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Variant of Bocard's Problem 
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#1
Dec212, 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^21[/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^2A[/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 nonsquare [itex]A[/itex] a finite number of solutions (usually 0). As an example, consider the solutions to [itex]n!=m^25[/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^25[/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^2s^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 nonsquare [itex]A[/itex]'s there is a finite number of solutions?  Edit: My program confirms that nonsquare 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. 


#2
Dec712, 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. 


#3
Dec812, 10:52 AM

P: 894




#4
Dec912, 05:46 PM

P: 29

Variant of Bocard's Problem



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