Let's break down the solution into **two steps** with explanations: --- ### **Step 1: Recall the definition of a pseudoprime to base \( a \)** A composite number \( n \) is called a *pseudoprime to base \( a \)* if: \[ a^{n-1} \equiv 1 \pmod{n} \] even though \( n \) is not prime. #### **Explanation:** For base \( a = 1 \), we have: \[ 1^{n-1} = 1 \] for any \( n \). For base \( a = -1 \), we have: \[ (-1)^{n-1} \] Now, if \( n \) is odd, \( n-1 \) is even, so \((-1)^{n-1} = 1\). --- ### **Step 2: Apply the definition to any odd composite integer \( n \) and both bases** - For base 1: \[ 1^{n-1} \equiv 1 \pmod{n} \] which is always true for any \( n \). - For base -1: \[ (-1)^{n-1} \equiv 1 \pmod{n} \] Since \( n \) is odd, \( n-1 \) is even, so this is always true. #### **Explanation:** Both congruences are satisfied for any odd composite \( n \), meaning **every odd composite integer is a pseudoprime to both base 1 and base -1**. --- ## **Final Answer** **Every odd composite integer \( n \) is a pseudoprime to both base 1 and base -1, because \( 1^{n-1} \equiv 1 \pmod{n} \) and \( (-1)^{n-1} \equiv 1 \pmod{n} \) for all such \( n \).**

Let's break down the proof into **two steps** with clear explanations: --- ### **Step 1: Understand the definition of a pseudoprime to a given base** A composite number \( n \) is called a *pseudoprime to base \( a \)* if: \[ a^{n-1} \equiv 1 \pmod{n} \] even though \( n \) is not prime. **Explanation:** This resembles Fermat's Little Theorem, which states that if \( n \) is prime and \( \gcd(a, n) = 1 \), then \( a^{n-1} \equiv 1 \pmod{n} \). Pseudoprimes are composite numbers that satisfy this congruence for certain bases. --- ### **Step 2: Show that every odd composite integer satisfies the congruences for bases 1 and -1** - For **base 1**: \[ 1^{n-1} = 1 \] which is trivially congruent to 1 modulo \( n \) for **any** \( n \). - For **base -1**: \[ (-1)^{n-1} \] Since \( n \) is odd, \( n-1 \) is even, so: \[ (-1)^{n-1} = 1 \] Therefore: \[ (-1)^{n-1} \equiv 1 \pmod{n} \] holds for any odd \( n \). **Explanation:** Because both \( 1^{n-1} \) and \( (-1)^{n-1} \) are equal to 1 when \( n \) is odd, the congruences are always satisfied. This means **every odd composite number is a pseudoprime to base 1 and base -1**. --- ## **Final Conclusion** **Every odd composite integer \( n \) is a pseudoprime to both base 1 and base -1 because:** \[ 1^{n-1} \equiv 1 \pmod{n} \] and \[ (-1)^{n-1} \equiv 1 \pmod{n} \] hold for all odd \( n \). --- **Hence, the statement is proven.**