Let $a$ be a natural number not divisible by $5$

Question

Let $a$ be a natural number not divisible by $5$

Prove that $8a^{8n}+a^{4n}-4$ is a multiple of $5$ for

all n natural number.

My try :

Using induction

Let : $ A=8a^{8n}+a^{4n}-4$

For n=0 then A=8+1-4=0=0\mod 5$ $->$ true

For : n+1

$A=8a^{8n+8}+a^{4n+4}-4$

But I don't know how I complete

Answer 1

Hint : \begin{eqnarray*} a^4 \equiv 1 \pmod{5} \end{eqnarray*} if $a \not \equiv 0 \pmod{5}$.

Answer 2

Note that $$8a^{8n}+a^{4n}-4=5a^{8n}+(a^{4n}-1)(3a^{4n}+4)$$ Proof that $5\vert(a^{4n}-1)$

Other way, Suppose that $a=5p+q$ with $p=0,1,2,3,\cdots $ and $q=1,2,3,4$. Then $$a^4=625 p^4 + 500 p^3 q + 150 p^2 q^2 + 20 p q^3 + q^4$$ How $q=1,2,3,4\implies q^4=1,16,81,256$ then $a^4=5k+1$ then $a^{4n}=(5k+1)^n\implies a^{4n}-1=(5k+1)^n-1=(5k+1-1)[(5k+1)^{n-1}+(5k+1)^{n-2}+\cdots+1]=5k[(5k+1)^{n-1}+(5k+1)^{n-2}+\cdots+1]$