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May 4th, 2010, 05:15 PM  #1 
Joined: Dec 2006 Posts: 132 Thanks: 0  Solve a Integer equation system
Let Find smallest 
May 5th, 2010, 06:23 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 13,399 Thanks: 249 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic  Re: Solve a Integer equation system
No solutions below 10^45. Edit: No solutions below 10^55. 
May 5th, 2010, 06:53 AM  #3 
Joined: Apr 2010 Posts: 65 Thanks: 0  Re: Solve a Integer equation system
when is square? (b=2k) or (c=2n+1) 
May 5th, 2010, 11:12 AM  #4 
Joined: Dec 2006 Posts: 132 Thanks: 0  Re: Solve a Integer equation system
The Chinese Remainder theorem shows that for some I don't know if this is useful. Seems a very tough problem. 
May 5th, 2010, 12:23 PM  #5  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 13,399 Thanks: 249 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic  Re: Solve a Integer equation system Quote:
If you wanted to incorporate my earlier search, you could say that for some , but that would be silly.  
May 5th, 2010, 01:39 PM  #6 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 13,399 Thanks: 249 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic  Re: Solve a Integer equation system I think it can be proven that my method can't be extended to any further primes  the proof should be easy with the amount of checking I did to remove special cases  but I won't bother writing it out. This above result (the behavior of d mod 3622080) can be used to show that c is 316675 mod 329280. This lets me check even further, showing that d > 10^70. I imagine that if I took into account behavior mod other primes (where there will be several residue classes, but hopefully not too many) I could push the verification to at least 10^80. But I have a feeling that there's a much better way to do all of this. 
May 6th, 2010, 07:39 AM  #7 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 13,399 Thanks: 249 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic  Re: Solve a Integer equation system
Working mod 372118412930880 I can show that d > 10^100. This actually wasn't nearly as hard as I expected  it took about 5 minutes (including about 1 second for preprocessing) and 23 kB. The method should scale pretty well; I'm not yet sure if it is memorybound or CPUbound. (I've picked the most efficient primes to use so far; further choices will be less good, so it won't scale as easily as it has so far.) Edit: Working mod 391343029186433681945280 took only 100 seconds (including about 18 seconds for preprocessing) and 71 MB to show that d > 10^125. It's definitely memorybound, at least as currently implemented  I could substitute time for memory to some degree if I can do fast CRTs. Also, due to a flaw in my implementation*, my actual memory use is slightly more than double the quoted  I'm essentially using two copies of a large array. * Flaw: I'm too lazy to do it the right way. 
May 6th, 2010, 03:50 PM  #8 
Joined: Dec 2006 Posts: 132 Thanks: 0  Re: Solve a Integer equation system
This is amazing! Would you please explain your algorithm? thanks

May 6th, 2010, 04:24 PM  #9  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 13,399 Thanks: 249 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic  Re: Solve a Integer equation system Quote:
Notice that 391343029186433681945280 = 2^6 * 3 * 5 * 7^3 * 11 * 31 * 61 * 211 * 331 * 421 * 1321 * 1471. 3622080 = 2^6 * 3 * 5 * 7^3 * 11 is the constant from earlier; the other primes {31, 61, 211, 331, 421, 1321, 1471} are ones which I calculated to have few appropriate residues. Here's a short example. I'll use d = 8 mod 30 rather than d = 1053698 mod 3622080 so the numbers are smaller. Consider the values of d mod 30 * 31 (where 31 is one of the primes dividing the large number above; I could have used 61 or 211 but again I'm going for size here). They are 8,38,68,98,128,158,188,218,248,278,308,338,368,398 ,428,458,488,518,548,578,608,638,668,698,728,758,7 88,818,848,878,908 But none of 68/2,158/2,278/2,308/2,368/2,398/2,458/2,488/2,518/2,548/2,668/2,728/2,788/2,818/2,848/2 are squares mod 30*31/2 (by quadratic reciprocity  use the Legendre symbol). So only 8,38,98,128,188,218,248,338,428,578,608,638,698,75 8,878,908 are possible. Of those, none of (382)/3,(1282)/3,(2182)/3,(2482)/3,(3382)/3,(4282)/3,(5782)/3,(6382)/3,(8782)/3 are cubes mod 30*31/3. Thus, c can be only 8,98,188,608,698,758,908 mod 30*31. You can use the Sun Zi's theorem (the CRT) to lift these to 1053698,37274498,73495298,55384898,91605698,408965 78,26408258 mod 112284480 = 3622080*31, which means that instead of checking 1 number out of every 30, you're checking 7 numbers out of every 112284480.  
May 7th, 2010, 10:32 AM  #10 
Joined: Dec 2006 Posts: 132 Thanks: 0  Re: Solve a Integer equation system
Thanks a lot CRGreathouse. I got lots to learn.


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