This is the rest of the question Show that every element of $S$ is divisible by $d$. I don’t know where to start in this problem. Here is a hint: let $n$ be an element of $S$. Then there exist integers $q$ and $r$, with $0$ less than or equal to $r$ less than $d$, such that $n = qd + r$. Using the special nature of $n$ and $d$, argue that $r = 0$.
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So, $a,b$ are fixed and integers? And $x,y$ vary amongst integers? – Thomas Andrews Jan 28 '20 at 15:37
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Welcome to Mathematics Stack Exchange. Consider the remainder when an element of $S$ is divided by $d$ – J. W. Tanner Jan 28 '20 at 15:37
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1As a rule, the title should not be the beginning of the question, but a useful summary of the question. The full question should be in the body of the question. – Thomas Andrews Jan 28 '20 at 15:39
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Yes Thomas that is correct – Jan 28 '20 at 15:39
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If didn’t fit so I had to do it like that – Jan 28 '20 at 15:40
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Presumably, $S$ is the set of all positive integers of this form, or there is no such least $d.$ – Thomas Andrews Jan 28 '20 at 15:41
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S has to be divisible by d – Jan 28 '20 at 15:42