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Old 2020-09-10, 11:38   #14
Alberico Lepore
 
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May 2017
ITALY

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Quote:
Originally Posted by retina View Post
That is your equation, you invented it. So you should be the one to understand it. I have no idea what you are trying to do, it all looks like nonsense to me.

It's up to you to show us how you factor things, not the other way around. Your claim, you prove it.
So maybe I found when it's true:
(a + b) mod 3 = 0
in two cases

M=[(a+b)/2-((a+b)/6-1)/2]*[2*[(a+b)/2-((a+b)/6-1)/2]-3]
and
M=[(a+b)/2-((a+b)/6+1)/2]*[2*[(a+b)/2-((a+b)/6+1)/2]+3]



so I tried to bring back a generic number (a + b) mod 3 = 0

in

M=[(a+b)/2-((a+b)/6-1)/2]*[2*[(a+b)/2-((a+b)/6-1)/2]-3]

but I didn't get any useful results

Example

N=161
,
2*(N+(n/2)^2-((a+b)/6-1)^2)+2*a^2+((b-a)/2)^2=((3*a+b)/2)^2
,
a*b=(N+(n/2)^2-((a+b)/6-1)^2)
,
2*(N+(n/2)^2-((a+b)/6-1)^2)+2*1^2+((a+b)/2+1)^2-((3*a+b)/2)^2=0

but I will continue to study

Last fiddled with by Alberico Lepore on 2020-09-10 at 11:42
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