Arthur Engel Number Theory Divisibility

Question

Question is straightforward (Arthur Engel's Problem Solving Strategies Chapter 6 Problem 12, Page 131), I was stuck for a while then read the provided solution:

Show $$641 \vert 2^{32} + 1. \text{ No calculator allowed!}$$

The provided solution is

$$641 = 5^4 + 2^4 = 5 \cdot 2^7 + 1 \text{ divides both } 5^4 \cdot 2^{28} + 2^{32} \text{ and } 5^4 \cdot 2^{28} - 1.$$ $$\text{Then, it also divides their difference } 2^{32} + 1.$$

I was wondering what the key point of this question is and the main takeaway is supposed to be. For example, in the previous questions (i.e. If $n$is not a prime, then $2^{n} - 1$ is not prime, the key takeaway is that $x - 1 \vert x^{n} - 1$). Is there something special about the number $641$ that I am suppose to know? Or is it something about $x^{2^{k}} + 1$? I am a bit confused as to what kind of tools I use take away from this problem for math contests. Or can someone give me another problem with the same takeaway lesson from this? Thanks!

Answer 1

It's an ad hoc method of computing (a multiple of) the order of $\,a\bmod n\,$ by elimination when a nice "splitting" is obvious: $\, 1 + a^j\:\! b\equiv 0\equiv \color{#c00}{a^\ell+b^k}\pmod{\!n},\, $ yielding

$\quad\ \bmod\ n\!:\ \ \ \ \ {-}1\ \equiv\:\, a^{\large j}\, b,\,$ so by Power & Product Rules
$\quad\ \ {\rm above}^{\large k}\, \Rightarrow\, \pm1^{\phantom{|^{|^|}}}\!\! \equiv\ a^{\large jk} \color{#c00}{b^{\large k}} \equiv\ a^{\large jk}(- \color{#c00}{a^{\large \ell}})\equiv\, -a^{jk+\ell},\ $ i.e. succinctly

$\quad\ \bmod \ n\!:\ \ \ \ \ {-}1\, \equiv\:\, a^{j} b\Rightarrow \pm1\equiv a^{jk}\color{#c00}{b^k} \equiv -a^{jk}\color{#c00}{a^\ell}\equiv -a^{jk+\ell}$

$\quad\ \bmod 641\!:\ \ {-}1\, \equiv\:\, 2^{7} 5\,\Rightarrow\, 1\equiv 2^{28}\color{#c00}{5^4} \equiv -2^{28}\color{#c00}{2^4}\equiv -2^{32}\ \ \ \ \rm [OP]$

$\quad\ \bmod 23\!:\ \ \ \ \ \ \ 1\, \equiv\:\, 2^{3} 3\,\Rightarrow\, 1\equiv 2^9\color{#c00}{3^3} \equiv 2^9\color{#c00}{2^2}\equiv 2^{11}\qquad \rm [others]$

$\quad\ \bmod 73\!:\ \ \ \,{-}1\, \equiv\:\, 2^{3} 3\,\Rightarrow\, 1\equiv 2^6\color{#c00}{3^4} \equiv 2^6\color{#c00}{2^3}\equiv 2^{9}$

$\quad\ \bmod 47\!:\ \ \ \ \ \ 1\ \equiv\:\, 2^{4} 3\,\Rightarrow\, 1\equiv 2^{20}\color{#c00}{3^5} \equiv 2^{20}\color{#c00}{3^3}\equiv 2^{23}$

$\quad\ \bmod 97\!:\ \ \ \ \ \ 1\, \equiv\:\, 2^{5} 3\,\Rightarrow\, 1\equiv 2^{20}\color{#c00}{3^4} \equiv -2^{20}\color{#c00}{2^4}\equiv -2^{24}$

$\quad\ \bmod 251\!:\ {-}1\, \equiv\ 2\, 5^3\!\Rightarrow\! {-}1\equiv 2\ \color{#c00}{5}^3 \equiv 2(\color{#c00}{2^8})^3\equiv 2^{25}$

