Re: A REAL TOUGH PROBLEM
- To: mathgroup at smc.vnet.net
- Subject: [mg15368] Re: [mg15343] A REAL TOUGH PROBLEM
- From: BobHanlon at aol.com
- Date: Sat, 9 Jan 1999 23:58:24 -0500
- Sender: owner-wri-mathgroup at wolfram.com
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