How to show $2l+1\equiv (2l+1)^{4n+1}\pmod{5}, \forall l$ in the integers?

Question

How would I go about showing $2l+1\equiv (2l+1)^{4n+1}\pmod{5}, \forall l$ in the integers?

I tried this, but got stuck. \begin{align*} 2l+1&\equiv x\pmod{5}\\ (2l+1)^{4n+1}\pmod{5} &\equiv x^{4n+1}\pmod{5} \end{align*}

I'm not sure what to do from here.

Should I do even/odd cases for $l$?

Answer 1

HINT

We have that

$$(2l+1)^{4n+1}=[(2l+1)^{4}]^n(2l+1)$$ then refer to Fermat's little theorem

$$a^{p-1}\equiv 1 \pmod p$$

Answer 2

Hint $\bmod 5\!:\,\ \color{#c00}a(\color{#0a0}a^{\large\color{#0a0}4n}\!-\!1)\equiv 0\,$ if $\,\color{#c00}{a\equiv 0},\,$ and also if $\,a\not\equiv 0\,$ by $\,\color{#0a0}{a^{\large 4}\equiv 1}\,$ by Fermat

Answer 3

if $2l+1=5k=>5k\equiv (5k)^{4n+1}\pmod{5}$.$$$$ if $2l+1=5k+1=>5k+1\equiv (5k+1)^{4n+1}\pmod{5}<=>1\equiv1\pmod 5$$$$$ if $2l+1=5k+3=>5k+3\equiv (5k+3)^{4n+1}\pmod{5}<=>3\equiv3^{4n+1}\equiv 3\cdot 81^n\equiv3\cdot 1^n\pmod 5$.