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(Question): If an arithmetic progression of positive integers $a, a+d, a+2d, \dots$ contains a perfect square, then it must contain a perfect square strictly less than $a+2d\sqrt{a}+d^2$.

I noticed that \begin{equation} (m-d)^2=m^2-d(2m-d) \end{equation} But I have no clue on how to prove the "strictly less" part. Thank you in advance for your comment/answer!

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    $\begingroup$ well, you haven't said what $a$ and $d$ might be. $\endgroup$
    – Will Jagy
    Feb 5, 2021 at 5:14
  • $\begingroup$ Pls tell what do you mean by a and d. $\endgroup$
    – user876009
    Feb 5, 2021 at 5:23
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    $\begingroup$ Presumably $a$ is the first term and $d$ is the common difference. $\endgroup$
    – Toby Mak
    Feb 5, 2021 at 5:24
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    $\begingroup$ Possibly more relevant that the given limit is $(\sqrt a+d)^2$. But I think you probably need to show how many steps from one square in the AP to the next one. $\endgroup$
    – Joffan
    Feb 5, 2021 at 5:28
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    $\begingroup$ Deep apologies, and yes, $a$ is the first term and $d$ is the common difference. I have edited the question. :) $\endgroup$
    – Ciel
    Feb 5, 2021 at 14:49

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I assume the arithmetic progression is non-degenerate, i.e., $d \neq 0$. Also, since the progression is of positive integers, then $a \gt 0$ and $d \gt 0$.

Let $a + kd$, for some integer $k \ge 0$, be the smallest element of the progression which is a perfect square. Thus, there's a positive integer $e$ such that

$$a + kd = e^2 \tag{1}\label{eq1A}$$

You're asking to prove

$$e^2 \lt a + 2d\sqrt{a} + d^2 \tag{2}\label{eq2A}$$

As Joffan's question comment states, the right side is equivalent to $(\sqrt{a} + d)^2$. Thus, \eqref{eq2A} is equivalent to trying to prove

$$e \lt \sqrt{a} + d \tag{3}\label{eq3A}$$

Assume, instead, that

$$e \ge \sqrt{a} + d \implies e - d \ge \sqrt{a} \implies (e - d)^2 \ge a \tag{4}\label{eq4A}$$

Note, though, that

$$\begin{equation}\begin{aligned} (e - d)^2 & = e^2 - 2ed + d^2 \\ & = (a + kd) - 2ed + d^2 \\ & = a + (k - 2e + d)d \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

This is also a perfect square smaller than $e^2$ (due to $e \gt e - d \gt 0$ using \eqref{eq4A}), and of the form $a + jd$ for an integer $j = k - 2e + d$, so it's a member of the arithmetic progression if $j \ge 0$. However, since $e^2$ is the smallest perfect square which is a member of the progression, this means $(e - d)^2$ cannot be a member, so this requires $j \lt 0$ and

$$(e - d)^2 \lt a \tag{6}\label{eq6A}$$

Note this contradicts \eqref{eq4A}, so the assumption it is correct is false. This means \eqref{eq3A} and, thus, \eqref{eq2A}, must be true instead.

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