Does equation $xy(x+y)(x-y)=10z^2$ have nonzero integer solutions?

Question

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.

Answer 1

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.

Answer 2

Sort of interesting. I replaced the $10$ with $2730$ in order to get at least one solution, it suddenly gave many:

Thu Dec  7 15:53:49 PST 2017
        10         3                2730  =   2 3 5 7 13
        13         7               10920  =   2^3 3 5 7 13
        13         8               10920  =   2^3 3 5 7 13
        14         1                2730  =   2 3 5 7 13
        15        13               10920  =   2^3 3 5 7 13
        21         5               43680  =   2^5 3 5 7 13
      1573       128        494892478080  =   2^7 3^5 5 7 11^2 13 17^2
      1694      1681        124939064250  =   2 3^3 5^3 7 11^2 13 41^2
      1701      1445       1979569912320  =   2^9 3^5 5 7 11^2 13 17^2
      3375        13        499756257000  =   2^3 3^3 5^3 7 11^2 13 41^2
      3610      3267      27819510288570  =   2 3^3 5 7^3 11^2 13 19^2 23^2
      6877       343     111278041154280  =   2^3 3^3 5 7^3 11^2 13 19^2 23^2
Thu Dec  7 15:55:10 PST 2017

===================================================

Multiplier set to $6$

Thu Dec  7 16:47:57 PST 2017
         2         1                   6  =   2 3
         3         1                  24  =   2^3 3
        25        24               29400  =   2^3 3 5^2 7^2
        49         1              117600  =   2^5 3 5^2 7^2
      2738       529      10452831898806  =   2 3^3 11^2 23^2 37^2 47^2
      3267      2209      41811327595224  =   2^3 3^3 11^2 23^2 37^2 47^2
Thu Dec  7 16:50:07 PST 2017

============================================================

Multiplier set to $7,$ and checked for $0<y < x < 11000$

Thu Dec  7 16:56:20 PST 2017
        16         9               25200  =   2^4 3^2 5^2 7
        25         7              100800  =   2^6 3^2 5^2 7
Thu Dec  7 16:58:08 PST 2017

================================================================

Multiplier set to $14$

Thu Dec  7 17:15:36 PST 2017
         8         1                 504  =   2^3 3^2 7
         9         7                2016  =   2^5 3^2 7
      4225      2016     117426776594400  =   2^5 3^2 5^2 7 13^2 47^2 79^2
      6241      2209     469707106377600  =   2^7 3^2 5^2 7 13^2 47^2 79^2

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