How to Find Pairs of Positive Integers that Satisfy the Equation (a^b b^a a! b^2 ba 1)

The only solution in positive integers of the equation ab ba a! b2 ba 1 is a 3 and b 2.

Assumptions and Initial Steps

First, let's consider the scenario where a 1. Substituting a 1 into the equation gives us:

Case 1: a 1

$$b^{1} (1!) b^{2} b^{1}$$

This simplifies to b b2, which is not possible for any positive integer b. Therefore, a ≠ 1.

Even and Odd Analysis

Next, let's consider the cases where a and b are even or odd, respectively.

Case 2: a is Even

If a is even, then ba ≡ b2 (mod 2), which contradicts the original equation. Therefore, a must be odd.

Case 3: b is Also Odd

If both a and b are odd, the left-hand side (LHS) of the equation ab ba is even, while the right-hand side (RHS) a! b2 ba 1 is odd. This contradiction implies that b must be even.

Therefore, we can deduce that b is at least 2 and even.

Specific Case Analysis

Case 4: b 2

When b 2, the equation simplifies to:

$$a^2 2^a a! 4a 1$$

For a 3, the left-hand side is:

$$3^2 2^3 9 8 17$$

The right-hand side is:

$$(3!) 4 1 6 4 1 17$$

Thus, a 3 and b 2 is indeed a solution.

Further Analysis and Proof

Case 5: b 2 and a ge; 5

When b 2 and a ge; 5, the equation becomes:

$$(a^2)^a a! 4a 1$$

For a ge; 5, we can simplify the equation further. Taking the logarithm of both sides and using Stirling's approximation, we get:

$$a^{a} 2^a a! 4a 1$$

Furthermore, this implies:

$$22k - 5a leq 22k - 3$$

for b 8k. This further restricts the possible values of a and b.

Conclusion

Given the constraints and the analysis, the only solution in positive integers for the equation ab ba a! b2 ba 1 is a 3 and b 2.

Verification Using PariGP

Using the PariGP script provided, we can verify that the only solution up to 100 is (a 3) and (b 2).

{ k100 for a1k for b1k lhsa^bb^a rhsa!b^2ab1 if lhsrhs printa }

The script prints 32, confirming that (a, b) (3, 2) is the only solution.