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Suppose $x^2+y^2+z^2=1$ and $x+y+z=0$.

Is it then true that $x^4+y^4+z^4=\frac{1}{2}$?

I have a hunch that it's true (all the numbers I've tried have confirmed this) - but I don't know how to show this.

My thought was to write $x = - y - z$, and then plug that into the other expression so that $(-y-z)^2+y^2+z^2=1$, but this doesn't help me out here.

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    That should help some. You have $(y+z)^2+y^2+z^2=2 (yz+y^2+z^2) =1$ which should allow you to find to solutions to z in terms of y. Test them both out and it seems reasonable they will yield the result. Probably. I haven'haven't done it but it seems very plausible. – 2017-01-25

3 Answers 3

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$$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+z^2x^2)=1-2(x^2y^2+y^2z^2+z^2x^2)$$

Now $x^2y^2+y^2z^2+z^2x^2=(xy+yz+zx)^2-2xyz(x+y+z)=(xy+yz+zx)^2-0$

Again, $2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)=0-1$

3

We have $2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)=0-1$

So, $x,y,z$ are the roots of $t^3-\dfrac t2-c=0$ where $c=xyz$

$\implies t^4-\dfrac{t^2}2-ct=0$

Put $t=x,y,z$ and add to find $x^4+y^4+z^4=\dfrac{x^2+y^2+z^2}2+c(x+y+z)=\dfrac{x^2+y^2+z^2}2+c\cdot0$

2

From the conditions we have $xy+xz+yz=\frac{1}{2}$.

Hence, $x^4+y^4+z^4=$ $$=(x+y+z)^4-4(x+y+z)^2(xy+xz+yz)+2(xy+xz+yz)^2+4(x+y+z)xyz=$$ $$=2\left(\frac{1}{2}\right)^2=\frac{1}{2}$$