Proving that there are infinitely many composites in an arithmetic progression The question says

Consider $S$={$a, a+d, a+2d, ...$} where $a$ and $d$ are positive integers. Show that there are infinitely many composite numbers in $S$

The only argument I could think of was that primes aren't equally spaced. So, the rest of the numbers must be composite numbers, which are infinitely many in number. Is this a legit solution, or is there a more elegant way to prove this?
Would appreciate your help.
Thanks in advance
 A: Addendum added to cover the case of $(a=1)$.

There is a much easier answer than the ones provided in the link.
If $a,d$ is not relatively prime, then it is immediately game over.  That is, if $e|a$, and $e|d$, then none of the elements in the sequence can be prime.
So, assume that $a,d$ are relatively prime.  The following argument doesn't really require that assumption.  However, it does provide clarity to the situation.
Consider the terms in the positions $(a+1), (2a+1), (3a+1), \cdots.$
These terms will be $(a + ad), (a + 2ad), (a + 3ad), \cdots.$
Each of these terms will be divisible by $a$, and therefore composite.  Therefore, there will be an infinite number of composite terms.
The linked answer tackles a much tougher problem, and can therefore be regarded as overkill.

Addendum
Thanks to B. Goddard for finding the flaw in my analysis, re $(a = 1)$.  In that case, the term in position $r$ will be $1 + (r-1)d.$
Take $k \in \Bbb{Z^+}.$
Then the term in position $(d^{2k}+1)$ will be $1 + d^{2k}(d) = 1 + d^{2k+1}.$
Since $(2k+1)$ is odd, the number will be divisible by $(d+1)$.
Therefore, when $a=1$, there will still be an infinite number of composite terms.
