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Old 2017-02-03, 04:50   #4
carpetpool
 
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"Sam"
Nov 2016

4768 Posts
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Quote:
Originally Posted by Batalov View Post
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*x-q-d = 190  \ne y
Unfortunately, I didn't bother to go into depth with three integer parameters. (I also got the equation wrong, its r*y-q-d = 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*31-5-4 = 53

3,7,2

10*13 - 3*7 = 109
13*7 - 3*10 = 61

2*61-7-6 = 109

4,9,2

13*17 - 4*9 = 185
17*9 - 4*13 = 101

2*101-9-8 = 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*y-q-d = x

always true?

Last fiddled with by carpetpool on 2017-02-03 at 05:08
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