Prove that $5^{2^n}−1$ contains exactly $n+2$ factors $2$

Question

I have to prove that $5^{2^n}−1$ contains exactly $n+2$ factors $2$ and then I have to show that the order of $5$ in $(\mathbb{Z}/2^n\mathbb{Z})^*$ equals $2^{n−2}$ (for $n ≥ 2$).

I know that for the first part I can use $5^{2^{n+1}} −1 = (5^{2^{n}} −1)(5^{2^{n}} +1)$, but then I do not know to continue.

I don't understand the question. "contains exactly n+2 factors 2" what does that mean? — Ameet Sharma, Oct 21 '20 at 09:58
@Ameet Sharma We need to prove that $5^{2^n}-1$ is divisible by $2^{n+2}.$ — Michael Rozenberg, Oct 21 '20 at 10:00

Michael Rozenberg · Answer 1 · 2020-10-21T10:03:06.497

3

$5^{2^n}+1$ is even and not divisible by $4$.

Thus, it's enough to check that $5^2-1$ is divisible by $8$ and your statement is true by induction.

edited Oct 21 '20 at 10:03

answered Oct 21 '20 at 09:55

Michael Rozenberg

194,933

Or even $5^{2^0}-1$ divisible by $4$ ;) – metamorphy Oct 21 '20 at 09:55
@metamorphy It's divisible by $2^8$. It follows from the induction. – Michael Rozenberg Oct 21 '20 at 09:59
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Thankyou for the quick response. I shall try to prove it. – Mathlover Oct 21 '20 at 10:02
I mean $n=\color{red}{0}$ as the base. $5^{2^{\color{red}{0}}}-1$ is divisible by $4$ (because it is equal to $4$). – metamorphy Oct 21 '20 at 10:45
@metamorphy Oh yes of course! I just saw $5^{2^6}-1.$ I did not see $0$. Sorry. – Michael Rozenberg Oct 21 '20 at 10:47

score 1 · Answer 2 · answered Oct 21 '20 at 11:39

1

The OP can keep using the general identity $a^2 - 1 = (a+1)(a-1)$ to factor the integer.

For $n \ge -1$,

$\tag 1 5^{2^{n+1}} −1 = 4 \,{\displaystyle \prod _{k=0}^{n}\, (5^{2^{k}} +1)}$

answered Oct 21 '20 at 11:39

CopyPasteIt

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Prove that $5^{2^n}−1$ contains exactly $n+2$ factors $2$

2 Answers2