Variant of Bocard's Problem

  • Thread starter Someone2841
  • Start date
In summary: B for n at B\leq13. This is what leads me to wonder if this extends to infinity.So by your argument, given A, you need to find a modulus N such that A is not a square modulo N. If you can find such an N then you can repeat your argument replacing 3 with N. It might help you to read up on quadratic residues. http://en.wikipedia.org/wiki/Quadratic_residueAlso, you might check out http://en.wikipedia.org/wiki/Law_of_quadratic_reciprocityThat provides the fastest way to check whether or not A
  • #1
Someone2841
44
6
[All numbers as assumed to be integers]

Bocard's problem asks to find the solutions to the Diophantine equation [itex]n!=m^2-1[/itex]. The only known [itex](n,m)[/itex] pairs are [itex](4,5),(5,11),(7,71)[/itex], and it is conjectured that there are no more. A generalization of this problem would be [itex]n!=m^2-A[/itex]. 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 [itex]A[/itex] a finite number of solutions (usually 0).

As an example, consider the solutions to [itex]n!=m^2-5[/itex]. To show it has a finite number of solutions, consider the equation in terms of mod 3. Assuming [itex]n\geq3[/itex], then [itex]n!\bmod3=0[/itex], and if it can be shown that [itex]m^2 \equiv 2 \pmod{3}[/itex] has no solutions, then [itex]n!=m^2-5[/itex] cannot have a solution with [itex]n\geq3[/itex]. It is simple enough to show that the solution set to [itex]m^2 \equiv 2 \pmod{3}[/itex] is null. As [itex]3x[/itex], [itex]3x+1[/itex], and [itex]3x+2[/itex] include all integers, showing that [itex](3x)^2=9x\equiv 0 \pmod{3}[/itex], [itex](3x+1)^2=9x^2+6x+1 \equiv 1 \pmod{3}[/itex], [itex](3x+2)^2=9x+12x+4\equiv 1 \pmod{3}[/itex] demonstrates that any [itex]n^2\bmod3[/itex] will be 0 or 1, but never 2. Therefore, [itex]n[/itex] can only equal 1 or 2; since neither provides a solution, the equation at [itex]A=5[/itex] has no solutions.

Making a quick list of [itex]a^2\bmod b[/itex] cycles (I say cycle since the first [itex]a[/itex] terms will repeat ad infinitum over [itex]b[/itex]) 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, [itex]a^2\bmod 3[/itex] will never include 2. Therefore, for all integers [itex]k[/itex], [itex]A=3k+2[/itex] will have a finite solution set. It's easy enough to write a program to loop through the first several [itex]b[/itex]'s and it seems that only numbers in the form [itex]n!=m^2-s^2[/itex] 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 [itex]A[/itex]'s there is a finite number of solutions?

---
Edit: My program confirms that non-square numbers from [itex]A=[/itex] 2 to 224 have finite solutions with an upper bound [itex]B[/itex] for [itex]n[/itex] at [itex]B\leq13[/itex]. This is what leads me to wonder if this extends to infinity.
 
Physics news on Phys.org
  • #3
Someone2841 said:
[All numbers as assumed to be integers]

Bocard's problem asks to find the solutions to the Diophantine equation [itex]n!=m^2-1[/itex]. The only known [itex](n,m)[/itex] pairs are [itex](4,5),(5,11),(7,71)[/itex], and it is conjectured that there are no more. A generalization of this problem would be [itex]n!=m^2-A[/itex]. 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 [itex]A[/itex] a finite number of solutions (usually 0).

As an example, consider the solutions to [itex]n!=m^2-5[/itex]. To show it has a finite number of solutions, consider the equation in terms of mod 3. Assuming [itex]n\geq3[/itex], then [itex]n!\bmod3=0[/itex], and if it can be shown that [itex]m^2 \equiv 2 \pmod{3}[/itex] has no solutions, then [itex]n!=m^2-5[/itex] cannot have a solution with [itex]n\geq3[/itex]. It is simple enough to show that the solution set to [itex]m^2 \equiv 2 \pmod{3}[/itex] is null. As [itex]3x[/itex], [itex]3x+1[/itex], and [itex]3x+2[/itex] include all integers, showing that [itex](3x)^2=9x\equiv 0 \pmod{3}[/itex], [itex](3x+1)^2=9x^2+6x+1 \equiv 1 \pmod{3}[/itex], [itex](3x+2)^2=9x+12x+4\equiv 1 \pmod{3}[/itex] demonstrates that any [itex]n^2\bmod3[/itex] will be 0 or 1, but never 2. Therefore, [itex]n[/itex] can only equal 1 or 2; since neither provides a solution, the equation at [itex]A=5[/itex] has no solutions.

