Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given $n>2$, by calculation or otherwise deduce that $5^{2^{n-3}} \neq -1 \pmod {2^n}$

Note:The problem arose when I tried to deduce $\langle5\rangle \cap \langle2^n-1\rangle=\{1\}$ in the group $\mathbb{Z}^{*}_{2^n}$, I have showed $\operatorname{ord}(5)=2^{n-2}$

share|cite|improve this question
Notice that if you write (mod 2^n) in LaTeX, it looks like this: $(mod 2^n)$; whereas if you write \pmod{2^n}, it looks like this: $\pmod{2^n}$. Also, instead of $<5>$, you can write $\langle5\rangle$. I did those and a couple of other $\TeX$ improvements. Also you need backslashes in \{5\} in order to get this: $\{5\}$. – Michael Hardy Aug 14 '12 at 1:46
Thanks! Do you also know how to type "not congruent to"? I tried both "\not\equiv" and "\cancel\equiv" but neither works here. – user31899 Aug 14 '12 at 2:06
a\not\equiv b $a\not\equiv b$. Seems to work. Is it possible that you typed a\not\equivb, with no space between \equiv and b? If you do that, it sees "\equivb" rather than "\equiv" and then "b". – Michael Hardy Aug 14 '12 at 2:11
$a \not\equiv b$ – user31899 Aug 14 '12 at 2:14
up vote 3 down vote accepted

We show by induction that if $n \ge 3$, then $5^{2^{n-3}}\equiv 1+2^{n-1}\pmod{2^n}$. And it is clear that $1+2^{n-1}\not\equiv -1 \pmod{2^n}$ if $n \ge 3$.

The result holds when $n=3$. Now we do the induction step. Suppose that we know that for a certain $k$, we have $5^{2^{k-3}}\equiv 1+2^{k-1}\pmod{2^k}$. We show that $5^{2^{k-2}}\equiv 1+2^{k}\pmod{2^{k+1}}$.

By assumption, $5^{2^{k-3}}=1+2^{k-1} +t2^k$ for some integer $t$. Square both sides, and simplify modulo $2^{k+1}$. We get $$5^{2^{k-2}}\equiv (1+2^{k-1})^2=1+2^k+2^{2k-2}\pmod{2^{k+1}}.$$ But $2^{2k-2}$ is divisible by $2^{k+1}$, since $2k-2 \ge k+1$ when $k \ge 3$. The result follows.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.