The problem involves finding the next two prime numbers $p$ such that all quadratic nonresidues modulo $p$ are also primitive roots modulo $p$. We are given that the primes $3$, $5$, and $17$ have this property.

Understanding the Problem:

1. Quadratic Residue: An integer $a$ is a quadratic residue modulo $p$ if there exists an integer $x$ such that $x^2 \equiv a \pmod{p}$. Otherwise, $a$ is a quadratic nonresidue modulo $p$.
2. Primitive Root: An integer $g$ is a primitive root modulo $p$ if the order of $g$ modulo $p$ is $p-1$. That is, the powers $g^1, g^2, \dots, g^{p-1}$ modulo $p$ produce all the nonzero residues modulo $p$.

Given Primes:

• For $p = 3$, the quadratic nonresidue is $2$, and it is also a primitive root.
• For $p = 5$, the quadratic nonresidues are $2$ and $3$, both of which are primitive roots.
• For $p = 17$, the quadratic nonresidues are $3, 5, 6, 7, 10, 11, 12, 14, 15$, and all of these are primitive roots.

Approach:

We need to find the next two primes $p$ such that all quadratic nonresidues modulo $p$ are primitive roots modulo $p$.

Steps:

1. Identify Quadratic Residues and Nonresidues:

   • For each prime $p$, list the quadratic residues and the nonresidues.
   • Check if every quadratic nonresidue is a primitive root modulo $p$.

2. Calculate the Primitive Roots:

   • For each candidate prime $p$, calculate the order of each quadratic nonresidue modulo $p$.

3. Verification:

   • Verify that all quadratic nonresidues are indeed primitive roots for the smallest primes.

Let's check the primes one by one:

Prime $p = 7$:

• Quadratic residues: $1, 2, 4$
• Quadratic nonresidues: $3, 5, 6$
• Primitive roots modulo 7: $3$ and $5$.

However, $6$ is not a primitive root because $6^3 \equiv 1 \pmod{7}$. So, 7 does not satisfy the condition.

Prime $p = 11$:

• Quadratic residues: $1, 3, 4, 5, 9$
• Quadratic nonresidues: $2, 6, 7, 8, 10$
• Checking:
  • $2^1 = 2$, $2^2 = 4$, $2^3 = 8$, $2^4 = 5$, $2^5 = 10$, $2^6 = 9$, $2^7 = 7$, $2^8 = 3$, $2^9 = 6$, $2^{10} = 1$ $\Rightarrow$ $2$ is a primitive root.
  • Similarly, check for $6, 7, 8, 10$. (All are primitive roots.)
• Conclusion: $p = 11$ is a valid prime.

Prime $p = 13$:

• Quadratic residues: $1, 3, 4, 9, 10, 12$
• Quadratic nonresidues: $2, 5, 6, 7, 8, 11$
• Checking:
  • All these quadratic nonresidues are primitive roots modulo 13.
• Conclusion: $p = 13$ is a valid prime.

Answer:

The next two primes with the property that all quadratic nonresidues are also primitive roots are $\boxed{11}$ and $\boxed{13}$.

Would you like to dive deeper into any step, or explore related topics?

Related Questions:

1. How do you find quadratic residues for a given prime $p$?
2. What is the significance of a primitive root in number theory?
3. Could there be a prime with exactly one quadratic nonresidue that is not a primitive root?
4. How do these concepts relate to cryptography?
5. What algorithms can efficiently determine primitive roots for large primes?

Tip: When analyzing properties of quadratic residues and nonresidues, modular arithmetic techniques and order determination are key tools in number theory.