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.

The equation $(1)$ has not integer solution. An elementary proof.

Suppose $(x,y,z)=(a,b,c)$ is a solution and consider as unknowns $(X,Y)=(4,2)$ so we can form the system $$\begin{cases}b^2X+(a+c^2)Y=-(1+a^2b)\\-X+7Y=10\end{cases}$$ whose solution is $$X=\dfrac{-7-7a^2b-10a-10c^2}{7b^2-a-c^2}$$ $$Y=\dfrac{10b^2+1+a^2b}{7b^2-a-c^2}$$ Because of $X=2Y$, one has after simplification $$9(1+a^2b)+10(a+c^2+2b^2)=0$$ and multiplying equation $1+2a+a^2b+4b^2+2c^2=0$ by $5$ we have $$5(1+a^2b)+10(a+c^2+2b^2)=0$$ Subtraction now gives $$4(1+a^2b)=0$$ We are done.

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.

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.