hajimbrak wrote: Suppose $p,q,r$ are three distinct primes such that $rp^3+p^2+p=2rq^2+q^2+q$.Find all possible values of $pqr$. What is this ! $rp^3=2rq^2+q^2+q-p^2-p=0mod2$ so $r=2$ or $p=2$. If $p=2$ then because $p,q,r$ are dinstict primes $r,q>2$ $8r+6=2rq^2+q^2+q$ $r(8-2q^2)=q^2+2-6$ so obvious no solutions because left is negative. So $r=2$.... So the first equation will be , $2p^3+p^2+p=5q^2+q$ $p(2p^2+p+1)=q(5q+1)$ (1) Because $p,q$ are dinstict then $(p,q)=1$ so $p|5q+1$ and $q|2p^2+p+1$ so there are positive integers $x,y$ such as, $5q+1=px$ , $2p^2+p+1=qy$ 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) $2p^2+p+1=qx$ (3) From (2) we get that $q=\dfrac{px-1}{5}$ so we put this $q$ at (3) ... $2p^2+p+1=\dfrac{px-1}{5}x$ $10p^2+5p+5=px^2-x$ $p(x^2-10p-5)=x+5$ (4) From (4) we can see that because $x>0$ then $x^2-10p-5>0$ and $x^2-10p-5\leq{x+5}$ (5) Also from (4) we get that $p|x+5$ so $p\leq{x+5}$ (6) From(5),(6) we get that $x^2-10(x+5)-5\leq{x^2-10p-5}\leq{x+5}$ $x^2-11x-60\leq{0}$ so $x\leq{15}$ 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 $2*19^3+19^2+19=2*2*53^2+53^2+53=14098$ so is true.