- **carpetpool**
(*https://www.mersenneforum.org/forumdisplay.php?f=145*)

- - **Large Polynomial Equation Problem**
(*https://www.mersenneforum.org/showthread.php?t=21998*)

Large Polynomial Equation ProblemHelp, comments, suggestions appreciated. :smile:
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 |

[QUOTE=carpetpool;452107]Help, comments, suggestions appreciated. :smile:
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[/QUOTE] 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. |

[QUOTE=carpetpool;452107]Help, comments, suggestions appreciated. :smile:
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[/QUOTE] 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 [TEX] \ne[/TEX] y |

[QUOTE=Batalov;452131]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 [TEX] \ne[/TEX] y[/QUOTE] 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? |

[QUOTE=carpetpool;452137]
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?[/QUOTE] 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. |

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