Solving linear congruencies (modulus) My book asks me to solve these different modular equations:
$$x + \bar 7 = \bar 3 \space \rm in \space \Bbb Z _{15}$$
$$x + x = \bar 3 \space \rm in \space \Bbb Z _7$$
$$ x + x + x = 5 \space \rm in \space \Bbb Z_7$$
where $\bar 7$, for example, means the congruence class of $7$ (mod 7 I think). So what's the diference in the $3$ equations?
For example, I know that a congruence have some properties. How can I transform the equations in congruencies so I can use these properties? These notations aren't clear for me. Could you help me to solve?
 A: In the first equation you work $\mod 15$ (this is what $\Bbb Z / 15 \Bbb Z = \Bbb Z _{15}$ means): you want to find a class $x$ such that added to the class $\bar 7$ you should get $\bar 3$. This means that $x = \bar 3 - \bar 7 = \overline {3 - 7} = \overline {-4} = \overline {15 - 4} = \overline {11}$. Note that because you are working $\mod 15$, $\overline {15} = \bar 0$, which means that you may "stick" a $15$ wherever you would normally put a $0$, and this is what I did under the long line above.
In the second and third equations you are required to work $\mod 7$. Note that $x + x = 2x$, but when working modulo some integer we put bars above numbers, so the second equation could be rewritten as $\bar 2 x = \bar 3$, which upon inverting $\bar 2$ gives $x = (\bar 2)^{-1} \space \bar 3$. Now, you cannot slip the power $-1$ under the bar, because this would lead to $\overline {(\frac 1 2)}$, and non-integer numbers do not have modular classes. In order to compute the inverse of $2$ in $\Bbb Z _7$ it is easiest to resort to Euler's theorem: the number of numbers coprime to $7$ is $\varphi(7) = 6$, so $\bar 2 ^{-1} = \bar 2 ^{\varphi(7) -1} = \bar 2 ^5 = \overline {2^5} = \overline {32} = \overline {4\cdot7 + 4} = \bar 4$ (because now $\bar 7 = \bar 0$), so $x = \bar 4 \cdot \bar 3 = \overline {12} = \bar 5$.
Similar to above, the third equation becomes $\bar 3 x = \bar 5$, so $x = (\bar 3) ^{-1} \space \bar 5 = \bar 3 ^{6-1} \space \bar 5 = \bar 3 ^5 \space \bar 5 = \overline {3^5 \cdot 5} = \overline {1215} = \overline {173 \cdot 7 + 4} = \bar 4$.
