`P=b^(2)c^(2)-ac-bd` where a, b, c and d, in that order, are four ocnsecutive natural numbers `(altb)`.
Which of the following statements is correct?

Options

A. `sqrt(P)` is always a prime number.
B. `sqrt(P)` is always a rational number, through not necessarily prime.
C. `sqrtP` can be rational or irretional depending on the numbers.
D. `sqrtP` is always an irrational number.

Answer

To solve the problem step by step, we will evaluate the expression \( P = b^2 c^2 - ac - bd \) where \( a, b, c, d \) are four consecutive natural numbers.

### Step 1: Define the consecutive natural numbers
Let’s define the four consecutive natural numbers:
- Let \( a = n \)
- Let \( b = n + 1 \)
- Let \( c = n + 2 \)
- Let \( d = n + 3 \)

### Step 2: Substitute the values into the expression
Now we substitute \( a, b, c, d \) into the expression for \( P \):
\[ P = (n + 1)^2(n + 2)^2 - n(n + 2) - (n + 1)(n + 3) \]

### Step 3: Expand the expression
Now we will expand each term:
1. Expand \( (n + 1)^2 \):
\[ (n + 1)^2 = n^2 + 2n + 1 \]
2. Expand \( (n + 2)^2 \):
\[ (n + 2)^2 = n^2 + 4n + 4 \]
3. Now multiply these two expansions:
\[ (n^2 + 2n + 1)(n^2 + 4n + 4) = n^4 + 6n^3 + 13n^2 + 8n + 4 \]

4. For \( -n(n + 2) \):
\[ -n(n + 2) = -n^2 - 2n \]

5. For \( -(n + 1)(n + 3) \):
\[ -(n + 1)(n + 3) = -n^2 - 4n - 3 \]

### Step 4: Combine all terms
Now we combine all the terms:
\[ P = (n^4 + 6n^3 + 13n^2 + 8n + 4) - (n^2 + 2n) - (n^2 + 4n + 3) \]
\[ P = n^4 + 6n^3 + 13n^2 + 8n + 4 - n^2 - 2n - n^2 - 4n - 3 \]
\[ P = n^4 + 6n^3 + (13n^2 - 2n - 4n - 2n) + (8n - 2n - 4n) + (4 - 3) \]
\[ P = n^4 + 6n^3 + 11n^2 + 2n + 1 \]

### Step 5: Evaluate \( P \) for specific values of \( n \)
Let’s evaluate \( P \) for the first few values of \( n \):

1. For \( n = 1 \):
\[ P = 1^4 + 6(1^3) + 11(1^2) + 2(1) + 1 = 1 + 6 + 11 + 2 + 1 = 21 \]
\(\sqrt{P} = \sqrt{21}\) (not prime)

2. For \( n = 2 \):
\[ P = 2^4 + 6(2^3) + 11(2^2) + 2(2) + 1 = 16 + 48 + 44 + 4 + 1 = 113 \]
\(\sqrt{P} = \sqrt{113}\) (not prime)

3. For \( n = 3 \):
\[ P = 3^4 + 6(3^3) + 11(3^2) + 2(3) + 1 = 81 + 162 + 99 + 6 + 1 = 349 \]
\(\sqrt{P} = \sqrt{349}\) (not prime)

### Conclusion
From the evaluations, we can see that \( \sqrt{P} \) does not consistently yield a prime number.