**Step 1: Understand the Fermat number and pseudoprime definitions.** A Fermat number is defined as \( F_m = 2^{2^m} + 1 \). A pseudoprime to the base \( a \) is a composite number \( n \) such that \( a^{n-1} \equiv 1 \mod n \). We are to show that if \( F_m \) is composite, then \( 2^{F_m - 1} \equiv 1 \pmod{F_m} \). **Step 2: Prove \( 2^{F_m - 1} \equiv 1 \pmod{F_m} \) for composite \( F_m \).** Recall that \( F_m = 2^{2^m} + 1 \). Therefore, \( F_m - 1 = 2^{2^m} \). Thus, \( 2^{F_m-1} = 2^{2^{2^m}} \). Now, notice that \( 2^{2^m} \equiv -1 \pmod{F_m} \), by the definition of Fermat numbers. So, \[ 2^{2^{2^m}} = (2^{2^m})^{2^{m}} \equiv (-1)^{2^{m}} \pmod{F_m} \] But \( 2^{m} \) is always even for \( m \geq 1 \), so \( (-1)^{2^m} = 1 \). Therefore, \[ 2^{2^{2^m}} \equiv 1 \pmod{F_m} \] which means \[ 2^{F_m-1} \equiv 1 \pmod{F_m} \] --- **Final Answer:** Every composite Fermat number \( F_m = 2^{2^m} + 1 \) satisfies \( 2^{F_m-1} \equiv 1 \pmod{F_m} \), so it is a pseudoprime to the base 2.