If the product of four consecutive natural numbers increased by a natural number p, is a perfect square, then the value of p is

Options

A. 8
B. 4
C. 2
D. 1

Answer

To solve the problem, we need to find the value of \( p \) such that the product of four consecutive natural numbers increased by \( p \) is a perfect square. Let's denote the four consecutive natural numbers as \( n, n+1, n+2, n+3 \).

### Step 1: Write the product of four consecutive natural numbers
The product of four consecutive natural numbers can be expressed as:
\[ P = n(n+1)(n+2)(n+3) \]

### Step 2: Analyze the expression
We need to find \( p \) such that:
\[ P + p = k^2 \]
for some integer \( k \). This means that \( P + p \) must be a perfect square.

### Step 3: Calculate the product for specific values of \( n \)
Let's calculate the product for some values of \( n \):

- For \( n = 1 \):
\[ P = 1 \cdot 2 \cdot 3 \cdot 4 = 24 \]
To make \( 24 + p \) a perfect square, we can check:
\[ 24 + 1 = 25 \quad (5^2) \quad \Rightarrow p = 1 \]

- For \( n = 2 \):
\[ P = 2 \cdot 3 \cdot 4 \cdot 5 = 120 \]
To make \( 120 + p \) a perfect square, we can check:
\[ 120 + 1 = 121 \quad (11^2) \quad \Rightarrow p = 1 \]

- For \( n = 3 \):
\[ P = 3 \cdot 4 \cdot 5 \cdot 6 = 360 \]
To make \( 360 + p \) a perfect square, we can check:
\[ 360 + 1 = 361 \quad (19^2) \quad \Rightarrow p = 1 \]

### Step 4: Generalize the result
From the calculations above, we observe that for any set of four consecutive natural numbers, the product \( P \) is always 1 less than a perfect square. Therefore, to make \( P + p \) a perfect square, we can conclude that:
\[ p = 1 \]

### Final Answer
Thus, the value of \( p \) is:
\[ \boxed{1} \]