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
>