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?

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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.