$\displaystyle \left\{ \begin{array}{l}a^{5} = 5b^{3} - 4c \\

b^{5} = 5c^{3} - 4a \\

c^{5} = 5a^{3} - 4b\end{array} \right\}$

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- Sep 1st 2007, 11:43 AMUnunuquantiumNIce one, who will do it first??
$\displaystyle \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 PMSoroban
Hello, Ununuquantium!

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

Quote:

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

b^5 \:= \:5c^3 - 4a \\

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

- Sep 1st 2007, 01:02 PMCaptainBlack
- Sep 5th 2007, 07:16 AMUnunuquantium
This problem is still unsolved, maybe someone will help?

- Sep 5th 2007, 08:37 AMjazzpianoMaybe 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 AMUnunuquantium
Yep, I know the same solution, but I cannot prove that there are not any other solution.....

- Sep 5th 2007, 12:51 PMtopsquark
- Sep 6th 2007, 06:48 AMjazzpianoArgh..