p|2qr+r =>p|(2q+1)r=> p|2q+1 or p|r Case 1: p|r p and r are prime numbers so p=r r|2pq+q but p=r so r|q=> r=q Case 2: p|2q+1 Similarly we have q|2r+1 and r|2p+1 therefore 2p+1>= p ,2r+1>=q, 2p+1>=r => 8p+7>=2q+1>=p but p|2q+1 so we can present 2q+1 by p similarly, we can present 2r+1 by q, 2p+1 by r So we can present p,q,r by each other From there we can solve the problem