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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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1 Answer

up vote 3 down vote accepted

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
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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
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