`P = {1,2,3,4…..20}`. How many integers n can be selected form the set P such that `(n^(2) + n^(3))` is a perfect square ?

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to find how many integers \( n \) can be selected from the set \( P = \{1, 2, 3, \ldots, 20\} \) such that \( n^2 + n^3 \) is a perfect square.

### Step-by-Step Solution:

1. **Rewrite the Expression**:
We start with the expression \( n^2 + n^3 \). We can factor out \( n^2 \):
\[ n^2 + n^3 = n^2(1 + n) \]

2. **Identify Conditions for Perfect Squares**:
For \( n^2(1 + n) \) to be a perfect square, both \( n^2 \) and \( 1 + n \) must be perfect squares. Since \( n^2 \) is always a perfect square for any integer \( n \), we only need to ensure that \( 1 + n \) is a perfect square.

3. **Set Up the Equation**:
Let \( 1 + n = k^2 \) for some integer \( k \). This implies:
\[ n = k^2 - 1 \]

4. **Determine Valid Values of \( n \)**:
We need to find values of \( k \) such that \( n = k^2 - 1 \) falls within the set \( P \) (i.e., \( 1 \leq n \leq 20 \)):
\[ 1 \leq k^2 - 1 \leq 20 \]
This simplifies to:
\[ 2 \leq k^2 \leq 21 \]
Taking square roots gives:
\[ \sqrt{2} \leq k \leq \sqrt{21} \]
Approximating the square roots, we find:
\[ 1.41 \leq k \leq 4.58 \]
Thus, \( k \) can take integer values \( 2, 3, 4 \).

5. **Calculate Corresponding \( n \) Values**:
- For \( k = 2 \):
\[ n = 2^2 - 1 = 4 - 1 = 3 \]
- For \( k = 3 \):
\[ n = 3^2 - 1 = 9 - 1 = 8 \]
- For \( k = 4 \):
\[ n = 4^2 - 1 = 16 - 1 = 15 \]

6. **List Valid \( n \) Values**:
The valid integers \( n \) from the set \( P \) are \( 3, 8, \) and \( 15 \).

7. **Count the Valid Integers**:
There are a total of **3 integers** \( n \) that satisfy the condition.

### Final Answer:
Thus, the number of integers \( n \) that can be selected from the set \( P \) such that \( n^2 + n^3 \) is a perfect square is **3**.

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