Quote:
Originally Posted by carpetpool
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*yqd = x
always true?

so x= (p+pr+1)*(pr+1+pr)p*(pr+1) = (p*r+p+1)*(2*p*r+1)(p^2)*rp =2*(p^2)*(r^2)+p*r+2*(p^2)*r+p+2*p*r+1(p^2)*rp = 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+(3r2)*d+r1 = 2d^2+3d+1+p^2+pd+p aka
(2*r+1)*d^2+(3r2)*d+r1  (2d^2+3d+1+p^2+pd+p) =0
(2r1)*d^2+(3rp2)*d+r2p^2 p =0
now it would be solving for p,d, and r to find a solution but I'm not that advanced.