Suppose an integer, a natural number n and a prime number p satisfy the equation `7x^2-44x + 12 = p^n` Find the largest value of p

Answer

To solve the equation \( 7x^2 - 44x + 12 = p^n \) where \( p \) is a prime number and \( n \) is a natural number, we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 7x^2 - 44x + 12 = p^n \]

### Step 2: Factor the quadratic expression
We can factor the quadratic expression on the left-hand side. We rewrite it as:
\[ 7x^2 - 42x - 2x + 12 = 0 \]
Grouping the terms, we have:
\[ (7x^2 - 42x) + (-2x + 12) = 0 \]
Factoring out common terms gives:
\[ 7x(x - 6) - 2(x - 6) = 0 \]
This can be factored as:
\[ (7x - 2)(x - 6) = 0 \]

### Step 3: Set each factor to zero
Setting each factor equal to zero gives us:
1. \( 7x - 2 = 0 \)
2. \( x - 6 = 0 \)

From \( 7x - 2 = 0 \):
\[ 7x = 2 \implies x = \frac{2}{7} \]
From \( x - 6 = 0 \):
\[ x = 6 \]

### Step 4: Evaluate \( p^n \) for each \( x \)
Now we will substitute these values of \( x \) back into the equation \( 7x^2 - 44x + 12 \) to find \( p^n \).

**For \( x = 6 \)**:
\[ 7(6^2) - 44(6) + 12 = 7(36) - 264 + 12 = 252 - 264 + 12 = 0 \]
This does not yield a valid \( p^n \).

**For \( x = \frac{2}{7} \)**:
\[ 7\left(\frac{2}{7}\right)^2 - 44\left(\frac{2}{7}\right) + 12 = 7\left(\frac{4}{49}\right) - \frac{88}{7} + 12 \]
Calculating each term:
\[ = \frac{28}{49} - \frac{88}{7} + 12 = \frac{28}{49} - \frac{616}{49} + \frac{588}{49} = \frac{28 - 616 + 588}{49} = \frac{0}{49} = 0 \]
This also does not yield a valid \( p^n \).

### Step 5: Check for integer values of \( x \)
Since \( x \) must be an integer, we can check integer values around 6 to see if they yield a prime number.

**Try \( x = 7 \)**:
\[ 7(7^2) - 44(7) + 12 = 7(49) - 308 + 12 = 343 - 308 + 12 = 47 \]
Here, \( p^n = 47 \).

### Step 6: Conclusion
The largest prime \( p \) that satisfies the equation is:
\[ p = 47 \]
Thus, the answer is:
\[ \boxed{47} \]