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Old 2017-02-03, 12:48   #5
science_man_88
 
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Jul 2009
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Quote:
Originally Posted by carpetpool View Post

I don't want to go ahead making a statement saying that p, pr+1, r will always work because I am not sure of it's certainty.

p = p
q = pr+1
r = r

and then if

d = pr

x=(p+q)*(q+d) - p*q
y=(q+d)*q - p*(p+q)

is

r*y-q-d = x

always true?
so x= (p+pr+1)*(pr+1+pr)-p*(pr+1) = (p*r+p+1)*(2*p*r+1)-(p^2)*r-p =2*(p^2)*(r^2)+p*r+2*(p^2)*r+p+2*p*r+1-(p^2)*r-p = 2*(p^2)*(r^2)+p*r+(p^2)*r+2*p*r+1 taking p*r=d simplifies to 2*d^2+d+d*p +2*d+1 = 2*d^2+(3+p)*d+1


and y= (pr+1+pr)*(pr+1) - p*(p+(pr+1)) using pr=d we simplify to (d+1+d)*(d+1)+p*(p+d+1) = d^2+d+d+1+d^2+d+p^2+pd+p = 2d^2+3d+1+p^2+pd+p

so your claim is that :

2*r*d^2+(3r+d)*d +r - (2*d+1) = 2d^2+3d+1+p^2+pd+p
(2*r+1)*d^2+(3r-2)*d+r-1 = 2d^2+3d+1+p^2+pd+p aka

(2*r+1)*d^2+(3r-2)*d+r-1 - (2d^2+3d+1+p^2+pd+p) =0
(2r-1)*d^2+(3r-p-2)*d+r-2-p^2 -p =0

now it would be solving for p,d, and r to find a solution but I'm not that advanced.

Last fiddled with by science_man_88 on 2017-02-03 at 12:49
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