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 \( n \in \mathbb{Z} \), we can follow these steps:

### Step 1: Set up the equation
We start by setting the equation:
\[ m^3 = 3n^2 + 3n + 7 \]
where \( m \) is an integer.

### Step 2: Analyze modulo 3
Next, we analyze the expression \( 3n^2 + 3n + 7 \) modulo 3. Notice that:
\[ 3n^2 \equiv 0 \mod 3, \quad 3n \equiv 0 \mod 3, \quad \text{and} \quad 7 \equiv 1 \mod 3 \]
Thus, we have:
\[ 3n^2 + 3n + 7 \equiv 0 + 0 + 1 \equiv 1 \mod 3 \]
This implies:
\[ m^3 \equiv 1 \mod 3 \]

### Step 3: Determine possible values for \( m \)
Next, we find the values of \( m \) that satisfy \( m^3 \equiv 1 \mod 3 \). The possible residues of \( m \) modulo 3 are 0, 1, and 2:
- 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 \( m^3 \equiv 2 \mod 3 \).

Thus, \( m \) must be congruent to 1 modulo 3:
\[ m \equiv 1 \mod 3 \]

### Step 4: Substitute \( m \)
We can express \( m \) in terms of another integer \( k \):
\[ m = 3k + 1 \]
for some integer \( k \).

### Step 5: Expand \( m^3 \)
Now, we substitute \( m \) back into the equation:
\[ m^3 = (3k + 1)^3 = 27k^3 + 27k^2 + 9k + 1 \]

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

### Step 7: Rearrange the equation
Rearranging gives:
\[ 3n^2 + 3n + 6 = 27k^3 + 27k^2 + 9k \]

### Step 8: Divide by 3
Dividing the entire equation by 3 results in:
\[ n^2 + n + 2 = 9k^3 + 9k^2 + 3k \]

### Step 9: Analyze the left-hand side
Now, we analyze the left-hand side \( n^2 + n + 2 \) modulo 3. We can check the values of \( n \) modulo 3:
- If \( n \equiv 0 \mod 3 \), then \( n^2 + n + 2 \equiv 2 \mod 3 \).
- If \( n \equiv 1 \mod 3 \), then \( n^2 + n + 2 \equiv 4 \equiv 1 \mod 3 \).
- If \( n \equiv 2 \mod 3 \), then \( n^2 + n + 2 \equiv 8 \equiv 2 \mod 3 \).

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

### Step 10: Analyze the right-hand side
Now, we analyze \( 9k^3 + 9k^2 + 3k \) modulo 3:
\[ 9k^3 + 9k^2 + 3k \equiv 0 \mod 3 \]
This means that the right-hand side is always divisible by 3.

### Step 11: Conclusion
Since the left-hand side \( n^2 + n + 2 \) can never be congruent to 0 modulo 3, there cannot be an integer \( n \) such that \( 3n^2 + 3n + 7 = m^3 \). Therefore, we conclude that there is no integer \( m \) such that:
\[ m^3 = 3n^2 + 3n + 7 \]

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

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