Prime Decomposition of 5th Fermat Number/Proof 1

Proof
From Divisor of Fermat Number, if $2^{\left({2^n}\right)} + 1$ has a divisor, it will be in the form:
 * $k 2^{n + 2} + 1$

In the case of $n = 5$, a divisor of $2^{\left({2^n}\right)} + 1$ is then of the form:
 * $k 2^7 + 1 = k \times 128 + 1$

Further, such a number will (for small $k$ at least) be prime, otherwise it will itself have a divisor which is not of that form.

Thus we investigate:

We have that:

But
 * $2^8 + 1 = 257$

and so:
 * $2^{\left({2^5}\right)} - 1 \equiv 0 \pmod {257}$

which means:


 * $2^{\left({2^5}\right)} + 1 \equiv 2 \pmod {257}$

so there is no need to do a trial division of $2^{\left({2^5}\right)} + 1$ by $257$.

Now dividing $4 \, 294 \, 967 \, 297$ by $641$ we find:

6 700 417   --- 641 ) 4 294 967 297      3 846      -        448 9        448 7        -            267 2            256 4            -             10 89              6 41             -              4 487              4 487              =====

and the divisor has been found.