AMC 10 · 2005 · #24

Grade 8 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration
prime-numbersperfect-squaresdifference-of-squares convert-to-algebrasystematic-enumeration ↑ Prerequisites: prime-numbersperfect-squaresdifference-of-squares
📏 Medium solution 💡 3 insights
Problem
For each integer m greater than 1, P(m) means the greatest prime factor of m. Count the positive integers n for which P(n)=√(n) and, at the same time, P(n+48)=√(n+48).

Pick an answer.

(A)
0
(B)
1
(C)
3
(D)
4
(E)
5

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The condition P(n)=√(n) is a disguise. Unpacking it (Tool #4, Introduce a Variable) shows n must be the square of a prime: write n=p² and n+48=q² with p,q prime. Subtracting turns the whole problem into one clean equation q²-p²=48. Factoring the difference of squares gives (q-p)(q+p)=48, so the primes are pinned down by the factor pairs of 48. Tool #2 (Make a Systematic List) enumerates those pairs, and Tool #3 (Eliminate Possibilities) throws out the pairs whose p or q is not prime, leaving the exact count.

1STEP 1

Decode P(n)=√n

A greatest prime factor is a whole number, so n is a perfect square p². And P(p²)=p only when p itself is prime.

P(n)=√(n)⇔ n=p² with p prime
2STEP 2

Apply it to both numbers

The same decoding gives n+48=q² with q prime; subtracting the two equations removes n and leaves q²-p²=48.

n=p², n+48=q² → q²-p²=48
3STEP 3

Factor the difference of squares

A difference of squares factors, so (q-p)(q+p)=48 — the factor pairs of 48 now do the work instead of blind prime hunting.

q²-p²=(q-p)(q+p)=48
4STEP 4

Both factors must be even

Their sum 2q is even, so the factors share parity; an even product rules out both odd, leaving 2 and 24, 4 and 12, 6 and 8.

{q-p, q+p}={2,24},{4,12},{6,8}
5STEP 5

Solve each pair and keep only primes

Halving each sum and difference, only (2,24) yields two primes p=11, q=13; the other pairs are not prime, so n=121 stands alone.

(2,24)→ p=11,q=13 (✓); (4,12)→ 4,8 (×); (6,8)→ 1,7 (×) → 1 value=(B)
Answer
1
Check the survivor directly: n=121=11², whose greatest prime factor is 11=√(121), and n+48=169=13², whose greatest prime factor is 13=√(169). Both conditions hold, so 121 is genuinely valid. The two rejected pairs fail on primality, so exactly one n works — matching choice (B) =1.
💡Key takeaway

“Greatest prime factor equals the square root” secretly means “prime squared,” so set q²-p²=48, factor it as (q-p)(q+p), and only p=11,q=13 gives two primes.

  • Decode P(n)=√n
  • Apply it to both numbers
  • Factor the difference of squares
  • Both factors must be even
  • Solve each pair and keep only primes

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