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

Hello, Ununuquantium!

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

Originally Posted by Ununuquantium

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

This problem is still unsolved, maybe someone 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?

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

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

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...