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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:
$$\alpha(n) \equiv 1 \pmod x\quad \text{and}\quad \beta(n) \equiv 0 \pmod x$$
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$n$：
$$\alpha(n) \equiv 1 \pmod x\quad \text{and}\quad \beta(n) \equiv 0 \pmod x$$
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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