Alternative proof of Euler's result that $641$ divides $2^{32} + 1$

Question

I am looking for an alternative proof of Euler's result that $641$ divides $2^{32} + 1$. I've seen a solution so far, understood the solution, but unfortunately I don't know how to think in order to approach such a solution, so I'd like to see another solution easy to approach.

Here's the solution I have so far:

Observe that $641 = 2^7*5 + 1 = 2^4+5^4.$ Hence $2^7*5 \equiv -1 \space mod \space641.$

Now, $2^7*5 \equiv -1 \pmod{641}$ yields $5^4*2^{28}=(5*2^7)^4 \equiv 1 \space \pmod{641}.$ This last congruence and $5^4 \equiv -2^4 \pmod{641}$ yields $-2^4*2^{28} \equiv 1 \pmod{641}$, which means that $641$ divides $2^{32}+1.$

Answer 1

Computational. But simpler.

$$2^{32} + 1 \equiv (2^{16})^2 + 1 \equiv (65536)^2 + 1$$

$$\equiv (154)^2 + 1 \equiv 23717 \equiv 0 \pmod{641}$$

Answer 2

In the prime field $\mathbb F_{641}$ we have $2^{640}=(2^{32})^{20}=1\Rightarrow2^{32}=\pm 1$

Besides $2^{32}-1=(2^{16}+1)(2^{16}-1)=65537\cdot65535=155\cdot153=639\ne 0$

Hence $2^{32}+1=0\in \mathbb F_{641}\iff 641 \text{ divides}\space 2^{32}+1$