Let p be a prime such that $p\equiv 1(\mathrm{mod}3)$. Let c be a cubic residue $(\mathrm{mod}p)$ such that $c^{\frac{p-1}{3}}\equiv 1(\mathrm{mod}p)$. In this paper, we present a refinement of Müller's algorithm for computing a cube root of c [11], which also improves Williams' [14], [15] Cipolla-Lehmer type algorithms. Under the assumption that a suitable irreducible polynomial of degree 3 is given, Müller gave a cube root algorithm which requires $8.5\log p$ modular multiplications. Our algorithm requires only $7.5\log p$ modular multiplications and is based on the recurrence relations arising from the irreducible polynomial $h(x)=x^3+c{t}^{3}x-c{t}^3$ for some integer t.

中文翻译：

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Müller立方根算法的改进

令p为质数$p\equiv \mathrm{1个}（\mathrm{模}3）$。令c为立方残基$（\mathrm{模}p）$ 这样 $C^{\frac{p-\mathrm{1个}}{3}}\equiv \mathrm{1个}（\mathrm{模}p）$。在本文中，我们对计算c的立方根的Müller算法进行了改进[11]，这也改进了Williams [14]，[15] Cipolla-Lehmer型算法。假设给出了一个合适的3级不可约多项式，Müller给出了一个立方根算法，该算法要求$8.5\mathrm{日志}p$模乘法。我们的算法只需要$7.5\mathrm{日志}p$ 模乘法，并且基于不可约多项式产生的递归关系 $H（X）=X^3+C{Ť}^{3}X-C{Ť}^3$对于一些整数t。