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- Feb 14, 2012
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Let \(\displaystyle S_n=x^n+y^n+z^n\). If \(\displaystyle S_1=0\), prove that \(\displaystyle \frac{S_5}{5}=\frac{S_3}{3}\cdot\frac{S_2}{2}\).
Thanks for participating, MarkFL! And your method is neat and elegant!My solution:
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.
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$\small(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)$ $0=x^2+y^2+z^2+2(xy+yz+xz)$ $x^2+y^2+z^2=-2(xy+yz+xz)$ $x^2+y^2+z^2=-2(xy+z(x+y))$ $x^2+y^2+z^2=-2xy-2z(-z)$ $x^2+y^2-z^2=-2xy$ $xy=-\left(\dfrac{x^2+y^2-z^2}{2} \right)$ | $\small(x+y+z)^3=x^3+y^3+z^3+3(xy(x+y)+yz(y+z)+xz(x+z))+6xyz$ $0=x^3+y^3+z^3+3(xy(-z)+yz(-x)+xz(-y))+6xyz$ $0=x^3+y^3+z^3+3(xy(-z)+yz(-x)+xz(-y))+6xyz$ $x^3+y^3+z^3=3xyz$ |