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Old 2017-02-02, 21:53   #1
carpetpool
 
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"Sam"
Nov 2016

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Lightbulb Large Polynomial Equation Problem

Help, comments, suggestions appreciated.

For three integers p, q, r such that gcd(p, q) = 1, gcd(r, q) = 1, let d = pr.

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

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

Prove that

r*x-q-d = y
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Old 2017-02-02, 22:43   #2
science_man_88
 
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Quote:
Originally Posted by carpetpool View Post
Help, comments, suggestions appreciated.

For three integers p, q, r such that gcd(p, q) = 1, gcd(r, q) = 1, let d = pr.

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

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

Prove that

r*x-q-d = y
so x=p(q)+r(p^2)+q^2+q(p)(r) -p(q) = d(p)+q^2+q(d)
and y=q^2+q(d)-p^2-p(q)
therefore y+q+d = q^2+q(d)-p^2+p(q)+q+d = (q+1+p)(q)+(q+1)(d)-p^2

r(x) = (q+1+p)(q)+(q+1)(d)-p^2 = d^2+r(q^2)+r(q)(d)

anyways for now I'm bored I guess.
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Old 2017-02-03, 01:45   #3
Batalov
 
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Quote:
Originally Posted by carpetpool View Post
Help, comments, suggestions appreciated.

For three integers p, q, r such that gcd(p, q) = 1, gcd(r, q) = 1, let d = pr.

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

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

Prove that

r*x-q-d = y
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
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Old 2017-02-03, 04:50   #4
carpetpool
 
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Nov 2016

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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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Old 2017-02-03, 12:48   #5
science_man_88
 
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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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