1975 AHSME Problems/Problem 22

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

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1975 AHSME Problems/Problem 22