2005 AMC 10A Problem 24

First decode the strange condition $P(m) = \sqrt{m}$.

1. What does the condition mean? $P(m)$ is a prime, so $\sqrt{m}$ must be a prime $p$, and then $m = p^2$. And for $m = p^2$ the only prime factor is $p$, so the condition holds. For example, $P(49) = 7 = \sqrt{49}$, but $P(36) = 3$ while $\sqrt{36} = 6$. So the condition just says: $m$ is the square of a prime.
2. Restate the problem. We need primes $p$ and $q$ with $n = p^2$ and $n + 48 = q^2$. So
$$q^2 - p^2 = 48.$$
3. Factor. $q^2 - p^2 = (q - p)(q + p)$, so $48$ must be split as (small factor) $\times$ (big factor) with small $= q - p$ and big $= q + p$.
4. Both factors are even. The two factors differ by $2p$, which is even, so they are both even or both odd. Their product $48$ is even, so both are even. The even pairs for $48$ are $2 \times 24$, $4 \times 12$, $6 \times 8$.
5. Solve each pair. $q$ is half the sum and $p$ is half the difference:

1. Answer. Only $n = 11^2 = 121$ works, and $121 + 48 = 169 = 13^2$.

The answer is $\boxed{\textbf{(B)}\ 1}$.

The fancy notation hides a simple idea: "largest prime factor equals the square root" means "square of a prime." Two squares with a fixed difference are handled by factoring the difference of squares, and the both-even rule cuts the list of factor pairs down to three.

Check small primes. If $p < q$ are primes, the gap $q^2 - p^2$ grows as the primes grow. For odd $p$, the next possible prime is at least $p + 2$, and $(p + 2)^2 - p^2 = 4p + 4$. That is more than $48$ once $p$ is bigger than $11$. So test only $p = 2, 3, 5, 7, 11$: adding $48$ to $4, 9, 25, 49, 121$ gives $52, 57, 73, 97, 169$. Only $169 = 13^2$ is a prime squared.

Decode the condition. If $P(m) = \sqrt{m}$, then $\sqrt{m}$ is an integer and is prime, say $\sqrt{m} = p$, so $m = p^2$. Conversely $P(p^2) = p = \sqrt{p^2}$. Thus the condition says exactly that $m$ is the square of a prime.

So we need primes $p, q$ with $n = p^2$ and $n + 48 = q^2$, i.e.
$$q^2 - p^2 = (q - p)(q + p) = 48.$$

The factors $q - p$ and $q + p$ differ by $2p$, an even number, so they have the same parity; since their product is even, both are even. The factor pairs of $48$ with both entries even are $(2, 24)$, $(4, 12)$, $(6, 8)$. Solving $q = \frac{\text{sum}}{2}$, $p = \frac{\text{difference}}{2}$:

• $(2, 24)$: $q = 13$, $p = 11$, both prime.
• $(4, 12)$: $q = 8$, not prime.
• $(6, 8)$: $q = 7$, $p = 1$, not prime.

Only $n = 11^2 = 121$ works (and indeed $121 + 48 = 169 = 13^2$).

The answer is $\boxed{\textbf{(B)}\ 1}$.