Let's solve the problem step by step: --- **Step 1: State the definition and property of Fermat numbers and pseudoprimes.** - Fermat numbers are of the form \( F_m = 2^{2^m} + 1 \). - A pseudoprime to the base 2 is a composite number \( n \) such that \( 2^{n-1} \equiv 1 \pmod{n} \). **Explanation:** We need to show that if \( F_m \) is composite, then \( 2^{F_m-1} \equiv 1 \pmod{F_m} \). --- **Step 2: Use properties of exponents and Fermat numbers to prove the condition.** - For Fermat numbers, \( F_m = 2^{2^m} + 1 \). - Therefore, \( F_m - 1 = 2^{2^m} \). - So, \( 2^{F_m-1} = 2^{2^{2^m}} \). - Notice that \( 2^{2^m} \equiv -1 \pmod{F_m} \). - Raising both sides to the power 2: \( (2^{2^m})^2 = 2^{2^{m+1}} \equiv 1 \pmod{F_m} \). - By repeated squaring, \( 2^{2^{2^m}} \equiv 1 \pmod{F_m} \). **Explanation:** This shows \( 2^{F_m-1} \equiv 1 \pmod{F_m} \), satisfying the pseudoprime condition for base 2. --- **Final Answer:** **Every composite Fermat number \( F_m = 2^{2^m} + 1 \) is a pseudoprime to the base 2 because \( 2^{F_m-1} \equiv 1 \pmod{F_m} \).**