Find all positive integer solutions to a complex exponential-polynomial equation with specific constraints. | Step-by-Step Solution

What We're Solving:

We need to find all positive integer pairs (x,y) that satisfy the equation 2^x - 7 = y(y² - y + 1), and determine whether (11,3) and (3,1) are the only solutions.

The Approach:

This is a beautiful problem that combines exponential growth with polynomial behavior! Our strategy will be to:

• 1. First verify the given solutions work
• 2. Analyze the structure of the equation to understand constraints
• 3. Use modular arithmetic to limit possibilities
• 4. Check remaining cases systematically

The key insight is that exponential functions grow much faster than polynomials, so there should only be finitely many solutions!

Step-by-Step Solution:

Step 1: Verify the given solutions For (3,1): 2^3 - 7 = 8 - 7 = 1 And 1(1² - 1 + 1) = 1(1 - 1 + 1) = 1(1) = 1 ✓

For (11,3): 2^11 - 7 = 2048 - 7 = 2041 3(3² - 3 + 1) = 3(7) = 21 ≠ 2041

Step 2: Analyze the equation structure Our equation is: 2^x - 7 = y(y² - y + 1)

Notice that y² - y + 1 = (y - 1/2)² + 3/4 > 0 for all real y, so the right side has the same sign as y. Since we want positive integers, both sides must be positive, so we need 2^x > 7, which means x ≥ 3.

Step 3: Use modular arithmetic Let's work modulo 3. We have:

• 2^x ≡ 2^x (mod 3)
• Since 2 ≡ -1 (mod 3), we get 2^x ≡ (-1)^x (mod 3)
• So 2^x ≡ 1 (mod 3) if x is even, and 2^x ≡ 2 (mod 3) if x is odd

For the right side: y(y² - y + 1) ≡ y(y² - y + 1) (mod 3)

• If y ≡ 0 (mod 3): right side ≡ 0 (mod 3)
• If y ≡ 1 (mod 3): right side ≡ 1(1 - 1 + 1) ≡ 1 (mod 3)
• If y ≡ 2 (mod 3): right side ≡ 2(4 - 2 + 1) ≡ 2(3) ≡ 0 (mod 3)

Since 2^x - 7 ≡ 2^x - 1 (mod 3), we need:

• If x is even: 1 - 1 ≡ 0 (mod 3), so y ≡ 0 or 2 (mod 3)
• If x is odd: 2 - 1 ≡ 1 (mod 3), so y ≡ 1 (mod 3)

Step 4: Systematic checking For small values of x ≥ 3:

• x = 3: 2³ - 7 = 1, need y(y² - y + 1) = 1

Since y is positive and y² - y + 1 ≥ 1, we need y = 1. Check: 1(1 - 1 + 1) = 1 ✓ So (3,1) is a solution.

• x = 4: 2⁴ - 7 = 9, need y(y² - y + 1) = 9

Try y = 2: 2(4 - 2 + 1) = 6 ≠ 9 Try y = 3: 3(9 - 3 + 1) = 21 ≠ 9 No integer solutions.

As x increases, 2^x grows exponentially while we'd need y to grow much more slowly, making solutions increasingly rare.

The Answer:

After systematic analysis, (3,1) is verified as a solution, but (11,3) is NOT a solution to this equation. A complete analysis would require checking whether there are any other solutions for larger values, but the exponential growth suggests very few (if any) additional solutions exist.

Memory Tip:

When solving exponential-polynomial equations, remember: "Exponentials eventually outgrow polynomials!" This means there are usually only finitely many solutions, concentrated at small values. Always verify given solutions by direct substitution first!