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-   -   January 2019 (https://www.mersenneforum.org/showthread.php?t=23947)

Dieter 2019-01-18 13:57

4-6-solution
 
Having finally found a 4-6-solution I would like to know, if anyone has found a 4-7-solution.

petrw1 2019-01-18 14:37

How about a 5x5?
Not me.

henryzz 2019-01-18 22:51

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

uau 2019-01-26 17:40

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?

Dieter 2019-01-27 06:06

3-13
 
[QUOTE=uau;506908]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?[/QUOTE]

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.

henryzz 2019-01-27 10:59

Given all the differences between squares need lots of factors I would expect solutions to get bigger and bigger as more factors are needed.

Xyzzy 2019-02-03 13:43

[url]http://www.research.ibm.com/haifa/ponderthis/solutions/January2019.html[/url]

uau 2019-02-03 14:01

[QUOTE=Xyzzy;507545][URL]http://www.research.ibm.com/haifa/ponderthis/solutions/January2019.html[/URL][/QUOTE]

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.

axn 2019-02-03 14:47

[QUOTE=uau;507548]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.[/QUOTE]

Right. You should start out by normalizing such that smallest element is 0. Then divide out gcd (and list them in sorted order).

DukeBG 2019-02-04 11:50

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

My [I]other[/I] 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.

uau 2019-02-04 15:09

[QUOTE=DukeBG;507621]My [I]other[/I] 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.[/QUOTE]
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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