20171003, 05:54  #45 
Aug 2006
3×1,993 Posts 
Sidenote: In this case you seem to be looking at the sum of a linear function, which is of course quadratic. If you want to know when two quadratics will be equal you may find this handy:
https://www.alpertron.com.ar/QUAD.HTM 
20171003, 07:46  #46 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2·11·449 Posts 
Obviously, you have a sum of two arithmetic progressions, which you equate and then you have to solve a diophantine equation, which is equivalent with factorization.
On your particular case where you sum odd numbers, this is the difference of squares factoring method. When you sum odd numbers to n, you get a square number. Summing them from m to n is just a difference of two squares. 
20171003, 11:11  #47  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:
Last fiddled with by science_man_88 on 20171003 at 11:14 

20171003, 14:55  #48  
"Ed Hall"
Dec 2009
Adirondack Mtns
4,201 Posts 
Quote:
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Thanks to all for your help in this. Off to study... 

20171003, 15:36  #49  
Aug 2006
3·1,993 Posts 
Quote:


20171003, 18:23  #50  
"Ed Hall"
Dec 2009
Adirondack Mtns
4,201 Posts 
Quote:
The high sequence is made up of the first square, 2704 and each sequential square upward. These are derived by adding the differences between each, which is a progression that increases by 2 for each step. The difference between 2704 and 2809 (53^{2}) is 105, the difference between 2809 and 2916 (54^{2}) is 107, etc. The low sequence is similar in that it uses sequential squares as well, but is offset by the composite. In this case, 2627+64 gives us the starting value of 2691. The series continues by adding the perfect square differences from 64 upward: 17, 19, 21, 23, etc. The intersection of the two series will be the value of the high square needed to factor 2627. In this case it is reached rather quickly at 2916. The low series has no more interest after is has intersected with the high series to provide the value 2691. 26912627=64. The fact that this was the original low square is coincidental for this sample. However, we now have the two squares needed to factor 2627. I have programmatically stepped and checked through progressions one element at a time to look for this coincidence and stepped through the high series only, checking for a difference that is a perfect square, (as suggested by fivemack, earlier). Both are exceptionally slow. So now I'm looking again at a way to calculate directly the intersection. I'm sure someone has studied and tossed this at some time in the past, but I'd like to visit it myself for a bit. Thanks much! 

20190719, 21:33  #51  
"Michael"
Aug 2006
Usually at home
2×41 Posts 
Quote:
The equation for Y^{2} where n + Y^{2} = X^{2} is x^{2} + bx + c = Y^{2} where b = 2m and c = m^{2}  n m = ceil(sqrt(n)) I looked at this equation but it is not very useful as x can be very large. It is only marginally better than Fermat's Method which is very slow for large Y^2. Last fiddled with by mgb on 20190719 at 21:35 

20190721, 14:35  #52  
Jul 2018
47_{8} Posts 
Quote:
=0 so: x=(10**22+117) and y=(10**22+193) an integer solve. https://www.alpertron.com.ar/QUAD.HTM a=3 b=5 c=7 d=11 e=13 f=5*10**441856*10**22190127 please click for mouse or finger for touchplate: 'solve' button: 112 different solve only 1 second, thank you dario, good work! but if digit > 5e2 then need many time, for example: 101*(10**776+1777)**2103*(10**776+1777)*(10**776+2037)+107*(10**776+2037)**2109*(10**776+1777)+127*(10**776+2037)105*10**1552402048*10**776390143971 =0 so: x=(10**776+1777) and y=(10**776+2037) an integer solve. a=101 b=103 c=107 d=109 e=127 f=105*10**1552402048*10**776390143971 a few hours, not any result. question: how can we solve 1e16 times faster? Last fiddled with by hal1se on 20190721 at 14:39 

20210129, 21:17  #53 
Dec 2014
3×5×17 Posts 
This thread is old, but I did not see any reference to quadratic residues to speed up Fermat's method.
See Knuth, Vol 2, Section 4.5.4, Algorithm D. 
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