20180111, 19:10  #1 
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
3·1,999 Posts 
The SquareSum problem
https://www.youtube.com/watch?v=G1m7goLCJDY
https://www.youtube.com/watch?v=7_ph5djCCnM How far can you prove upto 299 seems fairly easy to beat? Are cubes possible? 
20180111, 19:42  #2  
"Forget I exist"
Jul 2009
Dumbassville
20320_{8} Posts 
Quote:
Last fiddled with by science_man_88 on 20180111 at 20:09 Reason: Fixing math, counted 1 as a summable square. 

20180111, 21:26  #3  
"Forget I exist"
Jul 2009
Dumbassville
2^{4}·3·5^{2}·7 Posts 
Quote:
Last fiddled with by science_man_88 on 20180111 at 21:27 

20180112, 14:31  #4  
"Forget I exist"
Jul 2009
Dumbassville
2^{4}·3·5^{2}·7 Posts 
Quote:
Last fiddled with by science_man_88 on 20180112 at 15:30 

20180112, 19:21  #5 
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
3·1,999 Posts 
I have attached a graph of summing upto a cube complete to 124. No solution yet. Unless 108 is an end the solution is at least 235. It is at least all connected.

20180112, 19:37  #6 
"Forget I exist"
Jul 2009
Dumbassville
20D0_{16} Posts 
Nice work did my pm clue help ? There is still one other rule in my head, but it's more about number of possible connections versus actual connections.

20180112, 20:33  #7 
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
176D_{16} Posts 

20180112, 20:44  #8 
"Forget I exist"
Jul 2009
Dumbassville
2^{4}×3×5^{2}×7 Posts 
My clue by pm ,was that the endpoint sum parity is the same parity as the number of sums to the odd based powers. In Matt's first video we find endpoints 18 and 22 these sum to an even number, therefore there must be an even number of sums that sum to the odd squares ( aka the pairings of opposite parity, count 12) . The point about number of possible connections total, is simply that numbers that are half of an even based power have less connections, 18 only had 1 because it couldn't pair with itself and there are only two squares between it and twice it one that can't be made.

20180112, 23:58  #9 
"Robert Gerbicz"
Oct 2005
Hungary
3^{2}×5^{2}×7 Posts 
Nice problem, easily outperformed that in one day of work.
There exists a solution for n=15,16,17,23 and for all 25<=n<=1048576. (the last checked is n=2^20). To prove more we give a Hamiltonian cycle for 32<=n<=1048576. Download the "proof" at my drive https://drive.google.com/file/d/1S9E...ew?usp=sharing (8.9 MB compressed zip). There are only two cases in the algorithm: S: we give simply a valid sequence with length=n. F: we use the previous sequence with length=n1, and the two indexes i,j after the F symbol that is: S=a(1),...,a(n1) sequence flip the terms between i to j and insert n to the ith place (the indexes starts with one). We have binomial(n1,2) choices for i,j, and roughly n^(3/2) chance that this will be a good sequence, because we see only 3 new terms in the modified sequence: a(i1)+n, n+a(j) and a(i)+a(j+1) hence the probability that we can't find a solution for n is roughly (1n^(3/2))^(n^2)~exp(sqrt(n)) so we have not only good probability for a continuation, but this serie converges, so likely this will give a solution for each (large) n value. So for example if seq=[2,4,1,5,6,3] and we see n=7: F 2,4 then seq'=[2,7,5,1,4,6,3] To give a Hamiltonian cycle we use only 1<i<j<n1, because with this the first and the last term of the sequence won't change, so if it is a Hamiltonian path then also a cycle. And this happens in practice the largest n index for that we needed 'S' is at n=6109, The given solution is a Hamiltonian cycle for all 32<=n<=2^20. Computed these solutions in only 15 minutes. ps. there are much more such sequence transformations, but using only these we see only a few S, so needed to find a sequence from scratch in a few n cases. Double checking the file with brute force is still possible. 
20180113, 00:03  #10 
Aug 2006
5,987 Posts 

20180113, 01:42  #11 
"Forget I exist"
Jul 2009
Dumbassville
2^{4}·3·5^{2}·7 Posts 
I think factoring is actually an answer to this in a sense. The sum of powers is 2*T_n  endpoint sum. Modulo or factoring may have implications on any given case.
Last fiddled with by science_man_88 on 20180113 at 02:10 
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