# Understanding congruence equations

I'm trying to wrap my head around this:

$a \equiv b^{-1}\pmod{y}$

$x \equiv c^{-1} \pmod{y}$

I know all values but $x$ and $y$. Is it possible to solve this problem for $x$? Can anyone point out a good resource on this kind of problems?

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If $b,c$ have an inverse in $\pmod y$ then is valid :$$ab\equiv 1 \pmod y$$ and $$xc\equiv 1 \pmod y$$ From this you get: $$ab-1 = k_{1}y$$ $$xc-1 = k_{2}y$$

(For some $k_{1}$ and $k_{2}$ integers )

So you can find $x$ and $y$ from this ecuations system.

$Edit:$

For example,

If $y=5$, the elements with inverse are: $2,3$, also $b=2$ and $b^{-1}=3$.

Know that (with $a=8$ ):

$8 \equiv 3 \pmod 5$, is say , $a \equiv b^{-1} \pmod y$,

Note that also:

$16 =8 \cdot 2\equiv 1 \pmod 5$, is say , $ab \equiv 1 \pmod y$

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what are $k_1$ and $k_2$? –  Chris Nov 28 '10 at 18:49
Yocks: Just a LaTeX tip: for the congruence relation, it's better to use \pmod{x}, which takes an argument and automatically produces (mod x). –  Arturo Magidin Nov 28 '10 at 19:11
@Chris Remember that: $$a\equiv b \mod y \Leftrightarrow y\setminus(a-b)$$ –  Bryan Yocks Nov 28 '10 at 19:14
@Arturo : Thanks, I will use this –  Bryan Yocks Nov 28 '10 at 19:18