Thread: NIce one, who will do it first??

1. NIce one, who will do it first??

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

2. Hello, Ununuquantium!

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

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

3. 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

4. This problem is still unsolved, maybe someone will help?

5. 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?

6. Yep, I know the same solution, but I cannot prove that there are not any other solution.....

7. 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

8. Argh..

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...