Solving for the Largest Value of p

We are given a problem involving two natural numbers, $p$ and $q$, with the condition that $q > p$. We need to find the largest possible value for $p$ such that the expression $q^2 - 5p - 4$ is negative.

Understanding the Inequality

The core condition is that $q^2 - 5p - 4 < 0$. This is an inequality involving $p$ and $q$. Natural numbers are positive whole numbers, so $p, q \in \{1, 2, 3, ...\}$.

We can rewrite the inequality to isolate $q^2$: $q^2 < 5p + 4$

Analyzing the Constraints

We also know that $q > p$. Since $p$ and $q$ must be natural numbers, the smallest possible value for $q$ is $p+1$. We need to find the largest natural number $p$ for which there exists at least one natural number $q$ satisfying both $q > p$ and $q^2 < 5p + 4$.

Step-by-Step Analysis of Potential p Values

Let's test the possible values for $p$ provided in the options. We are looking for the largest value of $p$ that works. The options are 3, 4, 5, and 6.

Case 1: $p = 3$

• Condition: $q > 3$ and $q^2 < 5(3) + 4$.
• Simplified inequality: $q^2 < 15 + 4$, which means $q^2 < 19$.
• We need a natural number $q$ such that $q > 3$ and $q^2 < 19$.
• Let's check values for $q$ starting from $4$ (since $q > 3$):
  • If $q=4$, then $q^2 = 16$. Since $16 < 19$, this condition is met.
• Since we found a valid $q$ (e.g., $q=4$) for $p=3$, the value $p=3$ is a possible solution.

Case 2: $p = 4$

• Condition: $q > 4$ and $q^2 < 5(4) + 4$.
• Simplified inequality: $q^2 < 20 + 4$, which means $q^2 < 24$.
• We need a natural number $q$ such that $q > 4$ and $q^2 < 24$.
• Let's check values for $q$ starting from $5$ (since $q > 4$):
  • If $q=5$, then $q^2 = 25$. Since $25$ is NOT less than $24$, this condition is not met.
  • For any $q \ge 5$, $q^2$ will be $\ge 25$, so $q^2 < 24$ cannot be satisfied.
• Therefore, there is no natural number $q$ satisfying the conditions when $p=4$. So, $p=4$ is not a possible value.

Case 3: $p = 5$

• Condition: $q > 5$ and $q^2 < 5(5) + 4$.
• Simplified inequality: $q^2 < 25 + 4$, which means $q^2 < 29$.
• We need a natural number $q$ such that $q > 5$ and $q^2 < 29$.
• Let's check values for $q$ starting from $6$ (since $q > 5$):
  • If $q=6$, then $q^2 = 36$. Since $36$ is NOT less than $29$, this condition is not met.
  • For any $q \ge 6$, $q^2$ will be $\ge 36$, so $q^2 < 29$ cannot be satisfied.
• Therefore, there is no natural number $q$ satisfying the conditions when $p=5$. So, $p=5$ is not a possible value.

Case 4: $p = 6$

• Condition: $q > 6$ and $q^2 < 5(6) + 4$.
• Simplified inequality: $q^2 < 30 + 4$, which means $q^2 < 34$.
• We need a natural number $q$ such that $q > 6$ and $q^2 < 34$.
• Let's check values for $q$ starting from $7$ (since $q > 6$):
  • If $q=7$, then $q^2 = 49$. Since $49$ is NOT less than $34$, this condition is not met.
  • For any $q \ge 7$, $q^2$ will be $\ge 49$, so $q^2 < 34$ cannot be satisfied.
• Therefore, there is no natural number $q$ satisfying the conditions when $p=6$. So, $p=6$ is not a possible value.

Conclusion

By testing the possible values for $p$, we found that only $p=3$ allows for a natural number $q$ ($q>p$) such that $q^2 - 5p - 4 < 0$. Therefore, the largest value of $p$ that satisfies the given conditions is 3.