2005 AMC 10A Problem 24

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 $\boxed{\textbf{(B)}\ 1}$.

The disguised statement "$P(m) = \sqrt{m}$" is just "$m$ is a prime squared." Two squares with a fixed difference are handled by the difference-of-squares factoring, and the parity of the two factors cuts the factor pairs of $48$ from five to three. This factor-pair method is the standard attack on $x^2 - y^2 = N$.

Bound and check: consecutive odd primes $p < q$ have $q^2 - p^2 \geq (p+2)^2 - p^2 = 4p + 4$, which exceeds $48$ once $p > 11$. So only $p \in \{2, 3, 5, 7, 11\}$ need testing; $p^2 + 48$ gives $52, 57, 73, 97, 169$, and only $169 = 13^2$ is a prime square.

• Missing that the two factors must have the same parity, and then wasting time on pairs like 1 x 48 or 3 x 16, or accepting q = 8 or p = 1 as prime.
• Reading $P(n) = \sqrt{n}$ as "$n$ is a perfect square" and counting all square pairs differing by $48$ ($1$ and $49$, $16$ and $64$, $121$ and $169$), giving $3$.