- #1

- 1,426

- 3

Hi,

I just came across this problem some days ago, and I am tearing all my hair to solve it. I am nearly bald now...

The problem asks to solve for 2 unknowns x, and y:

[tex]\left\{ \begin{array}{ccc} x ^ 2 + y ^ 2 & = & 1 \\ x ^ {2006} + y ^ {2006} & = & 1 \end{array} \right.[/tex]

I know there should be two sets of roots:

[tex]\left\{ \begin{array}{ccc} x & = & 1 \\ y & = & 0 \end{array} \right. \quad \mbox{or} \quad \left\{ \begin{array}{ccc} x & = & 0 \\ y & = & 1 \end{array} \right.[/tex]

So, I rearranged the first equation to: x

I think there should be another easy way to solve the problem. Can anyone just give me a small crack at it?

Thanks in advance.

I just came across this problem some days ago, and I am tearing all my hair to solve it. I am nearly bald now...

The problem asks to solve for 2 unknowns x, and y:

[tex]\left\{ \begin{array}{ccc} x ^ 2 + y ^ 2 & = & 1 \\ x ^ {2006} + y ^ {2006} & = & 1 \end{array} \right.[/tex]

I know there should be two sets of roots:

[tex]\left\{ \begin{array}{ccc} x & = & 1 \\ y & = & 0 \end{array} \right. \quad \mbox{or} \quad \left\{ \begin{array}{ccc} x & = & 0 \\ y & = & 1 \end{array} \right.[/tex]

So, I rearranged the first equation to: x

^{2}= 1 - y^{2}, and substituted this to the second equation, leaving me an equation of power 2004. :yuck: :yuck: I must be missing something really big. :grumpy:I think there should be another easy way to solve the problem. Can anyone just give me a small crack at it?

Thanks in advance.

Last edited: