Prove that $5^{2n}-1$ is a multiple of 8 for all $n$ $\in$ $\mathbb{N}$.
Proof:
Using the Principle of Mathematical Induction:
Let $n=1$. If $n=1$, then $5^2-1=$ $25-1=$ $24$.
Since $24$ is divisible by $8$, the statement is true for $n=1$.
Assume the statement is true for $n=k$ where $k$ $\in$ $\mathbb{N}$.
Then the statement $5^{2k}-1$ is a multiple of $8$ is true. That is $5^{2k}-1=8m$ for some $m$ $\in$ $\mathbb{N}$.
We must prove that the statement is true for $n=k+1$. That is, we must prove that $5^{2(k+1)}-1$ is a multiple of 8.
The first few steps of the mathematical induction proof I understand. (Letting $n=1$ and assuming $n=k$ is true.) I even understand what that the end result of $n=k+1$ should look like in order for the statement to be true. What I'm having trouble with is the expansion of the exponent $5^{2(k+1)}-1$. I have an example and the solution from the back of the book, but I'm not following their solutions.
Solution from the back of the book:
$5^{2(k+1)}-1=$ $5^{2k+2}-1=$ $5^{2k+2}-5^2+5^2-1=$ $5^2(5^{2k}-1)+5^2-1$
The second solution I found was similar to the back of the book:
$5^{2(k+1)}-1=$
$5^{2k+2}-1=$
$5^{2k+2}-5^{2k}+5^{2k}-1$
$5^{2k}(25-1)+(5^{2k}-1)=$
$5^{2k}(24)+8m$
$\therefore$ $5^{2(k+1)}-1$ $\Rightarrow$ $8(5^{2k}(3)+m)$
I just don't understand the steps they took to expand the exponent. I'm not sure if I'm just forgetting the exponent rules or if I'm missing something else entirely. I don't know how to get from $5^{2k+2}-1$ to $8(5^{2k}(3)+m)$ on my own.
If anyone could advise me it'd be greatly appreciated.
Best Answer
I'll annotate the second solution, so hopefully it will make more sense.
$$5^{2(k+1)}-1 = 5^{2k+2}-1$$
The above step just expands $2(k+1) = 2k+2$ in the exponent.
$$=5^{2k+2}-5^{2k}+5^{2k}-1$$
This adds the value $5^{2k}$ and subtracts $5^{2k}$. Knowing to do this isn't obvious, but you should see that it is equivalent to the expression above.
$$=5^{2k}(25-1)+(5^{2k}-1)$$
This step did several things at once. First, it splits $5^{2k+2} = 5^{2k} \times 25$ so that we have $=5^{2k}(25) - 5^{2k} + 5^{2k} - 1$. Then we can factor out a $5^{2k}$ from $5^{2k} \times 25$ and $5^{2k} \times (-1)$ to get $5^{2k}(25-1)$. Again, it isn't obvious that this is the correct thing to do (how did we chose which of the positive or negative $5^{2k}$ to factor out?), but you should see again that it is equivalent to the statement above. After that, they just add parentheses around $5^{2k}-1$ to highlight that it is a pair (important for inductive step).
$=5^{2k}(24)+8m$
Here, we simplify the $(25-1) = 24$ (this is divisible by $8$!) and we use the fact that $5^{2k}-1 = 8m$ from our inductive statement, for some integer $m$.
$\therefore$ $5^{2(k+1)}-1$ $\Rightarrow$ $8(5^{2k}(3)+m)$
Now, we factor out the $8$. We know $5^{2k} \times 3 + m$ is an integer, so we are done.