences
 $\, w_n^2, x_n^2, y_n^2, z_n^2 \,$ are integer sequences.
