Let's break the solution into **2 steps** with explanations after each step, and provide the final answer at the end. --- ### **Step 1: Recall properties of Fermat numbers and the definition of pseudoprime to base 2** A **Fermat number** is of the form \( F_n = 2^{2^n} + 1 \). A composite number \( N \) is a **pseudoprime to the base 2** if \( 2^{N-1} \equiv 1 \pmod{N} \) but \( N \) is not prime. #### **Explanation:** We need to show that for any composite Fermat number \( F_n \), it holds that \( 2^{F_n-1} \equiv 1 \pmod{F_n} \). --- ### **Step 2: Prove that \( 2^{F_n-1} \equiv 1 \pmod{F_n} \) for any Fermat number \( F_n \)** Notice that for any Fermat number \( F_n = 2^{2^n} + 1 \), \[ F_n - 1 = 2^{2^n} \] So, \[ 2^{F_n-1} = 2^{2^{2^n}} \] By **Fermat's Little Theorem**, if \( p \) is a prime divisor of \( F_n \), then \( 2^{p-1} \equiv 1 \pmod{p} \). But for Fermat numbers, every prime divisor \( p \) of \( F_n \) satisfies \( p \equiv 1 \pmod{2^{n+1}} \). So \( 2^{2^{n+1}} \equiv 1 \pmod{p} \). But more generally, since \( F_n = 2^{2^n} + 1 \), we have \( 2^{2^n} \equiv -1 \pmod{F_n} \), so \[ 2^{2^{n+1}} = (2^{2^n})^2 \equiv (-1)^2 = 1 \pmod{F_n} \] Thus, \( 2^{k \cdot 2^{n+1}} \equiv 1 \pmod{F_n} \) for any integer \( k \). Now, \( 2^{F_n-1} = 2^{2^{2^n}} \). But \( F_n = 2^{2^n} + 1 \), so \( F_n - 1 = 2^{2^n} \). So, \[ 2^{F_n-1} = 2^{2^{2^n}} \] But since \( 2^{2^n} \equiv -1 \pmod{F_n} \), raising both sides to \( 2^{2^n} \) gives: \[ (2^{2^n})^{2^{2^n}} \equiv (-1)^{2^{2^n}} \pmod{F_n} \] Since \( 2^{2^n} \) is even, \( (-1)^{2^{2^n}} = 1 \). So, \[ 2^{F_n-1} \equiv 1 \pmod{F_n} \] #### **Explanation:** This proves that every Fermat number (whether prime or composite) satisfies the congruence for pseudoprimality to the base 2. --- ### **Final Answer** **Every composite Fermat number \( F_n = 2^{2^n} + 1 \) is a pseudoprime to the base 2, because \( 2^{F_n-1} \equiv 1 \pmod{F_n} \) holds for all Fermat numbers.**