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Difficulty: HardPrime Numbers and Prime Factorization

For how many positive integer values of nn is 28+211+2n2^8 + 2^{11} + 2^n equal to the square of an integer?

  1. A
    Zero
  2. OneAnswer
  3. C
    Two
  4. D
    Three
  5. E
    Four

Answer

There is exactly 1 positive integer value of nn (specifically, n=12n = 12).
The option stating 'One' is correct because factoring out 282^8 yields 28(9+2n−8)2^8(9 + 2^{n-8}). For this product to be a perfect square, 9+2n−89 + 2^{n-8} must equal k2k^2 for some integer kk. Rewriting this as (k−3)(k+3)=2n−8(k-3)(k+3) = 2^{n-8} forces both factors to be powers of 2 whose difference is 6. The unique solution to 2b−2a=62^b - 2^a = 6 is a=1a = 1 and b=3b = 3, which gives n−8=4n - 8 = 4, so n=12n = 12. No other positive integers n≤8n \le 8 produce a square.

Step-by-Step Solution

1
Analyze the expression for n>8n > 8 by factoring out 282^8.
28+211+2n=28(1+23+2n−8)=28(9+2n−8)2^8 + 2^{11} + 2^n = 2^8 (1 + 2^3 + 2^{n-8}) = 2^8 (9 + 2^{n-8}).
Since 28=(24)22^8 = (2^4)^2 is already a perfect square, the entire expression is a square if and only if 9+2n−89 + 2^{n-8} is a perfect square.
2
Set 9+2n−8=k29 + 2^{n-8} = k^2 for some integer k>3k > 3 and factor using prime power properties.
k2−9=2n−8  ⟹  (k−3)(k+3)=2n−8k^2 - 9 = 2^{n-8} \implies (k - 3)(k + 3) = 2^{n-8}.
The difference of squares allows us to express the product of two integers as a power of 2.
3
Solve for the prime factors of the terms (k−3)(k-3) and (k+3)(k+3).
Let k−3=2ak - 3 = 2^a and k+3=2bk + 3 = 2^b where a+b=n−8a + b = n - 8 and b>ab > a. Subtracting the two equations gives (k+3)−(k−3)=2b−2a=6  ⟹  2a(2b−a−1)=6=21⋅3(k + 3) - (k - 3) = 2^b - 2^a = 6 \implies 2^a(2^{b-a} - 1) = 6 = 2^1 \cdot 3.
The prime factorization of 6 uniquely dictates that 2a=21  ⟹  a=12^a = 2^1 \implies a = 1, and 2b−a−1=3  ⟹  2b−1=4  ⟹  b=32^{b-a} - 1 = 3 \implies 2^{b-1} = 4 \implies b = 3.
4
Determine nn and test values of n≤8n \le 8.
Since a=1a = 1 and b=3b = 3, n−8=1+3=4  ⟹  n=12n - 8 = 1 + 3 = 4 \implies n = 12. Testing n≤8n \le 8 reveals no other squares (e.g., for n=8n=8, 28(1+8+1)=10⋅282^8(1+8+1)=10 \cdot 2^8, not a square; for n=3n=3, 2312=23⋅1722312 = 2^3 \cdot 17^2, not a square).
This confirms that n=12n = 12 is the unique positive integer solution.

Key Concept

Prime factorization of differences of squares and prime power analysis
Estimated Time:2m 0s
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Topics

Prime Numbers and Prime Factorization

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