Prove by induction that $a^{4n+1}-a$ is divisible by $30$ for any $a$ and $n \geq 1$

Question

I don't know how to approach this problem since I don't know how to factor out $a$ which can be any number larger than $1$ in naturals, and that is the only approach I know for these kind of problems.

Answer 1

We do this by checking the divisibility by $2,3,5$ respectively.
First note that, if $a$ is divisible by any of the three, then so is $a^{4n+1}-a,$ so we shall assume that they are not divisible by none of the three.

For $2,$ notice that $a^{4n+1}\equiv a\pmod2$ so $a^{4n+1}-a\equiv a-a\equiv0\pmod2.$

For $3,$ observe that $a^4\equiv (a^2)^2\equiv1\pmod3$ by Fermat's little theorem, so $a^{4n+1}-a\equiv 1^n\times a-a\equiv0\pmod3.$

For $5,$ again using Fermat's little theorem, we have $a^4\equiv1\pmod5,$ thus $a^{4n+1}-a\equiv a(1^n-1)\equiv0\pmod5.$

Therefore $30\mid (a^{4n+1}-a).$

Hope this helps.

Answer 2

$\, n = 30 = 2\cdot 3\cdot 5 = \,$ product of primes $\rm \,p\,$ such that $\rm \ \color{#c00}{p\!-\!1\mid\ 4\mid (4n+1)\!-\!1}.\,$ Now apply

Theorem $\ $ For natural numbers $\rm\:a,e,n\:$ with $\rm\:e,n>1$

$\qquad\rm n\:|\:a^e-a\:$ for all $\rm\:a\:\iff n\:$ is squarefree, and prime $\rm\:p\:|\:n\,\Rightarrow\, \color{#c00}{p\!-\!1\mid e\!-\!1}$

Proof $\ $ See this answer for a simple proof.

Answer 3

Another way, distinct of induction, just to mention it.

It is easy to see that for all digit $a=1,2,3,4,5,6,7,8,9,0$ it is verified that $a^{4n+1}\equiv a\pmod{10}$ so that for all natural $a$ we have $a^{4n+1}-a$ is a multiple of $10$.

Besides $a^{4n+1}-a=a(a^{2n}+1)(a^{2n}-1)$ is clearly a multiple of $3$.

Thus $a^{4n+1}-a$ is a multiple of $30$.