Proof of divisibility using modular arithmetic: $5\mid 6^n - 5n + 4$

Question

Prove that:

$$6^n - 5n + 4 \space \text{is divisible by 5 for} \space n\ge1$$

Using Modular arithmetic. Please do not refer to other SE questions, there was one already posted but it was using induction, I want to use this number theory method.

Obviously we have to take $\pmod 5$

So:

$$6^n - 5n + 4 \equiv x \pmod 5$$

All we need to do prove is prove $x = 0$

How do we do that? I just need a hint, I am not sure how to solve congruences. Some ideas will be helpful.

Thanks!

Answer 1

Hint:- $6\equiv1 \pmod 5\implies 6^n\equiv1\pmod 5\tag{1}$

$-5(n-1)\equiv 0\pmod 5\tag{2}$

Solution:-

$(1)+(2)$ gives,$$6^n-5n+4\equiv0\pmod 5$$

Answer 2

Here are some hints. Working modulo $5$, we have $5 = 0$ and $6 = 1$.

Answer 3

Directly: $$6^{n}-5n+4=\left(6-1\right)\left(6^{n-1}+6^{n-2}+\cdots+6^{1}+6^{0}\right)-5n+5=$$$$5\left(6^{n-1}+6^{n-2}+\cdots+6^{1}+6^{0}-n+1\right)$$

Answer 4

You can go much simpler than that. Given any $n \in \mathbb{Z}^+$, we have $6^n \equiv 1 \pmod 5$ (look up "automorphic numbers" sometime). We can ignore $5n$ because $5n \equiv 0 \pmod 5$ regardless. So, working with modular arithmetic, we're essentially left with $1 + 4 = 0$.