Making a quick list of [itex]a^2\bmod b[/itex] cycles (I say cycle since the first [itex]a[/itex] terms will repeat ad infinitum over [itex]b[/itex]) 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, [itex]a^2\bmod 3[/itex] will never include 2. Therefore, for all integers [itex]k[/itex], [itex]A=3k+2[/itex] will have a finite solution set. It's easy enough to write a program to loop through the first several [itex]b[/itex]'s and it seems that only numbers in the form [itex]n!=m^2-s^2[/itex] 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 [itex]A[/itex]'s there is a finite number of solutions?

---
Edit: My program confirms that non-square numbers from [itex]A=[/itex] 2 to 224 have finite solutions with an upper bound [itex]B[/itex] for [itex]n[/itex] at [itex]B\leq13[/itex]. This is what leads me to wonder if this extends to infinity.
Your post is interesting and I thought it warranted comment but had none to offer. Glad to see that it is now getting a few comments.
 
  • #4
Vargo said:
So by your argument, given A, you need to find a modulus N such that A is not a square modulo N. If you can find such an N then you can repeat your argument replacing 3 with N. It might help you to read up on quadratic residues. http://en.wikipedia.org/wiki/Quadratic_residue

Also, you might check out http://en.wikipedia.org/wiki/Law_of_quadratic_reciprocity
That provides the fastest way to check whether or not A is a square modulo N.

Thanks for your reply! I was not familiar with quadratic residues before my original post, though I came across them just a few days ago. I think they are definitely key to the problem.

ramsey2879 said:
Your post is interesting and I thought it warranted comment but had none to offer. Glad to see that it is now getting a few comments.

Thanks :)
 
  • #5


I find this variant of Bocard's problem to be an interesting and challenging mathematical puzzle. Your approach of using modular arithmetic to determine the finite solutions for certain values of A is a valid and effective method. However, as you have mentioned, it is still an observed pattern without a general proof.

To prove that for all non-square A's there is a finite number of solutions, one possible approach could be to use algebraic number theory and properties of quadratic fields. This could potentially provide a deeper understanding of the relationship between n and m in the equation n!=m^2-A. Another approach could be to use advanced techniques from number theory, such as elliptic curves and modular forms, to establish a general proof for this problem.

In addition, it would be helpful to consider the specific cases where A is a perfect square, as these may have different solution sets compared to non-square A's. It may also be worth exploring whether there are any patterns or relationships between the values of A and the upper bound B for n.

Overall, your observations and approach to this problem are valuable contributions to the study of Bocard's problem and its variants. I encourage you to continue exploring this problem and possibly collaborate with other mathematicians to find a general proof for your conjecture.
 

What is Variant of Bocard's Problem?

Variant of Bocard's Problem is a mathematical problem that involves finding the number of different ways to arrange objects in a specific order. It is a variation of the original Bocard's Problem, which involves finding the number of ways to arrange objects without repetition.

What are the applications of Variant of Bocard's Problem?

Variant of Bocard's Problem has practical applications in fields such as computer science, genetics, and chemistry. It can be used to determine the number of possible combinations of genes in a genetic code or the number of ways to arrange elements in a chemical compound.

What is the formula for solving Variant of Bocard's Problem?

The formula for Variant of Bocard's Problem is n!/(n-r)!, where n is the total number of objects and r is the number of objects being arranged in a specific order. This formula is also known as the permutation formula.

What is the difference between Variant of Bocard's Problem and Bocard's Problem?

The main difference between Variant of Bocard's Problem and Bocard's Problem is that Variant of Bocard's Problem allows for repetition, while Bocard's Problem does not. This means that in Variant of Bocard's Problem, objects can be arranged in the same order more than once, while in Bocard's Problem, each object can only be used once in a particular arrangement.

How is Variant of Bocard's Problem related to other mathematical concepts?

Variant of Bocard's Problem is closely related to other mathematical concepts such as combinations and factorial. It can also be used to solve problems involving permutations with restrictions, such as when certain objects must be placed in a specific position in an arrangement.

Similar threads

  • Linear and Abstract Algebra
Replies
2
Views
705
  • Precalculus Mathematics Homework Help
Replies
4
Views
895
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
1
Views
1K
  • Precalculus Mathematics Homework Help
Replies
9
Views
1K
  • Precalculus Mathematics Homework Help
Replies
5
Views
872
  • Linear and Abstract Algebra
Replies
1
Views
730
  • Precalculus Mathematics Homework Help
Replies
3
Views
717
  • Precalculus Mathematics Homework Help
Replies
1
Views
747
  • Precalculus Mathematics Homework Help
Replies
2
Views
784
Back
Top