ord_101(b) = 3 & b^3 \equiv 1 \mod 101, b is a positive integer, does such a b exist?

Given that \( \text{ord}_{101}(b) = 3 \) and \( b^3 \equiv 1 \pmod{101} \), we need to determine if there exists a positive integer \( b \) that satisfies these conditions.

### Step 1: Understanding the conditions

1. **Order of \( b \) modulo 101**: The order of \( b \) modulo 101, denoted \( \text{ord}_{101}(b) \), is the smallest positive integer \( k \) such that \( b^k \equiv 1 \pmod{101} \). Given \( \text{ord}_{101}(b) = 3 \), it means \( b^3 \equiv 1 \pmod{101} \) and 3 is the smallest such exponent.
   
2. **\( b^3 \equiv 1 \pmod{101} \)**: This condition is automatically satisfied if the order of \( b \) is 3.

### Step 2: Checking the existence of such \( b \)

Since \( \text{ord}_{101}(b) = 3 \), \( b \) must be a generator of a cyclic subgroup of order 3 in the multiplicative group \( \mathbb{Z}_{101}^* \). The order of the multiplicative group \( \mathbb{Z}_{101}^* \) is \( 100 \) (since \( 101 \) is prime, and \( \mathbb{Z}_{101}^* \) has \( 101-1 = 100 \) elements). 

For \( \text{ord}_{101}(b) = 3 \) to hold, \( 3 \) must divide \( 100 \). Indeed, \( 100 = 2^2 \times 5^2 \), so 3 divides 100. Hence, there is a subgroup of order 3.

### Step 3: Finding \( b \) that satisfies the condition

To find such a \( b \), we need a number such that:
\[
b^3 \equiv 1 \pmod{101}
\]
but \( b \neq 1 \), since \( b = 1 \) would have order 1, not 3.

### Step 4: Constructing the solution

The subgroup of order 3 consists of the elements satisfying \( x^3 \equiv 1 \pmod{101} \), i.e., the roots of the equation:
\[
x^3 - 1 \equiv 0 \pmod{101}
\]
This can be factored as:
\[
(x-1)(x^2+x+1) \equiv 0 \pmod{101}
\]
The roots of \( x^2 + x + 1 \equiv 0 \pmod{101} \) are the elements we seek.

### Step 5: Solving the quadratic equation

We solve \( x^2 + x + 1 \equiv 0 \pmod{101} \). The discriminant of this quadratic equation is \( 1^2 - 4 \times 1 \times 1 = -3 \). We need to check if -3 is a quadratic residue modulo 101.

\[
x = \frac{-1 \pm \sqrt{-3}}{2} \pmod{101}
\]

Calculating:
- Find \( \sqrt{-3} \mod 101 \), or equivalently \( \sqrt{98} \mod 101 \).
- Verify whether this quadratic has a solution modulo 101.

If the discriminant has no square root modulo 101, there are no elements of order 3, meaning no \( b \) exists with \( \text{ord}_{101}(b) = 3 \).

### Conclusion

Upon examining this, we find that \( b \) exists if the discriminant is a quadratic residue. If not, then such a \( b \) does not exist. However, since we know the subgroup structure, such a \( b \) must exist. Thus, we conclude that **such a \( b \) does exist**.

Would you like details on the quadratic residue computation or further elaboration on any steps?

---

**Related Questions:**
1. How do you find the order of an element modulo a prime number?
2. What is the significance of the discriminant in determining quadratic residues?
3. How are cyclic subgroups of a given order identified in \( \mathbb{Z}_p^* \)?
4. How can we compute square roots modulo a prime number efficiently?
5. What is the relationship between the roots of unity and cyclic subgroups in modular arithmetic?

**Tip:** Always check the order of elements in cyclic groups modulo a prime to ensure you're working with the correct generators.

ord_101(b) = 3 & b^3 ≡ 1 mod 101, b is a positive integer, does such a b exist?

Explore whether there exists a positive integer b such that ord_101(b) = 3 and b^3 ≡ 1 mod 101. This detailed solution delves into modular arithmetic, cyclic groups, and quadratic residues, suitable for undergraduate students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation