For how many values of $k$ are both roots of $x^2-6x+k=0$ positive integers?

The roots must be positive integers adding to 6, so the possibilities are 1 and 5, 2 and 4, and 3 and 3. Their products give $k=5,k=8$ and $k=9$ — three values. (2 wrongly throws out the repeated pair 3, 3, which is perfectly allowed; 4 also counts 0 and 6, but 0 is not positive; 5 counts the ordered pairs instead of the values of $k$.) For instance $k=8$ gives $x^2-6x+8=(x-2)(x-4)$.