20170202, 21:53  #1 
"Sam"
Nov 2016
2^{2}×79 Posts 
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*xqd = y 
20170202, 22:43  #2  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:
and y=q^2+q(d)p^2p(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. 

20170203, 01:45  #3  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·2,281 Posts 
Quote:
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 

20170203, 04:50  #4  
"Sam"
Nov 2016
2^{2}×79 Posts 
Quote:
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? Last fiddled with by carpetpool on 20170203 at 05:08 

20170203, 12:48  #5  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:
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. Last fiddled with by science_man_88 on 20170203 at 12:49 

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