n乗された2つの数の和を基本対称式で表す
\(\alpha, \beta\)を解とする2次方程式を考える．
$$\begin{eqnarray}
\left(x-\alpha\right)\left(x-\beta\right)&=&0
\\x^2-\left(\alpha+\beta\right)x+\alpha\beta&=&0
\end{eqnarray}$$
\(\alpha,\beta\)は式の解なので，\(x\)に代入しても式は成り立つ．代入した2つの式を得る．
$$
\begin{eqnarray}
\left\{
\begin{array}{l}
\alpha^2-\left(\alpha+\beta\right)\alpha+\alpha\beta&=&0
\\\beta^2-\left(\alpha+\beta\right)\beta+\alpha\beta&=&0
\end{array}
\right.
\end{eqnarray}
$$
第一式には\(\alpha^{n-2}\)，第二式には\(\beta^{n-2}\)を両辺に掛ける．
$$
\begin{eqnarray}
\left\{
\begin{array}{l}
\alpha^{n-2} \cdot\left\{\alpha^2-\left(\alpha+\beta\right)\alpha+\alpha\beta\right\}&=&\alpha^{n-2}\cdot 0
\\\beta^{n-2}\cdot\left\{\beta^2 -\left(\alpha+\beta\right)\beta +\alpha\beta\right\}&=&\beta^{n-2} \cdot 0
\end{array}
\right.
\\
\\\left\{
\begin{array}{l}
\alpha^{n}-\left(\alpha+\beta\right)\alpha^{n-1}+\alpha\beta\alpha^{n-2}&=&0
\\\beta^{n}-\left(\alpha+\beta\right)\beta^{n-1}+\alpha\beta\beta^{n-2}&=&0
\end{array}
\right.
\end{eqnarray}
$$
両式を足し合わせ，n乗の和について解くことで．n乗の和を基本対称式で求める式を得る．
$$\begin{eqnarray}
&\alpha^{n}&-\left(\alpha+\beta\right)\alpha^{n-1}&+\alpha\beta\alpha^{n-2}&=&0
\\+)&\beta^{n}&-\left(\alpha+\beta\right)\beta^{n-1}&+\alpha\beta\beta^{n-2}&=&0
\\\hline
&\alpha^{n}+\beta^{n}&-\left(\alpha+\beta\right)\left(\alpha^{n-1}+\beta^{n-1}\right)&+\alpha\beta\left(\alpha^{n-2}+\beta^{n-2}\right)&=&0
\end{eqnarray}$$
$$\begin{eqnarray}
\alpha^{n}+\beta^{n}&=&\left(\alpha+\beta\right)\left(\alpha^{n-1}+\beta^{n-1}\right)-\alpha\beta\left(\alpha^{n-2}+\beta^{n-2}\right)
\end{eqnarray}$$
\(\alpha, \beta\)は任意の数でよいのでそれを\(x, y\)とすれば以下の式を得る．
$$\begin{eqnarray}
x^{n}+y^{n}&=&\left(x+y\right)\left(x^{n-1}+y^{n-1}\right)-xy\left(x^{n-2}+y^{n-2}\right)
\end{eqnarray}$$
問
original:
https://www.youtube.com/watch?v=KPT862KhxRM
$$\begin{eqnarray}
\left(\frac{1+\sqrt{13}}{2}\right)^7+\left(\frac{1-\sqrt{13}}{2}\right)^7の値を求めよ．
\end{eqnarray}$$
第一項,第二項の括弧内をそれぞれ\(\alpha, \beta\)とする．
$$\begin{eqnarray}
\alpha^7+\beta^7\;\cdots\;\alpha=\frac{1+\sqrt{13}}{2},\;\beta=\frac{1-\sqrt{13}}{2}
\end{eqnarray}$$
n乗の和を基本対称式で求める上記式を用い，2乗から7乗まで順次求めていく．
$$\begin{eqnarray}
\alpha^2+\beta^2&=&\left(\alpha+\beta\right)\left(\alpha^{2-1}+\beta^{2-1}\right)-\alpha\beta\left(\alpha^{2-2}+\beta^{2-2}\right)
\\&=&\left(\alpha+\beta\right)\left(\alpha+\beta\right)-\alpha\beta\left(\alpha^0+\beta^0\right)
\\&=&1\cdot1-(-3)2=1+6=7\;\cdots\;\alpha+\beta=1,\;\alpha\beta=-3,\;\alpha^0+\beta^0=1+1=2
\\
\\\alpha^3+\beta^3&=&\left(\alpha+\beta\right)\left(\alpha^{3-1}+\beta^{3-1}\right)-\alpha\beta\left(\alpha^{3-2}+\beta^{3-2}\right)
\\&=&\left(\alpha+\beta\right)\left(\alpha^{2}+\beta^{2}\right)-\alpha\beta\left(\alpha+\beta\right)
\\&=&1\cdot7-(-3)1=7+3=10
\\
\\\alpha^4+\beta^4&=&\left(\alpha+\beta\right)\left(\alpha^{4-1}+\beta^{4-1}\right)-\alpha\beta\left(\alpha^{4-2}+\beta^{4-2}\right)
\\&=&\left(\alpha+\beta\right)\left(\alpha^{3}+\beta^{3}\right)-\alpha\beta\left(\alpha^{2}+\beta^{2}\right)
\\&=&1\cdot10-(-3)7=10+21=31
\\
\\\alpha^5+\beta^5&=&\left(\alpha+\beta\right)\left(\alpha^{5-1}+\beta^{5-1}\right)-\alpha\beta\left(\alpha^{5-2}+\beta^{5-2}\right)
\\&=&\left(\alpha+\beta\right)\left(\alpha^{4}+\beta^{4}\right)-\alpha\beta\left(\alpha^{3}+\beta^{3}\right)
\\&=&1\cdot31-(-3)10=31+30=61
\\
\\\alpha^6+\beta^6&=&\left(\alpha+\beta\right)\left(\alpha^{6-1}+\beta^{6-1}\right)-\alpha\beta\left(\alpha^{6-2}+\beta^{6-2}\right)
\\&=&\left(\alpha+\beta\right)\left(\alpha^{5}+\beta^{5}\right)-\alpha\beta\left(\alpha^{4}+\beta^{4}\right)
\\&=&1\cdot61-(-3)31=61+93=154
\\
\\\alpha^7+\beta^7&=&\left(\alpha+\beta\right)\left(\alpha^{7-1}+\beta^{7-1}\right)-\alpha\beta\left(\alpha^{7-2}+\beta^{7-2}\right)
\\&=&\left(\alpha+\beta\right)\left(\alpha^{6}+\beta^{6}\right)-\alpha\beta\left(\alpha^{5}+\beta^{5}\right)
\\&=&1\cdot154-(-3)61=154+183=337
\end{eqnarray}$$
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