The question in the title arises from a problem in Stewart's "Galois Theory, Third Edition" (and possibly elsewhere) which has been bugging me for a few days since reading it:

Problem 19.5 (p. 224) asks:

```
Use the equations
$641 = 5^4+2^4 = 5\cdot 2^7+1$
to show that 641 divides $F_5$.
```

Now the latter expression is related to Euler's proof of his Theorem 8 in E134 and the ideas contained in the proof of that theorem is simple enough to lead to 641 being a candidate divisor of $F_5$ which can then be easily checked by hand/calculator.

However, the fact that Stewart includes the other expression as well intrigues me; I have been trying to use factorizations via sums of two squares to see how this expression might arise; for example since \[ F_5 = 65536^2+1^2 = 62264^2+20449^2 \] one can find this factor as \[ 641 = \gcd(65536*62264-1*20449, 65536*20449+1*62264) \]

But this approach is unsatisfactory since

1). Stewart does not mention the latter decomposition of $F_5$ as a sum of two squares

2). This approach makes no use of the decomposition of the potential factor as a sum of two squares/fourth powers

So, does anyone else have a clue as to what theorem/approach Stewart may have intended by including this decomposition. In particular, are there other less well-known theorems dealing with factoring Fermat numbers by expressing potential factors as Generalized Fermat numbers or some other similarly out of the hat approach? Or did Stewart include this expression for no good reason (which seems doubtful given the clarity of the approaches he takes throughout the rest of the book)?

BTW: As to why this is posted here, although the question asked in the book is certainly not research level, the intricacy of the other methods used to demonstrate 641 is a candidate factor indicate that if there is indeed a way to use the decomposition $5^4 + 2^4$ to factor $F_5$, such a method likely involves some deeper mathematics that is/was research-level.

Elementary methods in number theory: $F_5=2^{2^5}+1=(2^{32}+5^4\cdot 2^{28})-(5^4\cdot 2^{28}-1)=$ $2^{28}(2^4+5^4)-(5^2\cdot2^{14}+1)(5\cdot 2^7+1)(5\cdot 2^7-1)$. – Andrés E. Caicedo Sep 11 '13 at 3:22An introduction to the theory of numbers. It may be due to them, as there is no additional attribution.) – Andrés E. Caicedo Sep 11 '13 at 3:40