Points on Curve y² = x³ + 1 over ℝ_p

The problem asks about the size of the set \(E = \{(x, y) \in \mathbb{F}_p \times \mathbb{F}_p \mid y^2 = x^3 + 1\}\), where \(p\) is an odd prime and \(p \equiv 2 \pmod{3}\). \(\mathbb{F}_p\) is the finite field with \(p\) elements.

Understanding the Condition p ≡ 2 (mod 3)

The condition \(p \equiv 2 \pmod{3}\) for an odd prime \(p\) means that \(p-1\) is not divisible by 3. This is because if \(p \equiv 2 \pmod{3}\), then \(p-1 \equiv 1 \pmod{3}\). This condition is crucial because it tells us about the mapping \(x \mapsto x^3\) in the field \(\mathbb{F}_p\). The map \(x \mapsto x^k\) is a permutation (a bijection) of \(\mathbb{F}_p\) if and only if \(\gcd(k, p-1) = 1\). In our case, \(k=3\), so the map \(x \mapsto x^3\) is a bijection on \(\mathbb{F}_p\) if and only if \(\gcd(3, p-1) = 1\). Since \(p \equiv 2 \pmod{3}\), \(p-1 \equiv 1 \pmod{3}\), so \(\gcd(3, p-1) = 1\). Thus, the map \(x \mapsto x^3\) is a bijection on \(\mathbb{F}_p\).

Calculating the Size of Set E

The set \(E\) consists of pairs \((x, y) \in \mathbb{F}_p \times \mathbb{F}_p\) satisfying the equation \(y^2 = x^3 + 1\). To find the number of elements in \(E\), we need to sum, for each \(x \in \mathbb{F}_p\), the number of solutions \(y \in \mathbb{F}_p\) to the equation \(y^2 = x^3 + 1\). Let \(a = x^3 + 1\). For a fixed \(x\), we are looking for the number of solutions \(y\) to \(y^2 = a\). The number of solutions \(y\) in \(\mathbb{F}_p\) for \(y^2 = a\) is given by \(1 + \left(\frac{a}{p}\right)\), where \(\left(\frac{a}{p}\right)\) is the Legendre symbol. The total number of points in \(E\) is the sum over all possible values of \(x\): $|E| = \sum_{x \in \mathbb{F}_p} (1 + \left(\frac{x^3+1}{p}\right))$ We can split the sum: $|E| = \sum_{x \in \mathbb{F}_p} 1 + \sum_{x \in \mathbb{F}_p} \left(\frac{x^3+1}{p}\right)$ The first sum is simply \(p\): $\sum_{x \in \mathbb{F}_p} 1 = p$ Now consider the second sum: \(\sum_{x \in \mathbb{F}_p} \left(\frac{x^3+1}{p}\right)\). Since \(x \mapsto x^3\) is a bijection on \(\mathbb{F}_p\), the map \(x \mapsto x^3+1\) is also a bijection on \(\mathbb{F}_p\). This means that as \(x\) ranges over all elements of \(\mathbb{F}_p\), the expression \(x^3+1\) also ranges over all elements of \(\mathbb{F}_p\), taking on each value exactly once. Let \(a = x^3+1\). The sum becomes: $\sum_{a \in \mathbb{F}_p} \left(\frac{a}{p}\right)$ The sum of the Legendre symbol over all elements of \(\mathbb{F}_p\) is known to be 0: $\sum_{a \in \mathbb{F}_p} \left(\frac{a}{p}\right) = \left(\frac{0}{p}\right) + \sum_{a \in \mathbb{F}_p^*} \left(\frac{a}{p}\right) = 0 + 0 = 0$ (The sum over non-zero elements is zero because there are an equal number of quadratic residues and non-residues). Therefore, the total number of points in \(E\) is: $|E| = p + 0 = p$ So, the size of the set \(E\) is exactly \(p\).

Evaluating the Options

Based on the fact that \(|E| = p\), where \(p\) is an odd prime (\(p \ge 3\)), let's check the options:

1. E has atleast two elements: Since \(p\) is an odd prime, \(p \ge 3\). So \(|E|=p \ge 3\). This means \(|E|\) is indeed at least two. This statement is true.
2. E has atmost 2p elements: Since \(|E|=p\), and for any positive prime \(p\), \(p \le 2p\), this statement is true. \(|E| \le 2p\).
3. E can have p² elements: We found \(|E|=p\). Since \(p \ge 3\), \(p < p^2\). The set \(E\) contains exactly \(p\) elements, not \(p^2\). This statement is false.
4. E has atleast 2p elements: We found \(|E|=p\). Since \(p \ge 3\), \(p < 2p\). \(|E|\) is not at least \(2p\). This statement is false.

The true statements are option 1 and option 2.