Read: Finish Section 7.4.
Turn in: 7.101, 7.102, 7.103
- The statement of 7.103 appeared in the section on induction. This time, I want you to use the techniques of modular arithmetic (not induction) to prove it. The book says to separately consider when $n=0$, but you can ingnore that. If you’re having trouble geting started, try this: Corollary 7.87 says that $8 \mid 3^{2n} - 1 \iff [3^{2n} - 1]_8 = [0]_8$, so now use our recent techniques to show $[3^{2n} - 1]_8 = [0]_8$. And if you’re still stuck, note that $3^{2n} = 9^n$.
Extra practice: 7.104
- Hint for 7.104: first write down what it would mean for $4n+3$ to be a perfect square and then consider what this says modulo 4.
