Solving the Equation (x! 1 y^2) Over the Integers

The equation (x! 1 y^2) is a well-known problem in number theory. The goal is to find all integer solutions for (x) and (y) that satisfy this equation. Let's explore how to solve this problem step-by-step.

Step 1: Analyzing Small Values of (x)

To understand the behavior of the equation, let's start by examining small integer values of (x). We will check the values of (x) from 0 to 7:

For (x 0):
(0! 1 1 1 2) (not a perfect square)
For (x 1):
(1! 1 1 1 2) (not a perfect square)
For (x 2):
(2! 1 2 1 3) (not a perfect square)
For (x 3):
(3! 1 6 1 7) (not a perfect square)
For (x 4):
(4! 1 24 1 25 5^2) (perfect square)
For (x 5):
(5! 1 120 1 121 11^2) (perfect square)
For (x 6):
(6! 1 720 1 721) (not a perfect square, as (26^2 676) and (27^2 729))
For (x 7):
(7! 1 5040 1 5041) (not a perfect square, as (71^2 5041))
For (x 8):
(8! 1 40320 1 40321) (not a perfect square, as (200^2 40000) and (201^2 40401))

Step 2: General Analysis for Larger (x)

For (x geq 5), the factorial (x!) grows very quickly. Since (x!) for (x geq 5) is even, (x! 1) is odd. This means (y) must also be odd to ensure (y^2) is odd.

Step 3: Factorial Growth and Perfect Squares

As (x) increases, (x! 1) becomes increasingly unlikely to be a perfect square because the gaps between consecutive perfect squares increase.

Known Solutions

The known solutions for the equation (x! 1 y^2) are:

(x 4, y 5): (4! 1 5^2) (x 5, y 11): (5! 1 11^2) (x 7, y 71): (7! 1 71^2)

These are the only integer solutions known to date for the equation (x! 1 y^2). However, it is not known if there are any other solutions beyond these three.

Marius Overholt has shown that the equation has only finitely many solutions if the ABC conjecture is true, but at present, we don't even know if there are any other solutions beyond the known ones.

Additional Observations

For (x geq 6), there are no additional solutions. Large factorials have many zeros at the end. Hence, (x^2 - 1) has …001 as the last digit. Therefore, (y) must have 9 or 1 as the last digit. So, (y 10k 9) or (y 10k 1).

If (y 10k 1), then (x! 10k(10k - 1)). 10k(10k - 1) can have many zeros at the end if (k) has many zeros at the end, making it plausible for (x!) to match the required form.

If (y 10k 9), then (x! 10k(10k 10)). 10k(10k 10) can have one or two zeros at the end, but it is not sufficient to match large factorials after 20!.

This problem remains a fascinating challenge in number theory, and long arithmetics play a crucial role in exploring and validating potential solutions.