Prime Divisibility Rule

The problem states that P is a prime number and P divides $Q^2$. A fundamental property of prime numbers is that if a prime divides a product of two numbers, it must divide at least one of those numbers. Since $Q^2 = Q \times Q$, if P divides $Q^2$, it follows directly that P must divide Q.

Evaluating the Options

We need to identify which of the given options is NOT necessarily divisible by P, knowing that P divides Q.

• Option 1: $3Q$

  Since P divides Q, it will also divide any integer multiple of Q. Thus, P necessarily divides $3Q$.

• Option 2: $Q + 1$

  If P divides Q, Q can be written as $Q = k \times P$ for some integer k. Then $Q + 1 = k \times P + 1$. When $Q + 1$ is divided by P, the remainder is 1. Therefore, P does not necessarily divide $Q + 1$. For instance, let P = 5 and Q = 10. P divides $Q^2$ (5 divides 100) and P divides Q (5 divides 10). However, P does not divide $Q + 1$ (5 does not divide 11).

• Option 3: $2Q^2$

  Given that P divides $Q^2$, it must also divide any integer multiple of $Q^2$. Thus, P necessarily divides $2Q^2$.

• Option 4: $Q$

  As established earlier from the prime divisibility property, if P divides $Q^2$, P must divide Q.

Final Conclusion

Based on the analysis, the only expression that P does not necessarily divide is $Q + 1$.