$\quad\ \bmod 431\!:\, {-}1\, \equiv\ 2^{4} 3^3\,\Rightarrow\, 1\equiv 2^{16}(\color{#c00}{3^{4}})^3 \equiv 2^{16}(\color{#c00}{2^9})^3\equiv 2^{43}$

Generally we can eliminate $\,b\,$ from $\,a^i b^j \equiv \pm1 \equiv a^k b^l$ when they are independent, and this may yield a short certificate enabling efficient verification of some divisibility $\,n\mid a^m\pm1\,$ (e.g. analogous to Pratt primality certificates).

Historical Remark $ $ As mentioned in this old question, this method was used by Coxeter to verify the factor $641$ of the $5$'th Fermat number (which was mentioned in Hardy & Wright's popular number theory textbook)

Answer 2

Is there something special about the number $641$ that I am suppose to know? Or is it something about $x^{2^k}+1$?

What is special is the historical background. If $2^m + 1$ is prime then $m$ must be a power of $2$, and Fermat checked that $2^{2^k}+1$ is prime when $k = 0, 1, 2, 3, 4$. He claimed that $2^{2^k}+1$ is prime for all integers $k \geq 0$. About $100$ years later Euler determined that this is not true by showing $2^{2^5} + 1 = 4,\!294,\!967,\!297$ is divisible by $641$ and thus it is composite. That is what makes this example interesting. By the way, later calculations have led to no prime values of $2^{2^k}+1$ when $k \geq 5$ and it is now widely believed that $2^{2^k}+1$ is never prime when $k \geq 5$.

The way Euler was led to discover $641$ divides $2^{32}+1$ is described here. In modern language, the reasoning goes as follows. Let $p$ be a prime factor of $2^{32}+1$, so $2^{32} \equiv -1 \bmod p$ and thus $2^{64} \equiv 1 \bmod p$. That shows the order of $2 \bmod p$ divides $64$ but not $32$, so the order is $64$. Therefore $64 \mid (p-1)$ by Fermat's little theorem, so $p = 64n+1$. Run through $64n+1$ when $n = 1, 2, \ldots$ to check when it is prime, and when it is check whether it divides $4,\!294,\!967,\!297$. The number $64n+1$ is prime when $n = 3, 4, 7, 9, 10, 12, \ldots$ and we reach a prime factor of $2^{32}+1$ for the first time when $n = 10$, making $p = 640+1 = 641$: $$ 2^{2^5} + 1 = 4,\!294,\!967,\!297 = 641 \cdot 6700417 $$ So we only had to check $10$ numbers until we reached a proper factor of $2^{32}+1$, and the non-prime values of $64n+1$ when $n \leq 10$ are all easily detected because they are each divisible by $3$ or $5$.

Answer 3

COMMENT (Just for fun).-This problem, it seems, has to be solved with some artifice that is not easily perceptible. Here another way.

Prove without calculator that $2^{32}+1$ is divisible by $641$.

We have $\dfrac{2^{32}+1}{641}=\dfrac{16^8+1}{2\times16^2+8\times16+1}$. Consider the long division $f(x)=\dfrac{x^8+1}{2x^2+8x+1}$ so $$f(x)=\frac{x^6}{2}-2x^5+\frac{31x^4}{4}-30x^3+\frac{929x^2}{8}-\frac{899x}{2}+\frac{27839}{16}+R(x)$$ where the rest $R(x)=\dfrac{-13470x-\dfrac{27823}{16}}{2x^2+8x+1}$.

We must have $f(16)\in\mathbb N$ and for this it is clearly enough to have $$S=\frac{27839}{16}+R(16)\in\mathbb N$$ We can verify that $S=1401\in\mathbb N$.

A nice remark.- The IBM $1401$ was of the first computers for the public and it appeared in 10/05/1959 sixty five years ago.