
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Let \( x_{1}, x_{2} \) be the integer solutions of the equation \( x^{2}+p x+3q=0 \), where \( p \) and \( q \) are prime numbers. By Vieta's formulas:
\[ x_{1} + x_{2} = -p < 0 \, \text{and} \, x_{1} \cdot x_{2} = 3q > 0 \]

From this, we conclude that both roots must be negative integers. Given that \( q \) and \( 3 \) are primes, there are only two possibilities:
\[ x_{1} = -1, x_{2} = -3q \, \text{or} \, x_{1} = -3, x_{2} = -q \]

Consider the first case: if \( x_{1} = -1 \) and \( x_{2} = -3q \), then \( 1 + 3q = p \). Thus, \( p \) and \( q \) have different parities (one is even, the other is odd). The only even prime is 2. If \( p = 2 \), the equation \( 1 + 3q = 2 \) has no solutions in prime numbers, so this case is impossible. If \( p \) is odd, then \( q \) must be even, specifically 2. Here, \( p = 1 + 3 \cdot 2 = 7 \), which is prime. So, we have \( p = 7, q = 2 \), and \( x_{1} = -1, x_{2} = -6 \).

Consider the second case: if \( x_{1} = -3 \) and \( x_{2} = -q \), then \( 3 + q = p \). Again, \( p \) and \( q \) have different parities. If \( p = 2 \), the equation \( 3 + q = 2 \) has no solutions in prime numbers, so this case is impossible. If \( p \) is odd and \( q \) is even, specifically 2, then \( p = 3 + 2 = 5 \), which is prime. Thus, we have \( p = 5, q = 2 \), and \( x_{1} = -3, x_{2} = -2 \).

After evaluating all possible cases, there are no other solutions.

Present the answer in LaTex format: \boxed{Your answer}
user