Does there exist an integer such that its cube is equal to `3n^(2) + 3n + 7, n in I` ?

Answer

To determine whether there exists an integer \( m \) such that its cube is equal to \( 3n^2 + 3n + 7 \) for some integer \( n \), we can follow these steps:

### Step 1: Set up the equation
We start with the equation:
\[ m^3 = 3n^2 + 3n + 7 \]

### Step 2: Analyze \( m^3 \) modulo 3
We know that any integer \( m \) can be expressed in one of three forms when taken modulo 3:
- \( m \equiv 0 \mod 3 \)
- \( m \equiv 1 \mod 3 \)
- \( m \equiv 2 \mod 3 \)

Calculating \( m^3 \) for these cases:
- If \( m \equiv 0 \mod 3 \), then \( m^3 \equiv 0 \mod 3 \)
- If \( m \equiv 1 \mod 3 \), then \( m^3 \equiv 1 \mod 3 \)
- If \( m \equiv 2 \mod 3 \), then \( m^3 \equiv 2 \mod 3 \)

### Step 3: Analyze \( 3n^2 + 3n + 7 \) modulo 3
Now we simplify \( 3n^2 + 3n + 7 \) modulo 3:
\[ 3n^2 \equiv 0 \mod 3 \quad \text{and} \quad 3n \equiv 0 \mod 3 \]
Thus,
\[ 3n^2 + 3n + 7 \equiv 0 + 0 + 7 \equiv 1 \mod 3 \]

### Step 4: Set the congruences
From the previous steps, we have:
\[ m^3 \equiv 1 \mod 3 \]
This implies that \( m \equiv 1 \mod 3 \).

### Step 5: Substitute \( m \) in the equation
Let \( m = 3k + 1 \) for some integer \( k \). Then we compute \( m^3 \):
\[ m^3 = (3k + 1)^3 = 27k^3 + 27k^2 + 9k + 1 \]

### Step 6: Set the equation equal to \( 3n^2 + 3n + 7 \)
Substituting back into our original equation:
\[ 3n^2 + 3n + 7 = 27k^3 + 27k^2 + 9k + 1 \]
Rearranging gives:
\[ 3n^2 + 3n + 6 = 27k^3 + 27k^2 + 9k \]

### Step 7: Divide through by 3
Dividing the entire equation by 3:
\[ n^2 + n + 2 = 9k^3 + 9k^2 + 3k \]

### Step 8: Analyze \( n^2 + n + 2 \) modulo 3
Next, we analyze \( n^2 + n + 2 \) modulo 3. The possible values of \( n \mod 3 \) are:
- If \( n \equiv 0 \mod 3 \): \( n^2 + n + 2 \equiv 0 + 0 + 2 \equiv 2 \mod 3 \)
- If \( n \equiv 1 \mod 3 \): \( n^2 + n + 2 \equiv 1 + 1 + 2 \equiv 4 \equiv 1 \mod 3 \)
- If \( n \equiv 2 \mod 3 \): \( n^2 + n + 2 \equiv 4 + 2 + 2 \equiv 8 \equiv 2 \mod 3 \)

Thus, \( n^2 + n + 2 \) can be \( 1 \) or \( 2 \mod 3 \).

### Step 9: Conclusion
Since \( 9k^3 + 9k^2 + 3k \equiv 0 \mod 3 \), and \( n^2 + n + 2 \) can only be \( 1 \) or \( 2 \mod 3 \), there is no integer \( n \) such that \( n^2 + n + 2 \equiv 0 \mod 3 \).

Thus, we conclude that there does not exist an integer \( m \) such that \( m^3 = 3n^2 + 3n + 7 \).

### Final Answer
No, there does not exist an integer such that its cube is equal to \( 3n^2 + 3n + 7 \).