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I'm currently doing exercises focused on applying concepts of group theory, Lagrange theorem, orders, homeomorphism. I'm stuck on part e of exercise 4. I guess I'm supposed to use the mentioned concepts to solve this but I can't see how. Besides, since it's the last part of a bigger exercise I think the previous parts have some to do here.

e. i) Find remainder of $2^{20} \div 253$

e. ii) Knowing that $2^{55} \equiv -45$ $(mod$ $253)$, find order of $2$ in $U(253)$

Just a clarification, we use $U(n)$ for the multiplicative group of integers modulo n

Moo
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xtreyreader
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2 Answers2

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We work in the ring $\Bbb Z/253$. (So equalities are in this ring, we work modulo $253$.)

(i) $2^{20}=(2^{10})^2=1024^2=(1012+12)^2=(0+12)^2=144$.

(ii) $U(253)$ is the group of units $G=U(253)=(\Bbb Z/253)^\times$ in the ring $\Bbb Z/253$. The number of elements of $G$ is $n=\phi(253)=\phi(11\cdot 23)=\phi(11)\cdot \phi(23)=10\cdot 22=220$.

They give us the value for $2^{55}$. Then $2^{110}=(2^{55})^2=(-45)^2=2025=1$. So the order of $2$ divides $110$ by (a), but it does not divide $55$ by (b).

So $2$ is a divisor of $o(2)$. We check first the powers $110/5=22$ and $110/11=10$ of $2$. Well, $2^{10}=1024=12\ne 1$, and $2^{22}=2^{10}\cdot 2^{10}\cdot 2^2=12\cdot 12\cdot 4= 576 = 70$.

So the order of $2$ in $G=U(253)$ is $110$.

dan_fulea
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  • You are implicitly using the Order Test - something which should be made explicit (now rectified in my answer). Possibly the OP does not even know that yet. – Bill Dubuque Oct 23 '20 at 23:31
  • In (i), when $(1012+12)^2=(0+12)^2$, what did you do? I supposed that if $a\equiv b$ $(mod$ $c)$ and $d$ integer then $(a+d)^2 \equiv (b+d)^2$ $(mod$ $c)$ and I just proved it, but is there a more general reason? – xtreyreader Oct 24 '20 at 17:37
  • Nevermind, just proved it for $n$ using basic properties. – xtreyreader Oct 24 '20 at 18:32
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One possibility is that they intend to use the Order Test, i.e. $\,2\,$ has order $\,110\,$ iff $\,2^{110}\equiv 1,\,$ $\,2^{110/p}\not\equiv 1\,$ for all primes $\,p\mid 110,\,$ i.e. for $\,110/p\,\in\, \{ 110/2\! =\! 55,\ 110/5\! =\! 22,\ 110/11 \!=\! 10\}$.

Another possibility is they intend you to lift its order up from $\bmod 11,23\,$ to $\bmod 11\cdot 23\,$ as below. Note $\!\bmod Q\!=\!23\!:\ a\equiv 2\equiv 25\equiv 5^2\,$ so the Lemma applies.

Lemma $ $ Suppose $\,P=1+2p,\, Q = 1+2q\,$ for primes $\,P,Q,p,q\,$ with $\,(2p,q)=1$.
If $\bmod P\!:\ a^p\equiv -1,\, \color{#90f}{a\not\equiv \pm 1};\ \bmod Q\!:\ a\equiv b^2,\ a\not\equiv 1\,$ then $\,a\,$ has order $2pq\bmod PQ$

Proof $\bmod P\!:\ a^{2p}\equiv 1,\ a^{p},\color{#90f}{a^{2}\not\equiv 1}\Rightarrow\ \color{#c00}{o(a) = 2p}\,$ by little Fermat and the Order Test.

$\, $ also: $\ \bmod Q\!:\ a^q\equiv b^{2q}\equiv 1,\ a\not \equiv 1\Rightarrow\, \color{#0a0}{o(a) = q}\ $ similarly, hence

$\ \ PQ\mid a^k-1\!\iff\! P,Q\mid a^k-1\!\iff\! \color{#c00}{2p},\color{#0a0}q\mid k\!\iff\! 2pq\mid k,\, $ so $\,o(a) = 2pq\bmod{PQ}$.

Bill Dubuque
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  • The Lemma could be generalized further, but hopefully that form already suffices to convey the key idea(s) that they exercise intends to highlight. If you know quadratic reciprocity then you may find it helpful to view it from that perspective. – Bill Dubuque Oct 23 '20 at 23:33