Proving that $2^{n+2} \mid a^{2^{\large{n}}} - 1 $, for $a \in \mathbb{Z}$ odd, $n \in \mathbb{N}$

Question

How can I prove this? I was thinking of using strong induction on $n$, and I've managed to prove the base case using induction again on $a$,

$$2^3 \mid a^{2} - 1$$

but I couldn't work my way around the rest of the proof with the usual properties of divisibility (if $m \mid n$, then $a^m - 1 \mid a^n - 1$, etc).

Are you sure don't want $a$ odd? If $a$ is even, $a \pm 1$ are both odd, so $2^3 \not \mid a^2 - 1$ . . . Cheers! — Robert Lewis, May 01 '19 at 01:52
I am sure this is a duplicate. However, the first hit makes a bad duplicate target. — Jyrki Lahtonen, May 01 '19 at 08:36
Possible duplicate of Induction: prove using congruences. This looks better. — Jyrki Lahtonen, May 01 '19 at 08:39

score 2 · Answer 1 · answered May 01 '19 at 01:59

2

Hint: $$ a^{2^n} - 1 = (a^{2^{n-1}} - 1)(a^{2^{n-1}} + 1) $$

answered May 01 '19 at 01:59

Ethan Bolker

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We can also prove $a^{2^n}+1\equiv2\pmod8,$ hence herethe highest exponent of $2$ is $1$ – lab bhattacharjee May 01 '19 at 02:45

score 1 · Answer 2 · answered May 01 '19 at 02:30

Hint

If $t_m=a^{2^m}=1+b2^r,$. where $b,r$ are non-negative integers and $b$ is odd

$t_{m+1}=(1+b2^r)^2=1+b2^{r+1}(1+b2^{r-1})$

If $r-1>0,$ the highest power of $2$ that divides $t_{m+1}$ is $r+1$

$m=1\implies a^2=(2c+1)^2=8\cdot\dfrac{c(c+1)}2+1$ as $a$ is odd,$a=2c+1$(say)

$\implies2^3$ divides $a^2-1$

So, by induction the proposition can be established

score 1 · Answer 3 · answered May 01 '19 at 06:10

Induction.

The base case $n = 1$ reads

$2^3 \mid a^2 - 1; \tag 1$

this holds since, $a$ being odd, $a - 1$ and $a + 1$ are both even; since $a \pm 1$ are consecutive even integers, each is divisible by $2$ and one is divisible by $4 = 2^2$; this establishes (1).

We hypothesize

$2^{k + 2} \mid a^{2^k} - 1; \tag 2$

we have

$a^{2^{k + 1}} - 1 = (a^{2^k} - 1)(a^{2^k} + 1); \tag 3$

we also have

$2 \mid a^{2^k} + 1, \tag 4$

since $a$ is odd, implying $a^{2^k}$ is also odd; (2) and (4) imply

$\exists c, d \in \Bbb Z, \; 2^{k + 2}c = a^{2^k} - 1, \; 2d = a^{2^k} + 1; \tag 5$

thus, via (3),

$2^{(k + 1) + 2}cd = 2^{(k + 2) + 1}cd =(2^{k + 2}c)(2d) = (a^{2^k} - 1)(a^{2^k} + 1) = a^{2^{k + 1}} - 1; \tag 6$

and thus,

$2^{(k + 1) + 2} \mid a^{2^{k + 1}} - 1, \tag 7$

completing the induction and proving that

$2^{n + 2} \mid a^{2^n} - 1, \; \forall n \in \Bbb N. \tag 8$

Bach · Answer 4 · 2019-05-01T08:31:02.963

Let $\phi$ denote the Euler's totient function and by Euler's theorem we know that $$ a^{\phi(2^{n+2})}\equiv 1\pmod{2^{n+2}} $$ By the definition of Euler's totient fucntion we know that $$ \phi(2^{n+2})=2^{n+2}-2^{n+1}=2^{n+1} $$ so $$ (a^{2^{n}})^2\equiv 1\pmod{2^{n+2}} .$$ And we conclude that $$ a^{2^n}\equiv 1 \pmod{2^{n+2}} $$ since otherwise, $a$ would be either a primitive root mod $2^{n+2}$ contradicting the fact that No primitive root modulo $2^n$ for $n≥3$, or $ a^{2^k}\equiv1 $ for some $k\le n-1$, but then $a^{2^n}$ would be congrunt to $1$ mod $2^{n+2}$ too. Contradiction!

Proving that $2^{n+2} \mid a^{2^{\large{n}}} - 1 $, for $a \in \mathbb{Z}$ odd, $n \in \mathbb{N}$

4 Answers4