Consider the following statements:\n\n1. If $p > 2$ is a prime, then it can be written as $4n + 1$ or $4n + 3$ for a suitable natural number $n$.\n\n2. If $p > 2$ is a prime, then $(p - 1)(p + 1)$ is always divisible by 4.\n\nOf these statements,

### Concept & Logic This question tests fundamental properties of odd prime numbers, specifically their representation modulo 4 and the algebraic properties of consecutive even numbers. ### Step-by-Step Solution * **Analyze Statement 1:** "If $p > 2$ is a prime, then it can be written as $4n + 1$ or $4n + 3$..." Since $p > 2$ is prime, $p$ must be an odd number. When any odd integer is divided by 4, the remainder is always either 1 or 3. Therefore, every odd number can be represented as $4n + 1$ or $4n + 3$ (where $n$ is an integer $\ge 0$). Thus, Statement 1 is generally accepted as true in this context. * **Analyze Statement 2:** "If $p > 2$ is a prime, then $(p - 1)(p + 1)$ is always divisible by 4." Since $p$ is an odd prime, both $(p - 1)$ and $(p + 1)$ must be even numbers. They are consecutive even numbers (e.g., if $p=5$, they are 4 and 6). In any pair of consecutive even numbers, one is a multiple of 2, and the other is a multiple of 4. Their product will always be a multiple of $2 \times 4 = 8$. Since the product is always divisible by 8, it is inherently always divisible by 4. Statement 2 is true. ### Exam Strategy & Shortcut Instead of proving the algebra, simply test the statements with the smallest odd prime, $p = 3$. Statement 1: $3 = 4(0) + 3$. (True, fitting the $4n+3$ format). Statement 2: $(3 - 1)(3 + 1) = 2 \times 4 = 8$. 8 is divisible by 4. (True). Test $p = 5$: Statement 1: $5 = 4(1) + 1$. (True). Statement 2: $(5 - 1)(5 + 1) = 4 \times 6 = 24$. 24 is divisible by 4. (True). Both hold up immediately. ### Common Pitfall Getting tripped up by the technical definition of a "natural number" in Statement 1. If natural numbers are strictly $\{1, 2, 3...\}$ (excluding 0), then $p=3$ cannot be formed. However, in aptitude exams, this statement is testing the $modulo\ 4$ property of odd primes, and $n$ is routinely treated as a non-negative integer. Overthinking the $n=0$ edge case will lead you to pick the wrong option. ### Final Answer **Therefore, the correct answer is (1) and (2) are true.**