2002 AMC 10B Problem 6

Factor the quadratic: $n^2 - 3n + 2 = (n-1)(n-2)$.

The factors $n - 2$ and $n - 1$ are consecutive integers, so one of them is even and the product is always even. The only even prime is $2$, so we need
$$(n-1)(n-2) = 2.$$
Two consecutive integers with product $2$ are $1$ and $2$ (giving $n = 3$) or $-2$ and $-1$ (giving $n = 0$, which is not positive). Only $n = 3$ survives, and indeed $9 - 9 + 2 = 2$ is prime.

The answer is $\boxed{\textbf{(B)}\ \text{one}}$.

A quadratic that factors over the integers is a product of two integers, and a product is prime only when one factor is $\pm 1$. Here parity does even better: the product of consecutive integers is always even, so the only candidate prime is $2$, and the search collapses to a single equation. Factor first, then ask what the factors must be.

Without parity: for $n \ge 4$ both factors $n-1 \ge 3$ and $n - 2 \ge 2$ exceed $1$, so the product is composite. Check $n = 1, 2, 3$ directly: values $0, 0, 2$. Exactly one prime.

• Checking n = 1, 2, 3, 4 (values 0, 0, 2, 6) and guessing that more primes show up for larger n, or counting n = 0 as a positive integer.
• Treating $0$ or $1$ as prime, which would add $n = 1$ and $n = 2$ to the count.