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Old 2019-01-18, 13:57   #45
Dieter
 
Oct 2017

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Default 4-6-solution

Having finally found a 4-6-solution I would like to know, if anyone has found a 4-7-solution.
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Old 2019-01-18, 14:37   #46
petrw1
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"Wayne"
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How about a 5x5?
Not me.
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Old 2019-01-18, 22:51   #47
henryzz
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"David"
Sep 2007
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If ai+b is a square and aj-ai>0 is less than 2*sqrt(ai+b)+1 then aj+b can't be a square as it is less than the next square (sqrt(ai+b)+1)^2
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Old 2019-01-26, 17:40   #48
uau
 
Jan 2017

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I improved my search program a bit and have found 8 distinct 4+6 solutions. Looks like 4+7 or 5+5 solutions would have to be pretty huge. It's not obvious whether arbitrarily large solutions can be expected to exist at all. Has anyone tried to analyze that?
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Old 2019-01-27, 06:06   #49
Dieter
 
Oct 2017

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Quote:
Originally Posted by uau View Post
I improved my search program a bit and have found 8 distinct 4+6 solutions. Looks like 4+7 or 5+5 solutions would have to be pretty huge. It's not obvious whether arbitrarily large solutions can be expected to exist at all. Has anyone tried to analyze that?
Using your approach I have found a 3-13-solution - but I fear that this doesn‘t help very much. I continue to search for 4+7, but without analyzing.
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Old 2019-01-27, 10:59   #50
henryzz
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Given all the differences between squares need lots of factors I would expect solutions to get bigger and bigger as more factors are needed.
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Old 2019-02-03, 13:43   #51
Xyzzy
 
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http://www.research.ibm.com/haifa/po...nuary2019.html
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Old 2019-02-03, 14:01   #52
uau
 
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Quote:
Originally Posted by Xyzzy View Post
This lists the same 4+6 solution many times as a "different" one. If you multiply all the numbers by a second power, all the sums obviously stay squares (square times square is a square). So to tell whether solutions are truly distinct, you should make sure to divide out any such common multiples. That obviously wasn't done when writing this solution page.
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Old 2019-02-03, 14:47   #53
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Quote:
Originally Posted by uau View Post
This lists the same 4+6 solution many times as a "different" one. If you multiply all the numbers by a second power, all the sums obviously stay squares (square times square is a square). So to tell whether solutions are truly distinct, you should make sure to divide out any such common multiples. That obviously wasn't done when writing this solution page.
Right. You should start out by normalizing such that smallest element is 0. Then divide out gcd (and list them in sorted order).
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Old 2019-02-04, 11:50   #54
DukeBG
 
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My solution is not listed, as far as I can tell. Though I didn't try to normalize the listed ones.

[0, 36295, 233415, 717255] & [93^2, 267^2, 501^2, 1059^2]
the second set expanded [8649, 71289, 251001, 1121481]

I also claim that this pair of sets is the 4-4 solution with the smallest possible largest element of the set with the zero. (Assuming both sets contain non-negative numbers, of course).
I believe, though don't claim, that it is also the smallest possible largest element of both sets.

My other 4-4 solution is [0, 259875, 475875, 1313091] & [15^2, 447^2, 895^2, 1695^2]. In case anyone wants to make a registry of normalized solutions.

I ended up not bothering to find 4-5 or larger solution.

Last fiddled with by DukeBG on 2019-02-04 at 11:50
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Old 2019-02-04, 15:09   #55
uau
 
Jan 2017

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Quote:
Originally Posted by DukeBG View Post
My other 4-4 solution is [0, 259875, 475875, 1313091] & [15^2, 447^2, 895^2, 1695^2]. In case anyone wants to make a registry of normalized solutions.

I ended up not bothering to find 4-5 or larger solution.
I doubt anyone would bother with a list of 4+4 solutions, or at least not one maintained by hand. I found over two thousand different 4+5 solutions, and 4+4 ones are more common (I didn't directly count those). Currently found 4+6 solutions could be listed by hand, but 4+5 and smaller are too common for that.
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