<s>[INST] 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.
 [/INST] Let us take your equation $7x=15k+3$ and multiply both sides by $13$.
We get $91x=195k+39$. From where $(6\cdot 15+1)x=15\cdot 13k+2\cdot 15+9$, or what is the same $x=15(13k-6x+2)+9$. Since $k$ was certain integer we could say that $K:=13k-6x+2$ is certain integer and get $x=15K+9$.
So, all solutions leave the same remainder $9$ after division by $15$.
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