# If an arithmetic progression contains a perfect square, then it must contain a perfect square strictly less than ...

(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!

• well, you haven't said what $a$ and $d$ might be. Feb 5, 2021 at 5:14
• Pls tell what do you mean by a and d.
– user876009
Feb 5, 2021 at 5:23
• Presumably $a$ is the first term and $d$ is the common difference. Feb 5, 2021 at 5:24
• 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. Feb 5, 2021 at 5:28
• Deep apologies, and yes, $a$ is the first term and $d$ is the common difference. I have edited the question. :)
– Ciel
Feb 5, 2021 at 14:49

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

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

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

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

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

$$e \ge \sqrt{a} + d \implies e - d \ge \sqrt{a} \implies (e - d)^2 \ge a \tag{4}\label{eq4A}$$
\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
$$(e - d)^2 \lt a \tag{6}\label{eq6A}$$