Question

\[\begin{align}
x^1+y^1+z^1&=0\\
x^3+y^3+z^3&=3\\
x^5+y^5+z^5&=15\\\\
x^2+y^2+z^2&=\text{?}
\end{align}\]

Answer 1

(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?

Answer 2

\(6\), I think.

Answer 3

Working?

Answer 4

Mr. Math, how did you get that number?

Answer 5

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 \(x^2+y^2+z^2=6\).

Answer 6

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

Answer 7

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

Answer 8

I will recheck too! :)

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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!!!!! :))

Answer 13

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