# NIce one, who will do it first??

• Sep 1st 2007, 11:43 AM
Ununuquantium
NIce one, who will do it first??
$\left\{ \begin{array}{l}a^{5} = 5b^{3} - 4c \\
b^{5} = 5c^{3} - 4a \\
c^{5} = 5a^{3} - 4b\end{array} \right\}$
• Sep 1st 2007, 12:16 PM
Soroban
Hello, Ununuquantium!

Didn't you Preview your work before you posted it?

Quote:

$\begin{Bmatrix}a^5 \:=\: 5b^3 - 4c \\
b^5 \:= \:5c^3 - 4a \\
c^5 \:= \:5a^3 - 4b\end{Bmatrix}$

• Sep 1st 2007, 01:02 PM
CaptainBlack
Quote:

Originally Posted by Ununuquantium
$\left\{ \begin{array}{l}a^{5} = 5b^{3} - 4c \\
b^{5} = 5c^{3} - 4a \\
c^{5} = 5a^{3} - 4b\end{array} \right\}$

If you are setting problems to challenge people put them in one of the Lounge groups.
These groups are for people who want help with stuff they don't know the answer to.

RonL
• Sep 5th 2007, 07:16 AM
Ununuquantium
This problem is still unsolved, maybe someone will help?
• Sep 5th 2007, 08:37 AM
jazzpiano
Maybe it will help?
If you add these equations you get:

(a-2)(a-1)a(a+1)(a+2) + (b-2)(b-1)b(b+1)(b-2) + (c-2)(c-1)c(c+1)(d+1) = 0

One solution is: a=b=c=2 or 1 or 0 or -1 or -2

I don't know if there are other solutions, maybe someone can proof that a,b,c must be equal?
• Sep 5th 2007, 09:38 AM
Ununuquantium
Yep, I know the same solution, but I cannot prove that there are not any other solution.....
• Sep 5th 2007, 12:51 PM
topsquark
Quote:

Originally Posted by jazzpiano
If you add these equations you get:

(a-2)(a-1)a(a+1)(a+2) + (b-2)(b-1)b(b+1)(b-2) + (c-2)(c-1)c(c+1)(d+1) = 0

One solution is: a=b=c=2 or 1 or 0 or -1 or -2

I don't know if there are other solutions, maybe someone can proof that a,b,c must be equal?

Clever!

How about any combination of a = 0, c = -b where b is any real number.

-Dan
• Sep 6th 2007, 06:48 AM
jazzpiano
Argh..
Quote:

Originally Posted by topsquark
Clever!

How about any combination of a = 0, c = -b where b is any real number.

-Dan

I'm afraid it won't work...

Despite the fact, that in equation I wrote (a-2)(a-1).... you can pick eg. a=1 b=2 c=-1 and it seems to be correct,when you put that numbers to original equations it's incorrect...