Suppose p,q,r are three distinct primes such that rp3+p2+p=2rq2+q2+q. Find all possible values of pqr.
Problem
Source: India Postal Coaching 2014 Set 3 Problem 1
Tags: number theory unsolved, number theory
29.11.2014 23:00
hajimbrak wrote: Suppose p,q,r are three distinct primes such that rp3+p2+p=2rq2+q2+q.Find all possible values of pqr. What is this ! rp3=2rq2+q2+q−p2−p=0mod2 so r=2 or p=2. If p=2 then because p,q,r are dinstict primes r,q>2 8r+6=2rq2+q2+q r(8−2q2)=q2+2−6 so obvious no solutions because left is negative. So r=2.... So the first equation will be , 2p3+p2+p=5q2+q p(2p2+p+1)=q(5q+1) (1) Because p,q are dinstict then (p,q)=1 so p|5q+1 and q|2p2+p+1 so there are positive integers x,y such as, 5q+1=px , 2p2+p+1=qy so put these at the equation (1) and we have, p∗qy=q∗px so x=y so we want to solve the following system, 5q+1=px (2) 2p2+p+1=qx (3) From (2) we get that q=px−15 so we put this q at (3) ... 2p2+p+1=px−15x 10p2+5p+5=px2−x p(x2−10p−5)=x+5 (4) From (4) we can see that because x>0 then x2−10p−5>0 and x2−10p−5≤x+5 (5) Also from (4) we get that p|x+5 so p≤x+5 (6) From(5),(6) we get that x2−10(x+5)−5≤x2−10p−5≤x+5 x2−11x−60≤0 so x≤15 From (3) we can see that x=0mod2 so x=2,4,6,8,10,12,14 I checked one time and i found that only for x=14 has solution and the systems (2),(3) easy give the solutions (p,q)=(19,53) So finally the only slution is (p,q,r)=(19,53,2) We check the solution so 2∗193+192+19=2∗2∗532+532+53=14098 so is true.