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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 $l$?

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HINT

We have that

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

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

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  • $\begingroup$ How do we show gcd$(a,p)=1$? $\endgroup$ – kaisa Nov 16 '18 at 9:30
  • $\begingroup$ What about the identity if $p\mid 2l+1$? $\endgroup$ – user Nov 16 '18 at 11:30
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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

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

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