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Let's say I have set S and T being the set of all integer solutions to $ax+by=c$ and $ax+by=nc$ respectively, and set S* might be the same as set T.

S* = $\{ (n x_0 + n y_0) | (x_0, y_0) \in S\}$

How would I prove that S* $\subseteq$ T for all values of a,b,c,n $\in \mathbb{Z}$

To be honest, I don't even understand the question. I know that $ax+by=c$ is a diophantine equation, and that there exists a complete integer solution where $x = x_0+ \frac{b}{d}n$, $y = y_0+ \frac{a}{d}n$ $\forall n \in\mathbb{Z}$. Sadly that's as far as I can go before asking for help.

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    @AndréNicolas: I am slowly beginning to understand this question as well as LDEs with both of your help. When you made$(3,7)$a solution, was it because of the property that if$ax+by=c$has a solution, then it has infinitely many solutions? I know you the example should be quite obvious but I still do not understand why nc is no longer of the form $n x_0, n y_0$2012-01-31

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The statement that $S^* \subseteq T$ is equivalent to showing that if $(x_0,y_0)$ is an integer solution to $ax+by=c$, then $(nx_0,ny_0)$ is an integer solution to $ax+by=nc$.

This is easy: if $ax_0+by_0 = c$, then $a(nx_0) + b(ny_0) = n(ax_0+by_0) = nc$. And of course, if $x_0$, $y_0$, and $n$ are integers, then so are $nx_0$ and $ny_0$.

However, it is not the case that every solution to $ax+by=nc$ is necessarily $n$ times a solution to $ax+by$, as André Nicolas shows in the comments with an explicit example.