I'm currently grappling with a problem related to proving a statement using mathematical induction, for every integer n≥1, the expression 4^{2n} - 1 is divisible by 15.
So far, I understand the basic principle behind mathematical induction, which involves proving the statement for a base case (typically n=1) and then assuming it holds for an arbitrary positive integer k to prove it for k+1. However, I'm struggling with how to apply these steps specifically to show that 4^{2n} - 1 is indeed divisible by 15.
For the base case, I evaluated 4^{2(1)} - 1 = 15, which is divisible by 15, so it seems to hold up for n=1. My main challenge lies in the induction step, where I need to demonstrate that if the statement is true for n=k, then it must also be true for n=k+1. I have a rough idea that it involves some manipulation or factorization of the expression 4^{2(k+1)} - 1, but I'm not quite sure how to structure this argument or which properties of numbers to apply.
Could anyone guide me through the inductive step or point out any conceptual misunderstandings I might have? Also, if there's an alternative way to approach this problem that might provide a clearer or more intuitive proof, I'd be eager to learn about that as well.
You will find a number a direct good answers at How can I show that $4^{2n}-1$ is divisible by $15 $ for all $n$ greater or equal to $1$
In you really want to use induction, you already have the initial step for $n=1$: $4^{2\times 1} = 16 = 15 + 1$.
The induction step is not that much harder.
Let us assume that for $k$ between $1$ and $n$, $4^{2k} - 1$ is divisible by $15$. Then $\exists q \in \mathbb{N}, 4^{2n} = 15\times q + 1$.
It directly gives:
$$ \begin{aligned} 4^{2\left(n+1\right)} &= 16\times 4^{2n} = \left(15 + 1\right)\left(15\times q + 1\right)\\ &=15\times 15\times q + 15 + 15 \times q + 1\\ &= 15\left(15q +1+q\right) + 1 \end{aligned}$$
Which is enough to prove that $4^{2\times \left(n + 1\right)}-1$ is divisible by $15$.