* x^4 = y + z
* y^4 = x + z
* z^4 = x + y

I'd truly appreciate a pointer or a suggestion, even a broad one.

2. ## Re: Three-equations twelfth-grade system

Originally Posted by AndrewTahoe
* x^4 = y + z
* y^4 = x + z
* z^4 = x + y

I'd truly appreciate a pointer or a suggestion, even a broad one.

Subtract the second equation from the first to get x^4- y^4= y- x. Factor the left side.

3. ## Re: Three-equations twelfth-grade system

Thank you very much!

Yes, I had tried it before. You get:

x^3 + xy^2 + x^2*y + y^3 + 1 = 0

What about now? I tried to obtain two lower-grade factors with Ruffini's rule, but I can't find a root for the expression.

4. ## Re: Three-equations twelfth-grade system

Don't forget that you divided by x- y to get that. So x= y is a separate solution. If x= y, what must z be?

Also, if you subtract the third equation from the first, x- z= 0 or $x^3+ zx^2+ xz^2+ z^4+ 1= 0$. Use that with your equation in x and y.

5. ## Re: Three-equations twelfth-grade system

Sorry, I don't get it. When you divide by x-y, you state that x-y != 0 -> x != y. How can x=y be a solution?

Also, can you be a bit more specific in the other approach?

Thanks a lot for your attention!

6. ## Re: Three-equations twelfth-grade system

Originally Posted by HallsofIvy
Subtract the second equation from the first to get x^4- y^4= y- x. Factor the left side.
When you divide by y - x you do not get 0 on the RHS, you get 1. I think this is causing some of your confusion.

Originally Posted by AndrewTahoe
Sorry, I don't get it. When you divide by x-y, you state that x-y != 0 -> x != y. How can x=y be a solution?

Also, can you be a bit more specific in the other approach?

Thanks a lot for your attention!
x -y is a solution, but not the only solution.

Let's try factoring instead of division:

First let's put things in an easier format:
$\displaystyle x^4 - y^4 = y - x$

$\displaystyle (x^2 - y^2)(x^2 + y^2) = y - x$

$\displaystyle (x - y)(x + y)(x^2 + y^2) = y - x$

Note the change I'm making on the RHS:
$\displaystyle (x - y)(x + y)(x^2 + y^2) = -(x - y)$

$\displaystyle (x - y)(x + y)(x^2 + y^2) + (x - y) = 0$

Now factor:
$\displaystyle (x - y) \left [ (x + y)(x^2 + y^2) + 1 \right ] = 0$

So we know that either
$\displaystyle x - y = 0$

or
$\displaystyle (x + y)(x^2 + y^2) + 1 = 0$

The first equation gives y = x. The other equation? I'm not telling. Give it a try.

-Dan

7. ## Re: Three-equations twelfth-grade system

Thanks a lot! I'll give it a try.