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Problem 752

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Powers of $1+\sqrt{7}$

When $(1+\sqrt{7})$ is raised to an integral power, $n$, we always get a number of the form $(a+b\sqrt{7})$.
We write $(1+\sqrt{7}{)}^n=\alpha (n)+\beta (n)\sqrt{7}$.

For a given number $x$ we define $g(x)$ to be the smallest positive integer $n$ such that:
$$\alpha (n)\equiv 1(\mathrm{mod}x)\text{and}\beta (n)\equiv 0(\mathrm{mod}x)$$
and $g(x)=0$ if there is no such value of $n$. For example, $g(3)=0$, $g(5)=12$.

Further define
$$G(N)={\displaystyle \sum _{x=2}^{N}g(x)}$$
You are given $G({10}^2)=28891$ and $G({10}^3)=13131583$.

Find $G({10}^6)$.

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$1+\sqrt{7}$的幂

对于任意正整数$n$，$(1+\sqrt{7})$的$n$次幂总能写成$(a+b\sqrt{7})$的形式。
我们记$(1+\sqrt{7}{)}^n=\alpha (n)+\beta (n)\sqrt{7}$。

给定正整数$x$，我们记$g(x)$为满足下列条件的最小正整数$n$：
$$\alpha (n)\equiv 1(\mathrm{mod}x)\text{and}\beta (n)\equiv 0(\mathrm{mod}x)$$
若不存在这样的$n$，则$g(x)=0$。例如，$g(3)=0$，$g(5)=12$。

进一步记
$$G(N)={\displaystyle \sum _{x=2}^{N}g(x)}$$
已知$G({10}^2)=28891$，$G({10}^3)=13131583$。

求$G({10}^6)$。

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