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.
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