Re: A REAL TOUGH PROBLEM
- To: mathgroup at smc.vnet.net
- Subject: [mg15394] Re: A REAL TOUGH PROBLEM
- From: "Rajesh Ram" <rajeshram at lucent.com>
- Date: Tue, 12 Jan 1999 03:14:55 -0500
- Organization: Lucent Technologies, Columbus, Ohio
- References: <779d7l$bhd@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Bob, I have restated the problem with more constraints. Best of luck. NN Here is a problem for your thoughts : Let P(m) = a1^m + b1^m + c1^m Q(m) = a2^m + b2^m + c2^m R(m) = a3^m + b3^m + c3^m S(m) = a4^m + b4^m + c4^m Z(m) = (P(m)^2 - Q(m)^2) / (R(m)^2 - S(m)^2) Y(m) = (P(m)^2 + Q(m)^2) / (R(m)^2 + S(m)^2) where a1,a2,a3,a4,b1,b2,b3,b4,c1,c2,c3,c4 are all Integers. Find the solution for a1,a2,a3,a4,b1,b2,b3,b4,c1,c2,c3,c4 when 1) Z(3) = Z(5) = Z(7) 2) Y(3) = Y(5) = Y(7) 2) a1, a2, a3, a4, b1, b2, b3, b4, c1,c2,c3,c4 are non-zero 3) No two of a1, a2, a3, a4, b1, b2, b3, b4, c1,c2,c3,c4 are equal Thanks NN BobHanlon at aol.com wrote in message <779d7l$bhd at smc.vnet.net>... > >In a message dated 1/8/99 9:36:46 AM, root at nntpb.cb.lucent.com writes: > >>Let >> >>P(m) = a1^m + b1^m + c1^m >>Q(m) = a2^m + b2^m + c2^m >>R(m) = a3^m + b3^m + c3^m >>S(m) = a4^m + b4^m + c4^m >> >>Z(m) = (P(m)^2 - Q(m)^2) / (R(m)^2 - S(m)^2) >> >>where a1,a2,a3,a4,b1,b2,b3,b4,c1,c2,c3,c4 are all Integers. >> >>Find the solution for a1,a2,a3,a4,b1,b2,b3,b4,c1,c2,c3,c4 when >> >>Z(3) = Z(5) >> > >One obvious solution would be for Z[m] = 0, i.e., P[m] = Q[m] and R[m] ><> S[m] > >Clear[P, Q, R, S, Z, a1, a2, a3, a4, b1, b2, b3, b4, c1, c2, c3, c4]; > >a2 = a1; b2 = b1; c2 = c1; >P[m_] := a1^m + b1^m + c1^m; >Q[m_] := a2^m + b2^m + c2^m; >R[m_] := a3^m + b3^m + c3^m; >S[m_] := a4^m + b4^m + c4^m; >Z[m_] := (P[m]^2 - Q[m]^2)/(R[m]^2 - S[m]^2) > >Z[m] > >0 > >Then the values for {a1, a3, a4, b1, b3, b4, c1, c3, c4} are arbitrary >integers as long as R[m] <> S[m]. >Consequently, there are infinitely many solutions. You would need to >have more constraints than just Z[3] = Z[5]. > >Bob Hanlon >