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Following from questions 2090325 and 2090607, is it possible to show that any AP with first term $a$ and common difference $d$, where $a,d\in\mathbb N$ must contain a square and, following from question 2090325, infinitely many squares? If not, then what conditions can be placed on $a$ or $d$ such that this is true?

As a partial answer to the last part, from question 2090325, we can specify that if $a$ is a perfect square then there are infinitely many perfect squares in the AP.

Addendum

Following from comments by @lulu below, here's the Wikipedia entry for Quadratic Residues.

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    This is clearly false. Take $a=3$ and $d=9$. Every term is divisible by $3$ but not by $9$. – 2017-01-10
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    As a different sort of counterexample, let $a = 2$ and $d=5$. Note that $2+5k$ can never be a square as $2$ is not a square $\pmod 5$. – 2017-01-10
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    @lulu - thanks, good counterexamples. Any thoughts on the second part? – 2017-01-10
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    My second counterexample is the general answer. That is, there is some (hence infinitely many) squares in the progression iff $a$ is a quadratic residue $\pmod d$. That's more or less the definition of a quadratic residue. – 2017-01-10
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    @lulu - ok so that means that the only condition for the AP to have a square (and infinitely many squares) is that $a$ must be $0,1 \mod d$? – 2017-01-10
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    Of course not. $2$ is a square $\pmod 7$, say. Indeed $2+7\times 1=9$, a square. For $d=p$ a prime there are $\frac {p-1}2$ squares $\pmod p$. – 2017-01-10
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    @lulu - Great. Would you care to combine your comments into an answer? – 2017-01-10

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There is an obstruction $\pmod d$. Specifically, the goal is equivalent to the statement "$a$ is a square $\pmod d$". Indeed, $$a+dk=x^2\implies x^2\equiv a \pmod d$$ Conversely, solving $x^2\equiv a \pmod d$ means there is some integer $k$ with $a+dk=x^2$.

Thus, for example, you can not have a square in the progression $\{3,8,13,18,\cdots\}$ because $3$ is not a square $\pmod 5$.