Prove by induction $5^{2n} − 1$ is divisible by $8$ for all $n ∈ N$.

Question

From this link: https://www.math.purdue.edu/~catlind/MA341-spring2018/quizzes/Quiz1.pdf

Prove that $5^{2n} − 1$ is divisible by $8$ for all $n ∈ N$ .Solution: To prove by induction, we must

1. Show it is true for $n = 1$
2. Assume true for $n$ and prove it is true for $n + 1$.

Let $P(n)$ be the statement $5^{2n} − 1$ is divisible by $8$. When $n = 1$, $5^2 − 1 = 24$, which is divisible by $8$. This proves $P(1)$.

Suppose that $P(n)$ is true, i.e., $5^{2n} − 1$ is divisible by $8$. Then $5^{2(n+1)} − 1 = 5^2 · 5^{2n} − 1 = 5^2(5^{2n} − 1) + (5^2 − 1)$. Note that $5^{2n} − 1$ is divisible by $8$ using $P(n)$ and also $5^2 − 1 = 24$ is divisible by $8$. Thus we have proved that $P(n + 1)$ is true. We conclude that $P(n)$ is true for all $n ∈ N$

My questions:

1. $5^2 · 5^{2n} − 1 = 5^2(5^{2n} − 1) + (5^2 − 1)$. How is this calculation done? I know $5^2 · 5^{2n} − 1$ has been manipulated to come up with $(5^2 − 1)$ and $(5^{2n} − 1)$, two terms which we know the value of. But what's the process/intuition for this manipulation? Because instinctively when I look at $5^2 · 5^{2n} − 1$ I feel there can't be anymore simplification due to the $-1$ term.

2. Proof by induction according to this link is "show $n=1$ is true, assume $n$ is true, and then show $n+1$ is true". I now know the formula for proof by induction, but I still don't understand what it intrinsically means or how it guarantees $5^{2n} − 1$ is divisible by $8$ for all $n ∈ N$. Why is it a precondition to show the first term $n=1$ to be true? What's the purpose of this? Why is the $n$ term assumed to be true? Wouldn't it be better to have to show that as well? And then why choose $n+1$? Why not $n+2$ or $n+n$? And why must this be shown to be true instead of being assumed? As you can see I'm very confused here.

Answer 1

We assume $P(n)$ is true and then prove for $P(n+1)$. So, if $P(1)$ is true then so is $P(2)$. If $P(2)$ is true then so is $P(3)$. And so on.

We therefore need to find an expression for $P(n+1)$ that involves $P(n)$.

$5^25^{2n}-1=5^2(5^{2n}-1)+24$ also does the trick.

Answer 2

Noting that $5^{2n}-1=25^n-1$, we can let the proposition $$P(n):(25^n-1) \textrm{ is divisible by }8 \textrm{ for }n\in N. $$ Obviously, $8|25-1=24 \Rightarrow P(1)$ is true.

Assume $P(k)$ is true for some natural number $k $, i.e. $25^k-1=8m$ for some natural number $m$.

Now when $n=k+1$, $25^{k+1}-1=25\times 25^k-1=25\times(8m+1)-1=8(25m+3), $ which is divisible by $8$ and hence $P(k+1)$ is also true. By the principle of Mathematical Induction, we can conclude that $5^{2n}-1 $ is divisible by $8$ for all $n\in N.$

Answer 3

Write $5^{2n}-1=(5^n-1)(5^n+1)$

Clearly, $(5^n+1)$ is even and $(5^n-1)$ is divisible by $4$. It can be shown using induction.

Hence, we get what we are looking for.