## across Group Title \begin{align} x^1+y^1+z^1&=0\\ x^3+y^3+z^3&=3\\ x^5+y^5+z^5&=15\\\\ x^2+y^2+z^2&=\text{?} \end{align} 2 years ago 2 years ago

1. AsianDuck Group Title

(x+y+z)(x+y+z) = 0 -> x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 0 -> x^2 + y^2 + z^2 = -2(xy + xz + yz) Is this even in the right direction?

2. Mr.Math Group Title

$$6$$, I think.

3. AsianDuck Group Title

Working?

4. across Group Title

Mr. Math, how did you get that number?

5. Mr.Math Group Title

As Asian started $$x^2+y^2+z^2=-2(xy+xz+yz)$$ $$(a)$$ We have from the first and the second equation that $$x^3+y^3-(x+y)^3=3$$ $$\implies x^2y+xy^2=-1$$, and by symmetry we also have $$x^2z+zx^2=-1$$ and $$y^2z+yz^2=-1$$. We can also write $(x^2+y^2+z^2)(x^3+y^3+z^3)=3(x^2+y^2+z^2)$ $(x^5+y^5+z^5)+xy(xy^2+yx^2)+xz(xz^2+zx^2)+yz(yz^2+zy^2)$ $$\implies 3(x^2+y^2+z^2)=15-(xy+xz+yz)$$ $$(b)$$ By solving (a) and (b), it's easy to find $$x^2+y^2+z^2=6$$.

6. apoorvk Group Title

is MrM. right across?? cause i seem to be getting '4'??

7. apoorvk Group Title

i 'll recheck, i can make a lot of calculating errors. :

8. Mr.Math Group Title

I will recheck too! :)

9. Mr.Math Group Title

There's a missing equal sign there. It should be $(x^2+y^2+z^2)(x^3+y^3+z^3)=3(x^2+y^2+z^2)$ $$=(x^5+y^5+z^5)+xy(xy^2+yx^2)+xz(xz^2+zx^2)+yz(yz^2+zy^2)$$.

10. across Group Title

Mr. Math is right. :) This can be solved using Newton's identities*. Since $$x+y+z=0$$, they are the roots of $$t^3+at-b=0$$. Newton's identities say that$x^3+y^3+z^3=3b$$x^4+y^4+z^4=2a^2$and$x^2+y^2+z^2=-2a$This implies that $$b=1$$ and $$2a^2=15$$. It is then possible to solve for $$a$$ and find the value of $$x^2+y^2+z^2=-2a$$. Note that $$x^3+y^3+z^3=3b$$ was not necessary. Mr. Math's approach is a pedestrian way of achieving this, however. * http://en.wikipedia.org/wiki/Newton%27s_identities

11. Mr.Math Group Title

I don't know Newton's identities, I will read the link you gave. Nice problem by the way :)

12. apoorvk Group Title

whoa...i did a very stupid mistake towards the end :/ and now realise that the answer should have been somewhat symmetric (as in a multiple of 3 or some way), 4 would be pretty odd. Nice problem @across, had no clue about "newton's identities" will read that. and @Mr.Math , thanks for doing it before me, and posting the solution!!!!! :))

13. FoolForMath Group Title

For Newton Identities this (http://www.artofproblemsolving.com/Resources/Papers/PolynomialsAK.pdf) is a good reference.