smc

The parabola is symmetric through $y=-\frac{4}{3}x$, and the common distance is $5$, so the directrix is the line through $(1,7)$ and $(-7,1)$, which is the line $$3x-4y=-25.$$ Using the point-line distance formula, the parabola is the locus $$x^2+y^2=\frac{{\left|3x-4y+25\right|}^2}{3^2+4^2}$$ which rearranges to $(4x+3y{)}^2=25(6x-8y+25)$. Let $m=4x+3y\in \mathbb{Z}$, $\left|m\right|\le 1000$. Put $m=25k$ to obtain $$25k^2=6x-8y+25$$$$25k=4x+3y.$$ and accordingly we find by solving the system that $x=\frac{1}{2}(3k^2-3)+4k$ and $y=-2k^2+3k+2$. One can show that the values of $k$ that make $(x,y)$ an integer pair are precisely odd integers $k$. For $\left|25k\right|\le 1000$ this is $k=-39,-37,-35,\dots ,39$, so $40$ values work and the answer is $\boxed{40}$.