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this is the exercise I have:

In the ring $(\mathbb{Z}_{21}, +, \cdot)$:
i) Find the invertible elements;
ii) Find the zero divisors;
iii) For each invertible element find the inverse.

I have done this:
i)
since the $gcd(a,n) = 1$ where $a \ne 0\in \mathbb{Z}_{21}$ and $n = 21$, then the element $a$ has an inverse.
Or also, since $21 = 7 \cdot 3$, if $7 \nmid a \mbox{ and } 3 \nmid a$, then $a$ has an inverse.
Considering the things above, I have found them by inspection: $1,2,4,5,8,10,11,13,16,17,19,20$.
I know that they are few elements, and by attempts it is a good way, but, is there exists any algorithm to calculate them in a more rapid way?

ii)
since the $gcd(a,n) \ne 1$, or also since $21 = 7 \cdot 3$, if $7 \mid a \mbox{ and } 3 \mid a$, then $a$ is a zero divisor. So, by attempts, the zero divisors are $3,6,7,9,12,14,15,18$
Also, here, is there any other rapid algorithm to determine them?

iii)
for each invertible element must hold the following: $$ax \equiv 1 \mbox{ (mod 21) }$$ so,
$1x \equiv 1 \mbox{ (mod 21) }$
hence $1$ is the inverse of $1$.
$2x \equiv 1 \mbox{ (mod 21) }$
it means to solve this equation in $\mathbb{Z}_{21}$:
$\begin{array}{rcl}[2] \odot [x] & = & [1] \\ [2 \cdot x] & = & [1] \\ [2 \cdot 11 ] & = & [1] \\ [22] & = & [1] \\ [1] & = & [1]\end{array}$
so here the inverse of $2$ is $x = 11$.

In my book there is a hint: check if these following numbers are congruent modulo 21, they will help you to find the inverses:
$22,43,64,85,106,127,148,169,190,211,399=21 \cdot 19$
but, I don't know how to use them. Can you tell me anything about them?

Please, can you give any suggestions? Thanks!

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    Note that $1\equiv 1+21\equiv 22$ and similarly $1\equiv 1+2\cdot 21 \equiv 43$ etc... so the list of numbers they give you are all elements of the equivalence class of $1$. The hint is giving you a (*partial*) list of possible elements you can look for as outcomes of multiplications for elements to find the inverses with. E.g. $4\cdot 37=148$ and $148$ is on the list so therefore... – 2017-02-17
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    @JMoravitz Can you add some other details as answer below? I would want to accept your answer as the best one. – 2017-02-19

1 Answers 1

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we need $2x\equiv1(mod 21)\Rightarrow (2\times11 )x\equiv1\times11(mod21)\Rightarrow x\equiv11(mod 21) $

$4x\equiv1(mod 21)\Rightarrow (4\times5 )x\equiv1\times5(mod21)\Rightarrow -x\equiv5(mod 21) \Rightarrow x\equiv -5(mod 21)\equiv16(mod 21)$

$5x\equiv1(mod 21)\Rightarrow (4\times5 )x\equiv1\times4(mod21)\Rightarrow -x\equiv4(mod 21) \Rightarrow x\equiv -4(mod 21)\equiv17(mod 21)$

$8x\equiv1(mod 21)\Rightarrow (2\times8 )x\equiv 1\times2(mod21)\Rightarrow -5x\equiv2(mod 21) \Rightarrow -20x\equiv 8(mod 21)\Rightarrow x\equiv8(mod 21)$

and other elements in same way

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    it is not clear ... why after $-20x \equiv 8 \mbox{ (mod 21) }$ follows $\equiv 17$? – 2017-02-19
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    Soory, I copy last step , then edit and I forgot remove $\equiv 17$ . I edit my answer, $-20\equiv 1 (mod 21)$, not follows $\equiv 17$, – 2017-02-20