COMMENT.-A Desmos bug related to this issue.

  We know that if $(x,y,z)$ is an integer solution of $(1)$ then $x$ and $y$ are both odd so $x=-50$ cannot be part of a solution. However we expose here the false solution $(-50,-625,7)$ given by Desmos. 


Putting $z=7$ in $(1)$ we get the equation $$2x+x^2y+4y^2+99=0\quad\quad(4)$$ in which, by Desmos, the intersection of the line $x=-50$ and the curve $(4)$ gives the integer point $(x,y)=(-50,-625)$. But it is a false illusion that $(x,y,z)=(-50,-625,7)$ is an integer solution of the equation $(1)$ because the corresponding numerical value is $-1$ instead of $0$. 

The scale adopted cannot be the cause of this situation because, for example, we have real solutions of $(4)$, given by Desmos and close to integer solutions, $(x,y)= (-52,-676.002)$ and $(- 54,-729.003)$. As a perhaps useful detail, for even $x$ there are several examples in which $(x,y)$ is closer to an integer solution of $(4)$ than for odd $x$. 

I have tried to approximate zero with $y=-625.0001$ and  $y=-624.999$ but I get numerical values greater in absolute value than $1$.

I would really appreciate any comments on this.

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/ozTEF.png