Prove, for natural n, that 3 divides 4^n - 1 (by induction)
- Thread starter Tayeeba
- Start date
I am having trouble reading your attachment, but I think you are asking about logic something like this:
\(\displaystyle 4^k - 1 = 3j,\ where\ j \in \mathbb N^+ \implies 4^k = 3j + 1 \implies\)
\(\displaystyle 4 * 4^k = 4(3j + 1) \implies 4^{(k+1)} = 12j + 4 = 12j + 3 + 1 \implies\)
\(\displaystyle 4^{(k+1)} - 1 = 12j + 3 = 3(4j + 1),\ where\ 4j + 1 \in\ \mathbb N^+ \because j \in \mathbb N^+.\)
\(\displaystyle 4^k - 1 = 3j,\ where\ j \in \mathbb N^+ \implies 4^k = 3j + 1 \implies\)
\(\displaystyle 4 * 4^k = 4(3j + 1) \implies 4^{(k+1)} = 12j + 4 = 12j + 3 + 1 \implies\)
\(\displaystyle 4^{(k+1)} - 1 = 12j + 3 = 3(4j + 1),\ where\ 4j + 1 \in\ \mathbb N^+ \because j \in \mathbb N^+.\)
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