Show that prime $p=4n+1$ is a divisor of $n^{n}-1$

We know that $n$ is a quadratic residue modulo $p$, since $1 = \left( \frac{-1}{p}\right) = \left(\frac{4n}{p}\right) = \left(\frac{n}{p}\right)$. Let $k^2 \equiv n \pmod{p}$. Then

$$(k+2n)^2 = k^2 + 4nk + 4n^2 \equiv n -k - n \equiv -k \pmod{p},$$

so $n$ is in fact a fourth power, $n \equiv (k+2n)^4 \pmod{p}$. Hence

$$n^n \equiv (k+2n)^{4n} = (k+2n)^{p-1} \equiv 1 \pmod{p}.$$

We'll use this lemma which I'll leave for you to prove:

Ic $p=4n+1$ is prime, then $a^n\equiv 1\pmod {p}$ if and only if $a\equiv x^4\pmod p$ for some $x$ not divisible by $p.$

We'll show that $-4$ is always a fourth power, $\pmod p$.

If $n$ is even, then $p\equiv 1\pmod 8,$ so $2$ is a square, and hence $4$ is a fourth power. Also $(-1)^n\equiv 1\pmod p$, $-1$ is a fourth power modulo $p$ by our lemma, and so $-4$ is a fourth power.

If $n$ is odd, then $-1\equiv x^2$ where $x$ is not a square. This is because if $x$ were a square, then $x^4=1$ and $x^{(p-1)/2}=1$, and hence $x^2=1,$ since $\gcd\left(4,\frac{p-1}{2}\right)=2.$

Similarly, $2$ is not a square modulo $p,$ So $2x$ is a square, $\pmod p$. So $4x^2\equiv -4\pmod p$ is a fourth power.

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Knowing that $-4$ is a fourth power is all you need, since:

$$-4n\equiv 1\pmod p$$

So $n$ is a fourth power, too, so, again by our lemma, $n^n\equiv 1\pmod p$.

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An alternative approach to showing that $x^4+4\equiv 0\pmod{p}$ for some $x$ is to use that:

$$x^4+4 = (x^2+2)^2-(2x)^2=(x^2-2x+2)(x^2+2x+2)=((x-1)^2+1)((x+1)^2+1)$$

Thus we have that $x^4+4\equiv 0\pmod p$ when $(x\pm 1)^2\equiv -1\pmod p.$ But since $p\equiv 1\pmod{4},$ $-1$ is a square modulo $p$, we know such $x$ exists.

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Using that last approach as a jumping off point, we can start with $y$ such that $y^2\equiv -1\pmod{p}.$

Then $$\begin{align}(y+1)^4 &= y^4+4y^3+6y^2+4y+1 \\&\equiv 1+4y(-1)+6(-1)+4y+1\pmod{p}\\&=1+(-6)+1\\&=-4\pmod{p}\end{align}$$