$x^3-x=y^2-19y+98$
$\leftrightarrow$ $x(x-1)(x+1)=y(y-1)-18y+98$
$3|x(x-1)(x+1)$ and $3|18y$ and $98$$\equiv$$2$ $(mod 3)$
$\rightarrow$ $y(y-1)$ $\equiv$ $1$ $(mod 3)$
$\rightarrow$ $3$$\nmid$$y$ and $3$$\nmid$$y-1$
$\rightarrow$ $y$ $\equiv$ $1$ $(mod 3)$ or $y$ $\equiv$ $2$ $(mod 3)$
If $y$ $\equiv$ $1$ $(mod 3)$ then $(y-1)$ $\equiv$ $0$ $(mod 3)$ _ contradiction
If $y$ $\equiv$ $2$ $(mod 3)$ then $y(y-1)$ $\equiv$ $2$ $(mod 3)$ _ contradiction