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Find $a$ and $b$

(1) $a[(1+\sqrt{5})/2] + b[(1-\sqrt{5})/2] = 1$

(2) $a[(1+\sqrt{5})/2]^2 + b[(1-\sqrt{5})/2]^2 = 2$

Is there a smart method to solve this Simultaneous Linear Equations ?

Thanks you guys!

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Like for any linear system, Gaussian pivoting will do. Or, since it's a 2x2 system, Cramer's rule is fine. Also, don't forget there are nice relations with golden ratio $\phi=\frac{1+\sqrt{5}}2$. For instance, $\phi^2=\phi+1$, or $-\frac 1{\phi}=\frac{1-\sqrt{5}}2$ This will help simplify your computation. – Jean-Claude Arbaut Dec 19 '13 at 11:24
Also, are you trying to find coefficients when solving $u_{n+1}=u_{n}+u_{n-1}$ ? If it's so, you can also use the case $n=0$ in $u_n=a\phi^n+b(-1/\phi)^n$, not only $n=1$ and $n=2$ as you seem to do here. – Jean-Claude Arbaut Dec 19 '13 at 11:31
up vote 1 down vote accepted

Write $\varphi=\dfrac{1+\sqrt{5}}{2}$, so that $$ \frac{1-\sqrt{5}}{2}=1-\varphi $$ and your system becomes \begin{cases} \varphi a+(1-\varphi)b=1\\ \varphi^2 a+(1-\varphi)^2 b=2 \end{cases}

Multiply the first equation by $1-\phi$ and subtract; now $$ a(\varphi(1-\varphi)-\varphi^2)=1-\varphi-2 $$ shouldn't be difficult to solve.

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Let $\alpha=\frac{1+\sqrt5}{2}, \beta=\frac{1-\sqrt5}{2}$. Since we have ${\alpha}^2=1+\alpha, {\beta}^2=1+\beta,$ we have $$a(1+\alpha)+b(1+\beta)=2.$$ Then, you can get $a+b=1$. And?

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