Quote:
Originally Posted by Batalov
You can't prove that. It's false.
Take p,q,r = 2,5,3
d = p*r = 6
x=(p+q)*(q+d)  p*q = 67
y=(q+d)*q  p*(p+q) = 41
r*xqd = 190 y

Unfortunately, I didn't bother to go into depth with three integer parameters. (I also got the equation wrong, its r*yqd = x, yet Batalov's counterexample is still valid.) The point is the above congruence isn't always correct. I did, observe the following after continuously testing values of p, q, and r:
Pairs that work (but unproven): p, 2p+1, 2
2,5,2
7*9  2*5 = 53
9*5  2*7 = 31
2*3154 = 53
3,7,2
10*13  3*7 = 109
13*7  3*10 = 61
2*6176 = 109
4,9,2
13*17  4*9 = 185
17*9  4*13 = 101
2*10198 = 185
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?