Insight on the proof of $2^b-1$ does not divide $2^a+1$.

Question

Proposition. For any integer $a$ and $b$ where $b>2$, we never have $2^b-1|2^a+1$.

WLOG let $b<a$. The proof uses a critical fact that can be stated very simply as follows:

$2^a+1$ leaves a reminder of the form $2^r+1$ when divided by $2^b-1$, where $r\leq b$.

This again, directly follows by the following equation:

$$2^{r+kb}+1=2^r(2^{kb}-1)+(2^r+1).$$

I am digging some insight from these steps so that the above step is not just a “trick” to use. For example in the proposition the base $2$ can be replaced by any positive integer $x\geq 2$. And observe that the exponential function minus $1$(it is not true in general for $x^n+k$ where $k\neq -1$) satisfies the following functional equation:

One of the solution set for $f(x+y)=f(x)f(y)+f(x)+f(y)$ where $f:\mathbb Z_+\to \mathbb Z$ is the set of all exponential function minus $1$: $f(x)=x^n-1$, where the integer $x>1$ and $n>1$.

I am looking for a way to make the solution of the first line proposition not a “trick” to memorize, but a link to a theory or a more general result so the proof will follow naturally.

Answer 1

if $m|n$ then $m|(n-km)$

if $2^b - 1| 2^a + 1$ then $2^b - 1| 2^a + 1 - 2^{a-b}(2^b -1)$

Which we can simplify to say $2^b-1 | 2^{a-b} + 1$

We are just applying the Euclidean algorithm.

And we can repeat this as many times as we need to such that $2^b-1 | 2^{a-kb} + 1$

with $0 \le a-kb < b$

And if $b>2$ we have a remainder.

Answer 2

We can view it as $ $ q-analog of $\,\ \ \bbox[5px,border:1px solid #c00]{b\mid a\iff b\mid a\bmod b}\,\ $ [divisibility mod reduction]

i.e. for $\,f_n = 2^n-1\,$ we have $\, \bbox[5px,border:1px solid #c00]{ f_b\mid f_a\!\!\iff\!\! f_b\mid f_{\,a\bmod b}}\,\ $ [cf. proof here].

Here the OP sequence $\,f_n = 2^n-1\,$ has even further nice arithmetical structure - namely it is a strong divisibility sequence i.e. $\gcd(a_m,a_n) = a_{\gcd(m,n)}$ so we can further apply common euclidean gcd arguments to the indices (exponents).

This generalizes from $\,2^n$ to $\,a^n.\,$ The proofs exploit modular order reduction as below.

Theorem $ \ \ $ Suppose that: $\,\ \color{#c00}{a^{\large e}\equiv\, 1}\,\pmod{\! m}\ $ and $\, e>0,\ n,k\ge 0\,$ are integers. Then

$\qquad n\equiv k\pmod{\! \color{#c00}e}\,\Longrightarrow\,a^{\large n}\equiv a^{\large k}\pmod{\!m}.\ $ Further, $ $ conversely

$\qquad n\equiv k\pmod{\! \color{#c00}e}\,\Longleftarrow\,a^{\large n}\equiv a^{\large k}\pmod{\!m}\ \, $ if $\,a\,$ has order $\,\color{#c00}e\,$ mod $\,m$

Answer 3

Claim: If $a,b,r, n\in \mathbb N$ if $a \equiv r \pmod b$ then $n^a \equiv n^r\pmod{n^b -1}$.

Pf: wolog $r < a$ so there is a $k > 0; k\in \mathbb N$ so that $a= kb + r$.

So $n^a = n^{a-b}(n^b-1)+ n^{a-b}\equiv n^{a-b}\pmod{n^b-1}$.

and by induction, for any $a-kb \ge b$

$n^a \equiv n^{a-(k-1)b} = n^{a-kb}(n^b-1) + n^{a-kb}\equiv n^{a-kb}\pmod{n^b - 1}$.

And so $n^a\equiv n^{n-kb}=n^r\pmod {2^b -1}$

.....

And that's all. Just because something is "tricky" doesn't mean it isn't valid.

==== old answer =====

$2^a + 1 \equiv m \pmod {2^b -1}$ means there is a $k$ so that $2^a+1 =k*(2^b-1) + m$.

To find so possible values for $k, m$ (and if $0 \le m < 2^a + 1$ then then $k$ and $m$ are unique) we notice.

$k*(2^b-1) + m = 2^k-k + m$.

If $k = 2^{a-b}$ we get $2^{a} + 1 = 2^{a-b}(2^b -1) + m = 2^a - 2^{a-b} + m$ and so

$2^{a-b} + 1 = m$ and

That's that. I'm not sure why that seems like "trick".

Or we couls simply note:

$2^a + 1 = 2^{b-a}2^b + 1 = 2^{b-1}(2^b - 1) + 2^{b-a} + 1\equiv 2^{b-a} +1\pmod{2^b -1}$.

or example in the proposition the base 2 can be replaced by any positive integer x≥2.

Sure.

If $b < a$ then

$x^a+1 = x^{a-b}(x^{b}-1) + x^{a-b} + 1 \equiv x^{a-b} + 1\pmod {x^b-1}$.

We can make that a theorem (never "theory".... math doesn't have "theories").

Theorem: for $a, b \in \mathbb N$ then for any $k\in \mathbb Z$ so that $a + kb \ge 0$ then $n^{a}\equiv n^{a+kb}\pmod {n^b -1}$.

Pf: If $a > b$ then $n^a = n^{a-b}(n^b -1) + n^{a-b}\equiv n^{a-b}\pmod {n^b-1}$ and by induction $n^a \equiv n^{a-kb}\pmod {n^b-1}$ for all $k; a-kb \ge 0$. And $n^{a+kb} \equiv n^{a}\pmod{n^b -1}$ for all $k \ge 0$ by induction.

Answer 4

When looking at divisibility (or non-divisibility) by some number $m$, it is often useful to look at things mod $m$. In this case, $m=2^b-1$.

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Note that $$ 2^b\equiv1\pmod{2^b-1}\tag1 $$ It may prove helpful to use this equivalence to simplify things. For instance, multiplying $(1)$ by $2^{a-b}$ gives $$ 2^a\equiv2^{a-b}\pmod{2^b-1}\tag2 $$ For some $q,r\in\mathbb{Z}$, $0\le r\lt b$, we can write $$ a=qb+r\tag3 $$ and repeating the reasoning behind $(2)$, we get $$ \begin{align} 2^a &\equiv2^{a-qb}&\pmod{2^b-1}\\ &\equiv2^r&\pmod{2^b-1}\tag4 \end{align} $$ Thus, adding $1$ to $(4)$ gives $$ 2^a+1\equiv2^r+1\pmod{2^b-1}\tag5 $$ where $0\le r\lt b$. Thus, for $b\ge3$, $2^b-2^r\ge4$, and so $$ 2\le2^r+1\le2^b-3\lt2^b-1\tag6 $$ so that $2^a+1\equiv2^r+1\not\equiv0\pmod{2^b-1}$. That is, $$ 2^b-1\nmid2^a+1\tag7 $$