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I came across this observation while trying to answer this post using the Pell equation $x^2-2y^2=1$. Define,

$$P(m) = \frac{ (3+2\sqrt{2})^m+(3-2\sqrt{2})^m}{2}$$

$$Q(m) = \frac{ (3+2\sqrt{2})^m-(3-2\sqrt{2})^m}{2\sqrt{2}}$$

such that $P(m)^2-2Q(m)^2=1$. We have,

$$P(2^k) = \color{brown}{3, 17, 577, 665857, 886731088897,\dots}$$

$$Q(2^k) = \color{blue}{2, 12, 408, 470832, 627013566048,\dots}$$

Notice that,

$$\color{blue}{12} = 2^2\cdot\color{brown}3$$

$$\color{blue}{408} = 2^3\cdot\color{brown}{3\cdot17}$$

$$\color{blue}{470832} = 2^4\color{brown}{\cdot3\cdot17\cdot577}$$

and so on. In general, given the roots of $ax^2+bx+c=0$,

$$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

and let $D=b^2-4ac$.

Question: Is it true that,

$$\frac{2^{n}}{a}\prod_{k=0}^n \frac{x_1^{2^k}+x_2^{2^k}}{2}=\frac{x_1^{2^{n+1}}-x_2^{2^{n+1}}}{2\sqrt{D}}$$

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Note that $x^2-y^2=(x-y)(x+y)$ and by induction $$x^{2^{n+1}}-y^{2^{n+1}}=(x^{2^n}-y^{2^n})(x^{2^n}+y^{2^n})=(x-y)\prod_{k=0}^n(x^{2^k}+y^{2^k}).$$ For roots $x_1, x_2$ of a quadratic equation and $x_1\leq x_2$ we have $x_2-x_1 = \sqrt{D}/a$. This proves your identity.

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  • $\begingroup$ I didn't expect the reason would be so mundane. :) $\endgroup$ – Tito Piezas III Dec 25 '15 at 8:29

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