- #1
Someone2841
- 43
- 6
[All numbers as assumed to be integers]
Bocard's problem asks to find the solutions to the Diophantine equation n!=m^2-1. The only known (n,m) pairs are (4,5),(5,11),(7,71), and it is conjectured that there are no more. A generalization of this problem would be n!=m^2-A. I've been playing around with this formula, and while I do not have a general proof right now, it seems that one could prove there to be for any non-square A a finite number of solutions (usually 0).
As an example, consider the solutions to n!=m^2-5. To show it has a finite number of solutions, consider the equation in terms of mod 3. Assuming n\geq3, then n!\bmod3=0, and if it can be shown that m^2 \equiv 2 \pmod{3} has no solutions, then n!=m^2-5 cannot have a solution with n\geq3. It is simple enough to show that the solution set to m^2 \equiv 2 \pmod{3} is null. As 3x, 3x+1, and 3x+2 include all integers, showing that (3x)^2=9x\equiv 0 \pmod{3}, (3x+1)^2=9x^2+6x+1 \equiv 1 \pmod{3}, (3x+2)^2=9x+12x+4\equiv 1 \pmod{3} demonstrates that any n^2\bmod3 will be 0 or 1, but never 2. Therefore, n can only equal 1 or 2; since neither provides a solution, the equation at A=5 has no solutions.
Making a quick list of a^2\bmod b cycles (I say cycle since the first a terms will repeat ad infinitum over b) and determining which numbers mod b do not exist therein gives a method for finding linear equations for which numbers will have finite solutions. For example, a^2\bmod 3 will never include 2. Therefore, for all integers k, A=3k+2 will have a finite solution set. It's easy enough to write a program to loop through the first several b's and it seems that only numbers in the form n!=m^2-s^2 cannot be proven finite by this method; this is simply an observed pattern without a general proof.
Does anyone have any thoughts on this (definitely let me know if I've come to incorrect conclusions)? Is there a good way to go about proving that for all non-square A's there is a finite number of solutions?
---
Edit: My program confirms that non-square numbers from A= 2 to 224 have finite solutions with an upper bound B for n at B\leq13. This is what leads me to wonder if this extends to infinity.
Bocard's problem asks to find the solutions to the Diophantine equation n!=m^2-1. The only known (n,m) pairs are (4,5),(5,11),(7,71), and it is conjectured that there are no more. A generalization of this problem would be n!=m^2-A. I've been playing around with this formula, and while I do not have a general proof right now, it seems that one could prove there to be for any non-square A a finite number of solutions (usually 0).
As an example, consider the solutions to n!=m^2-5. To show it has a finite number of solutions, consider the equation in terms of mod 3. Assuming n\geq3, then n!\bmod3=0, and if it can be shown that m^2 \equiv 2 \pmod{3} has no solutions, then n!=m^2-5 cannot have a solution with n\geq3. It is simple enough to show that the solution set to m^2 \equiv 2 \pmod{3} is null. As 3x, 3x+1, and 3x+2 include all integers, showing that (3x)^2=9x\equiv 0 \pmod{3}, (3x+1)^2=9x^2+6x+1 \equiv 1 \pmod{3}, (3x+2)^2=9x+12x+4\equiv 1 \pmod{3} demonstrates that any n^2\bmod3 will be 0 or 1, but never 2. Therefore, n can only equal 1 or 2; since neither provides a solution, the equation at A=5 has no solutions.
Making a quick list of a^2\bmod b cycles (I say cycle since the first a terms will repeat ad infinitum over b) and determining which numbers mod b do not exist therein gives a method for finding linear equations for which numbers will have finite solutions. For example, a^2\bmod 3 will never include 2. Therefore, for all integers k, A=3k+2 will have a finite solution set. It's easy enough to write a program to loop through the first several b's and it seems that only numbers in the form n!=m^2-s^2 cannot be proven finite by this method; this is simply an observed pattern without a general proof.
Does anyone have any thoughts on this (definitely let me know if I've come to incorrect conclusions)? Is there a good way to go about proving that for all non-square A's there is a finite number of solutions?
---
Edit: My program confirms that non-square numbers from A= 2 to 224 have finite solutions with an upper bound B for n at B\leq13. This is what leads me to wonder if this extends to infinity.
