It is easy to prove when $a>1$. Each element in this set is of the form $a+nd$. Whenever n is a multiple of a, then $n=ak$, thus $a+nd=a+ak=a(1+k)$. Thus n is composite. But the difficulty comes when $a=1$. I cannot find a way. Any hint will be much appreciated.
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1Assuming you have proved it for $a>1$, the $a=1$ case is easily done by writing $S={A-d, A, A+d, A+2d, \cdots}$ where $A=a+d>1$. – ultralegend5385 Nov 13 '21 at 13:27
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Take $n = k^2d-2k$ where $k\in \mathbb{N}$. Then $$1+nd = 1+k^2d^2-2kd =(kd-1)^2$$
so there is also infinitely many squares in it.
nonuser
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For $a\geq 1, d \geq 1$ let $n=1+k(a+d)$, then $$a+nd=(kd+1)(a+d).$$ Notice that $a+d\geq 2$.
More generally, if you have a polynomial $f(x)$ over integers and $n=f(m)$ for integers $n,m$, then $f(m+kn)$ is a multiple of $n$ for any integer $k$ (see Proving a polynomial $f(x)$ composite for infinitely many $x$). In your case you have $f(x)=dx+a$, so we have $f(1)=d+a$ and $d+a\mid f(1+k(d+a))$.
Sil
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