Let $a,b,c,n$ be non-zero integers.
Let $S$ be the set of integer solutions to $ax+by=nc$ and
$T$={$(nx,ny):(x,y)$ is an integer solution to $ax+by=c$}.
I need to prove/disprove that $S \subseteq T, T \subseteq S, S=T$.
I am thinking $S \subsetneq T$ because there are some integer solutions $(x,y)$ such that $(x,y) \in S$ but $(x,y) \notin T$. So then I would need to prove $T \subseteq S.$
Does anyone know how I can go about solving this...?
Thanks in advance!
$T\subset S$ because $(n x, n y)\in T\implies a (n x)+b (n y)=n(a x+b y)=n c\implies (n x, n y)\in S.$
For an example where $T\ne S,$ let $a=b=n=2$ and $c=1.$ Then $T$ is empty but $(1,0)\in S.$
For an example where $T=S,$ let $n=1.$
For an example where $\phi \ne T\ne S, $ let $a=3, b=5, c=1,n=2. $ Then $(9,-5)\in S$ and $(9,-5)\not \in T$ but $(2,-1)\in T.$
More generally : Let $e=\gcd (a,b)$.If $T\ne \phi $ then $e|c$ because any $a x+b y$ is divisible by $e.$
Suppose $e|c$ and $|n|>1$. There are integers $x_0, y_0$ with $a x_0+b y_0=e.$ Then $a(n x_0 (c/e)+ b/e)+b( n y_0 (c/e)-a/e)=(n c/e)(a x_0+b y_0)+(a b/e-b a/e)=(n c/e)e=n c.$
Observe that $n x_0 (c/e)$ and $n y_0 (c/e)$ are integers because $e|c$, and that $b/e,\; a/e$ are integers by def'n of $e.$
So $(n x_0 (c/e)+b/e, n y_0 (c/e)-a/e)$ belongs to $S.$ But it does not belong to $T$ because otherwise $n|(b/e)$ and $n|(a/e),$ which is impossible. ( Because $|n|>1$ and $\gcd (a/e,b/e)=\gcd (a/\gcd (a,b), b/\gcd (a,b) )=1.$)