Can $S_5$ be written as a multiple of $S_3$ and $S_2$?

[anemone]

Let \(\displaystyle S_n=x^n+y^n+z^n\). If \(\displaystyle S_1=0\), prove that \(\displaystyle \frac{S_5}{5}=\frac{S_3}{3}\cdot\frac{S_2}{2}\).

 

[MarkFL]

Re: Prove S_5/5= S_3/3. S_2/2

My solution:

Spoiler

If we view $S_n$ as a recursive algorithm, we see that it must come from the characteristic equation:

\(\displaystyle (r-x)(r-y)(r-z)=0\)

\(\displaystyle r^3-(x+y+z)r^2+(xy+xz+yz)r-xyz=0\)

Since $x+y+z=S_1=0$, we obtain the following recursion:

\(\displaystyle S_{n+3}=-(xy+xz+yz)S_{n+1}+xyzS_{n}\)

Now, observing we may write:

\(\displaystyle (x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\)

\(\displaystyle 0=S_2+2(xy+xz+yz)\)

\(\displaystyle -(xy+xz+yz)=\frac{S_2}{2}\)

Also, we find:

\(\displaystyle (x+y+z)^3=-2\left(x^3+y^3+z^3 \right)+3\left(x^2+y^2+z^2 \right)(x+y+z)+6xyz\)

\(\displaystyle 0=-2S_3+6xyz\)

\(\displaystyle xyz=\frac{S^3}{3}\)

And so our recursion may be written:

\(\displaystyle S_{n+3}=\frac{S_2}{2}S_{n+1}+\frac{S_3}{3}S_{n}\)

Letting $n=2$, we then find:

\(\displaystyle S_{5}=\frac{S_2}{2}S_{3}+\frac{S_3}{3}S_{2}\)

\(\displaystyle S_{5}=\frac{5}{6}S_2S_{3}\)

\(\displaystyle \frac{S_5}{5}=\frac{S_3}{3}\cdot\frac{S_2}{2}\)

Shown as desired.

 

[anemone]

Re: Prove S_5/5= S_3/3. S_2/2

My solution:

Spoiler

If we view $S_n$ as a recursive algorithm, we see that it must come from the characteristic equation:

\(\displaystyle (r-x)(r-y)(r-z)=0\)

\(\displaystyle r^3-(x+y+z)r^2+(xy+xz+yz)r-xyz=0\)

Since $x+y+z=S_1=0$, we obtain the following recursion:

\(\displaystyle S_{n+3}=-(xy+xz+yz)S_{n+1}+xyzS_{n}\)

Now, observing we may write:

\(\displaystyle (x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\)

\(\displaystyle 0=S_2+2(xy+xz+yz)\)

\(\displaystyle -(xy+xz+yz)=\frac{S_2}{2}\)

Also, we find:

\(\displaystyle (x+y+z)^3=-2\left(x^3+y^3+z^3 \right)+3\left(x^2+y^2+z^2 \right)(x+y+z)+6xyz\)

\(\displaystyle 0=-2S_3+6xyz\)

\(\displaystyle xyz=\frac{S^3}{3}\)

And so our recursion may be written:

\(\displaystyle S_{n+3}=\frac{S_2}{2}S_{n+1}+\frac{S_3}{3}S_{n}\)

Letting $n=2$, we then find:

\(\displaystyle S_{5}=\frac{S_2}{2}S_{3}+\frac{S_3}{3}S_{2}\)

\(\displaystyle S_{5}=\frac{5}{6}S_2S_{3}\)

\(\displaystyle \frac{S_5}{5}=\frac{S_3}{3}\cdot\frac{S_2}{2}\)

Shown as desired.

Thanks for participating, MarkFL! And your method is neat and elegant!

We are given $S_n=x^n+y^n+z^n$ and $S_1=0$ which implies $x+y+z=0$.

From this given information we then have

1.	2.
$\small(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)$ $0=x^2+y^2+z^2+2(xy+yz+xz)$ $x^2+y^2+z^2=-2(xy+yz+xz)$ $x^2+y^2+z^2=-2(xy+z(x+y))$ $x^2+y^2+z^2=-2xy-2z(-z)$ $x^2+y^2-z^2=-2xy$ $xy=-\left(\dfrac{x^2+y^2-z^2}{2} \right)$	$\small(x+y+z)^3=x^3+y^3+z^3+3(xy(x+y)+yz(y+z)+xz(x+z))+6xyz$ $0=x^3+y^3+z^3+3(xy(-z)+yz(-x)+xz(-y))+6xyz$ $0=x^3+y^3+z^3+3(xy(-z)+yz(-x)+xz(-y))+6xyz$ $x^3+y^3+z^3=3xyz$

We're asked to prove $\dfrac{S_5}{5}=\dfrac{S_3}{3}\cdot\dfrac{S_2}{2}$.

We see that

$S_5=x^5+y^5+z^5$

$\;\;\;\;\;=x^5+y^5+(-x-y)^5$

$\;\;\;\;\;=x^5+y^5-(x+y)^5$

$\;\;\;\;\;=x^5+y^5-(x^5+5x^4y+10x^3y^3+10x^2y^3+5xy^4+y^5)$

$\;\;\;\;\;=-(5x^4y+10x^3y^3+10x^2y^3+5xy^4)$

$\;\;\;\;\;=-(5xy(x^3+y^3)+10x^2y^2(x+y))$

$\;\;\;\;\;=-5xy((x^3+y^3)+2xy(x+y))$

$\;\;\;\;\;=-5xy((x+y)(x^2-xy+y^2)+2xy(x+y))$

$\;\;\;\;\;=-5xy((x+y)(x^2-xy+y^2+2xy))$

$\;\;\;\;\;=-5xy((x+y)(x^2+xy+y^2))$

$\;\;\;\;\;=-5\left(\dfrac{(x^3+y^3+z^3}{3z} \right)(-z)(x^2-\left(\dfrac{x^2+y^2-z^2}{2} \right)+y^2))$

$\;\;\;\;\;=5\left(\dfrac{x^3+y^3+z^3}{3} \right)\left(\dfrac{x^2+y^2+z^2}{2} \right)$

and therefore we obtain

$\dfrac{S_5}{5}=\dfrac{S_3}{3}\cdot\dfrac{S_2}{2}$ and we're done.(Emo)