If $p$ and $q$ are primes and $x^2-px+q=0$ has distinct positive integral roots, then which of the following statements are true? 
$I.\ \text{The difference of the roots is odd.} \\ II.\ \text{At least one root is prime.} \\ III.\ p^2-q\ \text{is prime}. \\ IV.\ p+q\ \text{is prime}$
$\\ \textbf{(A)}\ I\ \text{only} \qquad \textbf{(B)}\ II\ \text{only} \qquad \textbf{(C)}\ II\ \text{and}\ III\ \text{only} \\ \textbf{(D)}\ I, II, \text{and}\ IV\ \text{only}\ \qquad \textbf{(E)}\ \text{All are true.}$
Since the roots are both positive integers, we can say that $x^2-px+q=(x-1)(x-q)$ since $q$ only has $2$ divisors. Thus, the roots are $1$ and $q$ and $p=q+1$. The only two primes which differ by $1$ are $2,3$ so $p=3$ and $q=2$. 
$I$ is true because $3-2=1$.
$II$ is true because one of the roots is $2$ which is prime.
$III$ is true because $3^2-2=7$ is prime.
$IV$ is true because $2+3=5$ is prime. 
Thus, the answer is $\textbf{(E)}$.
-brainiacmaniac31