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Variant of Bocard's Problem

by Someone2841
Tags: bocard, variant
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Someone2841
#1
Dec2-12, 04:50 AM
P: 29
[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.
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Vargo
#2
Dec7-12, 09:58 PM
P: 350
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_...ic_reciprocity
That provides the fastest way to check whether or not A is a square modulo N.
ramsey2879
#3
Dec8-12, 10:52 AM
P: 894
Quote Quote by Someone2841 View Post
[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.

Someone2841
#4
Dec9-12, 05:46 PM
P: 29
Variant of Bocard's Problem

Quote Quote by Vargo View Post
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_...ic_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.

Quote Quote by ramsey2879 View Post
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 :)


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