Prove that $ax + by = a + c$ has solutions if and only if $ax + by = c$ has solutions

Question

I suppose that $ax + by = c$ so $ax \equiv c \pmod b$ and there exists a solution $d = mcd(a,b)$ if and only if $d|c$ and I don't what can I do. Can you guide me, please?

Answer 1

Well, if $x_s$ and $y_s$ solve $ax + by = c$, we have:

$ax + by = a + c = a + ax_s + by_s \Rightarrow a(x-(x_s+1)) + b(y-y_s) = 0 \Rightarrow$

$\begin{align}x_r=x_s+1\\y_r=y_s\phantom{....}\end{align}\;$ solve $ax+by=a+c$

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Now, if $x_r$ and $y_r$ solve $ax+by = a+c$, we have:

$ax + by = c = ax_r-a+by_r \Rightarrow a(x-(x_r-1))+b(y-y_r) = 0 \Rightarrow$

$\begin{align}x_s=x_r-1\\y_s=y_r\phantom{....}\end{align}\;$ solve $ax+by=a$

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I don't know if it is formal enough for you, but I hope it helps anyway. Good luck!