Incongruent solutions to $7x \equiv 3$ (mod $15$) I'm supposed to find all the incongruent solutions to the congruency $7x \equiv 3$ (mod $15$)
\begin{align*}
7x &\equiv 3 \mod{15} \\
7x - 3 &= 15k \hspace{1in} (k \in \mathbb{Z}) \\
7x &= 15k+3\\
x &= \dfrac{15k+3}{7}\\
\end{align*}
Since $x$ must be an integer, we must find a pattern for $k$ that grants this.  We know that $\frac{k+3}{7}$ must be equal to some integer, say $m$.  Solving for $k$, we have $k=4+7m$.
Substituting this into our value for $x$, we get:
\begin{align*}
x & = \dfrac{15(4+7m) + 3}{7}.\\
&= \dfrac{63}{7} + \frac{105m}{7}.\\
&= 9+15m.
\end{align*}
So, $x = 9+15m, m\in \mathbb{Z}.$
So, is this what I was looking for?  I'm not exactly sure what is meant by incongruent solutions.

$$\left ( 7,15 \right )=1 \Rightarrow$$ $$ x \equiv 3(7)^{^ \phi(15)-1 } (mod 15)$$$$x \equiv 3(7)^{^7}\equiv 3.13 \equiv 9 (mod15)$$, which is the unique solution by Euler-Fermat theorem.
