It is known that the remainder of the division by $7$ of a natural number $a$ is $6$
Prove by induction that for every $n\in N$ the remainder of the division of $a^{2n}$ by $7$ is $1$.
Basically, As I understood, It's possible to use Modular Arithmetic there. So we could implement the mod (≡) symbol, and use mod7 as the operation in the induction proof.
We found that it's true for $n=1$. So we see that that's our result: $(a)2≡(−1)2$.
Then we assume that it's true for $k\in N$.
And then the final step of induction, is that if it's true for $k\in N$, we check if it's true for $k+1$. $a^{2k+2}=a^{2(k+1)}≡1 (mod 7)$.
And we see that it's true for each $n \in N$.
Base info : $a \equiv 6~~ (\text{mod} 7) \rightarrow a \equiv -1~~ (\text{mod} 7)$.
Goal: $a^{2\cdot{\color{red}n}}\equiv1$ for all $n$.
Just do this:
$$(a)^{2\cdot{\color{red}k}} \equiv 1~~ (\text{mod} 7)~~~(\text{true by assumption})$$
$$(a)^{2\cdot{\color{red}k}}\cdot{\color{green}{(a)^{2\cdot{\color{green}1}}}} \equiv 1\cdot{\color{green}{(1)}}~~ (\text{mod} 7) ~~ (\text{mult by base case})$$
$$a^{2k+2} = a^{2(\color{red}{k+1})}\equiv 1~~(\text{mod 7})$$
So the fact that the property holds for $k$ implies that holds for $k+1$. For the principle of induction we have that, $a^{2\cdot{\color{red}n}}\equiv1$ holds for all $n$.
NOTE: $a\equiv b ~~(\text{mod c})$ means that if we divide $a$ by $c$ ($a/c$, we will ignore the quotient and save the remainder. this remainder is $b$, so in a few words, $a\equiv b ~~(\text{mod c})$ means $a = b + ck$ for some $k\in\mathbb{Z}$ and $0\le b<c$.
further notes: for example we can say that $7 \equiv 3~~ (\text{mod 4})$ because $7 = 3 + 4\cdot1$, but also, we can say that $7 \equiv -1~~ (\text{mod 4})$ because $7 = -1 + 4\cdot2$. For example, if $a\equiv 6~~ (\text{mod 7}) $, it is because $a = 6+7k$, so we can write $7k = 7(k+1)-7$ and we will have $a = 6+7(k+1)-7 $ which means that $a = -1+7(k+1) $ this means that $a\equiv -1 (\mod 7)$
If you understand this logic, you can re-write the proof with $a|b$ notation :D