How can I show that Fermat number $F_{5}=2^{2^5}+1$ is divisible by $641$.
To show that Fermat number $F_{5}$ is divisible by $641$.
2 Answers
If you know at least a few basic facts about congruences, it's not difficult to do this by hand, e.g. as follows:
$\newcommand{\kong}[3]{{#1}\equiv{#2}\pmod{#3}}$ $\kong{2^8}{256}{641}$
$2^{16} \equiv 256^2= 64\cdot4\cdot256 =1024\cdot64=102\cdot640+256 \equiv \kong{256-102}{154}{641}$
$2^{32} \equiv 154^2 = 14^2\cdot11^2 = 196\cdot121 = (3\cdot64+4)(2\cdot64-7)= 6\cdot64^2+8\cdot64-21\cdot64-28=(384+8-21)\cdot64-28 = 371\cdot64-28 = 37\cdot640+64-28\equiv \kong{-37+36}{-1}{641}$
The last line means that $641\mid F_5=2^{32}+1$.
We have just used $\kong{640}{-1}{641}$ a few times.
Hardy and Wright give in their book a different argument, see p.18.
If we notice that $641=5^4+2^4=5\cdot 2^7+1$ then we have that $641\mid 5^4\cdot2^{28}+2^{32}$ (we multiplied the first expression by $2^{28}$) and we also have $641\mid 5^4\cdot2^{28}-1$ if we use $x+1\mid x^4-1$ for $x=5\cdot2^7$.
If we subtract the above numbers, we get $641\mid 2^{32}+1=F_5$
They attribute this proof to Coxeter, see p.27.
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0Unfortunately this not generalize to any known method for larger Fermat numbers , so the Q of which regular polygons are constructible with unmarked ruler and a drawing compass is still not completely answered. – 2017-07-16
$\rm\begin{eqnarray} \bmod\, 641\!:\,\ {-}1\, &\equiv&\rm\:\, 5\,\cdot\, 2^{\large 7}\\[.2em] \rm above^{\large 4}\ \Rightarrow\:\ \ 1\, &\equiv&\rm\! \underbrace{\color{#c00}{5^{\large 4}}\!}_{\Large\!\!\!\!\! \color{#c00}{\equiv\,-2^{\Large 4}}}\:\! 2^{\large 28}\equiv\, -2^{\large 32}\ \ \ \ {\bf QED}\end{eqnarray}$
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0@user236182 We'll have to disagree about "the same". In any case, I've posted this numerous places over the years, going back to ancient sci.math days. It is my condensed version of a well-known classic proof that appears in various forms in some elementary number theory textbooks. – 2017-09-03