Generalized Shanks Algorithm for Quadratic Extension Field

This section explains how to compute square roots in the quadratic extension field $\mathbb{F}_{p^2}$, particularly when the base field $\mathbb{F}_p$ satisfies $p \equiv 3 \pmod 4$. The method is based on Algorithm 9 from ePrint 2012/685.

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Goal

Given $a \in \mathbb{F}_{p^2}$, find $x \in \mathbb{F}_{p^2}$ such that:

$$x^2 = a$$

where $x = x_0 + x_1 \cdot i$ and $i^2 = -1$.

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Algorithm

The algorithm proceeds as follows:

1. Compute $a_1 = a^{(p - 3)/4}$

2. Compute $\alpha = a_1^2 \cdot a = a^{(p-1)/2}$

3. Check whether $\alpha^{p + 1} = a^{(p^2-1)/2} = -1$. If so, then $a$ is a non-residue, and no square root exists.

4. Compute $\beta = a_1 \cdot a = a^{(p+1)/4}$

5. If $\alpha = -1$, return $x = \beta \cdot i = a^{(p+1)/4} \cdot i$

6. Otherwise:

   • Compute $b = (\alpha + 1)^{(p - 1)/2} = \left(1 + a^{(p-1)/2} \right)^{(p - 1)/2}$

   • Return $x = b \cdot \beta = \left( 1 + a^{(p-1)/2} \right)^{(p - 1)/2} \cdot a^{(p+1)/4}$

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Intuition Behind the Algorithm

To compute a square root of $a \in \mathbb{F}_{p^2}$, we can find an element $b \in \mathbb{F}_{p^2}$ and an odd integer $s$ such that:

$$b^2 \cdot a^s = 1$$

In this case, a square root of $a$ is given by:

$$x = b \cdot a^{(s + 1)/2}$$

because:

$$x^2 = \left( b \cdot a^{(s+1)/2} \right)^2 = b^2 \cdot a^{s+1} = (b^2 \cdot a^s) \cdot a = a$$

To find such $b$ and $s$, we set:

• $s = \frac{p - 1}{2}$

• $b = \left( 1 + a^{(p-1)/2} \right)^{(p - 1)/2}$

Case 1: $b \ne 0$

We can verify:

$$\begin{aligned} b^2 \cdot a^s &= \left(1 + a^{(p-1)/2}\right)^{(p-1)} \cdot a^{(p-1)/2} \\ &= \left(1 + a^{(p-1)/2}\right)^p \cdot \left(1 + a^{(p-1)/2}\right)^{-1} \cdot a^{(p-1)/2} \\ &= \left(1 + a^{p(p-1)/2}\right) \cdot \left(1 + a^{(p-1)/2}\right)^{-1} \cdot a^{(p-1)/2} \\ &= \left(a^{(p-1)/2} + 1\right) \cdot \left(1 + a^{(p-1)/2}\right)^{-1} = 1 \end{aligned}$$

So, $x = b \cdot a^{(p+1)/4}$ is a valid square root of $a$.

Explanations

This step works due to the properties of the Frobenius endomorphism, which states that:

$$\left(1 + a^{(p-1)/2}\right)^p = 1 + a^{p \cdot (p-1)/2}$$

since exponentiation by $p$ acts as an automorphism over $\mathbb{F}_{p^2}$, and in particular, fixes elements in the base field $\mathbb{F}_p$.

In addition, every nonzero element $a \in \mathbb{F}_{p^2}$ satisfies:

$$a^{p^2 - 1} = 1$$

because the multiplicative group $\mathbb{F}_{p^2}^*$ has order $p^2 - 1$. This ensures that all powers used in the algorithm, including $a^{(p^2 - 1)/2}$, are well-defined and lie within the group.

Case 2: $b = 0$

Then, by definition of $b$, we must have:

$$1 + a^{(p-1)/2} = 0 \quad \Rightarrow \quad a^{(p-1)/2} = -1$$

In this case, define $x = a^{(p+1)/4} \cdot i$, where $i^2 = -1$. Then:

$$\begin{aligned} x^2 &= \left( a^{(p+1)/4} \cdot i \right)^2 = a^{(p+1)/2} \cdot i^2 \\ &= a^{(p+1)/2} \cdot (-1) = a \cdot a^{(p-1)/2} \cdot (-1) = a \cdot (-1) \cdot (-1) = a \end{aligned}$$

Thus, $x$ is again a valid square root of $a$.

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Final Formula

We can summarize the square root of $a \in \mathbb{F}_{p^2}$, for $p \equiv 3 \pmod 4$, as:

$$x = \begin{cases} a^{(p+1)/4} \cdot i & \text{if } a^{(p-1)/2} = -1 \\ \left( 1 + a^{(p-1)/2} \right)^{(p - 1)/2} \cdot a^{(p+1)/4} & \text{otherwise} \end{cases}$$

This elegant formula generalizes the classic Shanks algorithm from $\mathbb{F}_p$ to quadratic extensions $\mathbb{F}_{p^2}$, and avoids iterative procedures entirely.

Written by ryan Kim from A41