Induction with extra variable and conditional proposition

Question

Could someone take a look at the following proposition I'm trying to prove? Note: I'm required to use induction, strong induction, or proof by smallest counter example. My approach was to use regular induction.

My proof is not complete but could someone see what I have so far and tell me if I'm on the right track? This may be way off base since up until now I've only been working proof by induction problems that start with something along the lines of "For every integer $n$..." or "If $n\in\mathbb{N}$..." etc.

$Proposition$ Suppose $a\in\mathbb{Z}$. Prove that $5\mid2^na$ implies $5\mid a$ for any $n\in\mathbb{N}$.

$Proof$. We will prove this using mathematical induction.

Base Case. Let $n=1$. Then $5\mid2^{(1)}a$ which simplifies to $5\mid2a$. This means $2a=5b$ where $b\in\mathbb{Z}$. Since the LHS of our equation is even, we know $b$ must be even as well. So $b=2c$ for some $c\in\mathbb{Z}$. Thus $2a=5(2c)=2(5c)$. Dividing this equation by $2$ gives $a=5c$. Therfore $5\mid a$.

Inductive Step. Let the natural number $n=k\ge1$. We need to show that if $5 \mid 2^ka$ implies $5\mid a$ then $5 \mid 2^{k+1}a$ implies $5\mid a$. We will use direct proof.

This is far as I've gotten.

[ADDED]

So then the Inductive Step would unfold as:

Inductive Step. Let the natural number $n=k\ge1$. We need to show that if $5 \mid 2^ka$ implies $5\mid a$ then $5 \mid 2^{k+1}a$ implies $5\mid a$. We will use direct proof.

Let us assume the statement $5\mid 2^ka$ implies $5\mid a$ is true. Now suppose $5\mid2^{k+1}a$. Observe that $2^{k+1}a=2\cdot2^ka=2a'$ where $a'=2^ka$. Since we know $5\mid2a'$ then it must be the case that $5\mid a'$ since $2$ is even. By our inductive hypothesis if $5\mid a'$ and $a'=2^ka$ then it follows that $5\mid a$.

But if $5\mid2^ka$ implies $5\mid a$, which in turn implies that $5\mid2^{k+1}a$ implies $5\mid a$, we know that if $5\mid2^na$ then $5\mid a$ for $a\in\mathbb{Z}$ and any $n\in\mathbb{N}$. $\square$

Answer 1

What you did so far is very good. Now suppose that holds for $k$ and suppose that $5\mid 2^{k+1}a$. Note that $2^{k+1}a=2\times 2^ka=2a'$ where $a'=2^ka$. You know that your proposition holds for $1$ and $k$...

Answer 2

I know that this is not was asked, but using induction to prove that result somehow seems wrong, like using a hammer to open a bottle.

So, I decided to see if I could come up with a proof of the "correct" theorem, and here's what I got:

Theorem:

Suppose $c | ab$ and $(c, b) = 1$. Prove that $c | a$.

Proof.

Since $(c, b) = 1$, there are $u, v$ such that $uc-bv = 1$.

Since $c | ab$, there is a $k$ such that $ab = ck$.

Since $uc=bv+ 1$, $uca=bva+ a$.

Since $ab = ck$, $uca=ckv+ a$, so that $a =uca-ckv =c(ua-kv) $, so that, finally, $c | a$.