modulo arithmetic

**aman_cc** · April 13th 2010, 03:56 AM

This question is inspired from a earlier post on this forum.

Prove that if

x.y = -1 mod 24

Then, prove

x = -y mod 24.

Is there a smart way to do this? All I could do was enlist the distinct values for x,y and prove the above for each such pair.

Thanks

**tonio** · April 13th 2010, 06:16 AM

Originally Posted by aman_cc

This question is inspired from a earlier post on this forum.

Prove that if

x.y = -1 mod 24

Then, prove

x = -y mod 24.

Is there a smart way to do this? All I could do was enlist the distinct values for x,y and prove the above for each such pair.

Thanks

The group of units modulo 24, better known as $\left(\mathbb{Z}\slash 24\mathbb{Z}\right)^{*}$ is the elementary abelian group $\left(\mathbb{Z}\slash 2\mathbb{Z}\right)^3$ , in which every non-unit element has order two (i.e., it's an involution)...

Tonio

**aman_cc** · April 14th 2010, 07:53 PM

Originally Posted by tonio

The group of units modulo 24, better known as $\left(\mathbb{Z}\slash 24\mathbb{Z}\right)^{*}$ is the elementary abelian group $\left(\mathbb{Z}\slash 2\mathbb{Z}\right)^3$ , in which every non-unit element has order two (i.e., it's an involution)...

Tonio

Thanks Tonio. I am afraid I did not understand everythign you said (I have limited exposure to Group Theory)

What you must be saying is that ans to my question is direct application of the fact you wrote. But is there a way to do it understand it without relying on this.

I will tell you my approach - I found all pairs of x,y (between 0 and 23) for which the first condition is satisfied. Then checked the result to be proved for each such case. And hence satisfied my self with it.

But I'm not too happy with this approach. What if instead of 24 - I had a a bigger number? Is 24 the only number that satisfies this - are there others? If yes - what is the criteria etc? (I know for sure there are a few more numbers like that). So, these are the questions I am struggling to answer.

Any direction / input will be great.

**chiph588@** · April 14th 2010, 08:35 PM

$xy\equiv-1\bmod{24}\implies x(-y)\equiv1\bmod{24}\implies -y\equiv x^{-1}\bmod{24}$

Now $\mathbb{Z}/24\mathbb{Z}$ is special in the sense that $x^{-1}\equiv x\bmod{24}$ for all $x$ with an inverse.

**Drexel28** · April 14th 2010, 11:06 PM

Originally Posted by chiph588@

$xy\equiv-1\bmod{24}\implies x(-y)\equiv1\bmod{24}\implies -y\equiv x^{-1}\bmod{24}$

Now $\mathbb{Z}/24\mathbb{Z}$ is special in the sense that $x^{-1}\equiv x\bmod{24}$ for all $x$ with an inverse.

.

**aman_cc** · April 15th 2010, 01:08 AM

Thanks but this is what I am not able to prove. I do not know / have never studied the group tonio mentioned in his post.

**chiph588@** · April 15th 2010, 07:10 AM

Originally Posted by aman_cc

Thanks but this is what I am not able to prove. I do not know / have never studied the group tonio mentioned in his post.

$\mathbb{Z}/24\mathbb{Z} = \{0,1,2,\ldots,23\}$

$(\mathbb{Z}/24\mathbb{Z})^\times = \{x\in \mathbb{Z}/24\mathbb{Z} \;|\; (x,24)=1 \} = \{1,5,7,11,13,17,19,23\}$

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