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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.

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  • $\begingroup$ Note that we have $5^{2n} - 1 = 25^n - 1 = (25 - 1)(1 + 25 + 25^2 + \dots + 25^{n-1})$. $\endgroup$ – Qiaochu Yuan Sep 30 '17 at 2:52
  • $\begingroup$ For alternate methods to prove this, see this list for methods to prove the similar question $4\mid 5^n-1$. Adapt those methods to prove the stronger statement that $24\mid 5^{2n}-1$ $\endgroup$ – JMoravitz Sep 30 '17 at 3:57
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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.

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  • $\begingroup$ The step $5^{2k+2}-1=$ $5^{2k+2}-5^{2k}+5^{2k}-1$ looks like it was factored. I say this because $-5^{2k}+5^{2k}$ would can out when added just like a polynomial. Was $5^{2k+2}-1$ just factored (un-factored)? If so, how? I understand everything else. Your solution was spot on. Thanks. $\endgroup$ – ErinA Sep 30 '17 at 4:58
  • $\begingroup$ By factor, I think you mean splitting up an expression into products. If so, then $5^{2k+2}-1 = 5^{2k+2} - 5^{2k} + 5^{2k} - 1$ was not factored, since there is no product - only sums. They simply added $5^{2k}$ and subtracted $5^{2k}$. This technique appears in some proofs sometimes (adding and subtracting by the same number) and it's not called factoring. Hope that clears it up. $\endgroup$ – user2825632 Oct 2 '17 at 5:04
  • $\begingroup$ It does. Thanks! $\endgroup$ – ErinA Oct 3 '17 at 16:11
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I believe the solution is not expanding the exponent, but simply adding and subtracting by the same number. $5^{2k+2} - 1 = 5^{2k+2} - 5^{2k} + 5^{2k} - 1$. Does this make sense? After this, they simply use the distributive property. Since $5^{2k+2} = (5^2)(5^{2k})$ by the power of a product rule, the expression can be factored into $5^{2k}(5^2 - 1) + (5^{2k} - 1)$.

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