I notice that the equation $x^2-y^2=17$ is a difference of squares, so I can factor it as $(x+y)(x-y)=17$. Since $x$ and $y$ are integers, this means that both $x+y$ and $x-y$ must be factors of 17. 17 is a prime number, so its only factors are $\pm 1$ and $\pm 17$. This gives me four possible pairs of factors: $(1,17)$, $(-1,-17)$, $(17,1)$, and $(-17,-1)$. For each pair of factors, I can solve for $x$ and $y$ by adding and subtracting the factors and dividing by 2. For example, if $x+y=1$ and $x-y=17$, then $x=\frac{1+17}{2}=9$ and $y=\frac{1-17}{2}=-8$. This gives me one lattice point on the hyperbola: $(9,-8)$. Similarly, if $x+y=-1$ and $x-y=-17$, then $x=\frac{-1-17}{2}=-9$ and $y=\frac{-1+17}{2}=8$. This gives me another lattice point on the hyperbola: $(-9,8)$. If $x+y=17$ and $x-y=1$, then $x=\frac{17+1}{2}=9$ and $y=\frac{17-1}{2}=8$. This gives me another lattice point on the hyperbola: $(9,8)$. Finally, if $x+y=-17$ and $x-y=-1$, then $x=\frac{-17-1}{2}=-9$ and $y=\frac{-17+1}{2}=-8$. This gives me another lattice point on the hyperbola: $(-9,-8)$. I have found four lattice points on the hyperbola: $(9,-8)$, $(-9,8)$, $(9,8)$, and $(-9,-8)$. I don't think there are any more lattice points on the hyperbola, because any other pair of factors of 17 would not be integers, and any other pair of integers would not be factors of 17. Therefore, the answer is 4. 
A lattice point in the $xy$-plane is a point both of whose coordinates are integers (not necessarily positive).  How many lattice points lie on the hyperbola $x^2-y^2=17$?