20190204, 20:24  #56  
Sep 2017
2^{2}·3·7 Posts 
Quote:


20190205, 09:57  #57 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2^{2}×1,433 Posts 
It seems to me that this problem should have a much easier solution than I have found. My code spends the majority of its time finding possible B2s and A3s(A1=0). I wish I could come up with an efficient way of finding all numbers of the form x^2b1 and x^2b2.

20190205, 13:02  #58  
Mar 2018
129_{10} Posts 
Quote:
Quote:
Last fiddled with by DukeBG on 20190205 at 13:05 

20190205, 17:11  #59  
Sep 2017
2^{2}·3·7 Posts 
Quote:
[0, 54432, 119392, 263872] & [324, 79524, 227529, 1258884] Last fiddled with by SmartMersenne on 20190205 at 17:19 Reason: Typo in third number of the second set. Fixed 

20190205, 18:27  #60 
Jan 2017
79 Posts 

20190205, 18:41  #61  
Jan 2017
79 Posts 
Quote:
If x^{2}+D=y^{2}, then D=y^{2}x^{2}=(yx)(y+x). If you generate all divisors k of D, then you can solve for possible values of x from (yx)=k, (y+x)=D/k > x = (D/k  k)/2. 

20190205, 20:05  #62  
Mar 2018
129_{10} Posts 
Quote:
However I mistakenly ignored even D's. The (...)/2 part requires that the two divisors have the same evenness (oddity? translating math terms I'm used to into english is sometimes a pain), so 2 as a prime divisor cannot be in power 1. If present in D, it should be present in both sides of the "split" into two divisors. (My mistake was that I decided it shouldn't be present at all because the solutions with it on both sides could be simplified into a solution without it. Nevermind it, wrong conclusion) Quote:
Raises the question of how to normalize these solutions then... Although that doesn't help in any way in finding bigger sets. 

20190206, 10:48  #63 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2^{2}×1,433 Posts 
I was calculating a list of x^2b1 storing it and then reusing the filtered list when looking for a b2.
The thing I hadn't thought about was using difference of squares trick. I need to think on this further. Using my current method I think can do an exhaustive search upto 1e9 in around 24 hours. 
20190206, 12:55  #64 
Jan 2017
79 Posts 
Exhaustive search of what? You'll find all solutions that have what parameter under 1e9? All the numbers in both sets are below that?
Last fiddled with by uau on 20190206 at 12:56 
20190206, 14:26  #65 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2^{2}×1,433 Posts 

20190222, 13:04  #66 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
1664_{16} Posts 
Only just noticed that all the 4,6 solutions at the end are wrong. 1899+1899 is not square.
@uau What were your search limits for this puzzle? Did you search for 5,5 solutions? After pinching your difference of squares idea, I have just started a run where a1=0, b1 and a2 < 1e9 and everything else < 1e12 
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