Quote:
Originally Posted by retina
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)/61)/2]*[2*[(a+b)/2((a+b)/61)/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)/61)/2]*[2*[(a+b)/2((a+b)/61)/2]3]
but I didn't get any useful results
Example
N=161
,
2*(N+(n/2)^2((a+b)/61)^2)+2*a^2+((ba)/2)^2=((3*a+b)/2)^2
,
a*b=(N+(n/2)^2((a+b)/61)^2)
,
2*(N+(n/2)^2((a+b)/61)^2)+2*1^2+((a+b)/2+1)^2((3*a+b)/2)^2=0
but I will continue to study