Where are the Squares?

Alternative 1

We claim that $a_n=b_n^2+(-1)^n$ , where $b_n=2b_{n-1}+b_{n-2}$ , and $b_0=0$ and $b_1=2$ . Note that $a_n$ has the solution $a_n=\dfrac12\left(3-2\sqrt{2}\right)^n+\dfrac12\left(3+2\sqrt{2}\right)^n,$ and $b_n$ has the solution $b_n=\dfrac{1}{\sqrt{2}}\left(1+\sqrt{2}\right)^n-\dfrac{1}{\sqrt{2}}\left(1-\sqrt{2}\right)^n$ (both of which can be obtained by solving their respective characteristic equations). So we must prove that $\begin{aligned} \dfrac12\left(3-2\sqrt{2}\right)^n+\dfrac12\left(3+2\sqrt{2}\right)^n &= \left(\dfrac{1}{\sqrt{2}}\left(1+\sqrt{2}\right)^n-\dfrac{1}{\sqrt{2}}\left(1-\sqrt{2}\right)^n\right)^2+(-1)^n \\ &=\dfrac{\left(3+2\sqrt{2}\right)^n+\left(3+2\sqrt{2}\right)^n-2\left(1-\sqrt{2}\right)^n\left(1+\sqrt{2}\right)^n}{2}+(-1)^n \\ &=\dfrac12\left(3-2\sqrt{2}\right)^n+\dfrac12\left(3+2\sqrt{2}\right)^n, \end{aligned}$ which is indeed true. So our claim is true, and hence $a_n=b_n^2+(-1)^n$ , so $a_n$ is always $1$ away from a perfect square. But the only two perfect squares that have a difference of $1$ are $0$ and $1$ , and thus, due to the increasing nature of $a_n$ , we know that $a_0=1=1^2$ is the only perfect square in the sequence. Hence, the answer is $\boxed{1}$ .

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Alternative 2

Consider Pell's equation $x^2-2y^2=1$ (where $n=2$ ); it's fundamental solution is $(x_1,y_1)=(3,2)$ , since this is the smallest solution where both $x$ and $y$ are positive integers. The complete set of solutions for $x$ and $y$ can then be described as $\begin{aligned} x_n &= 3x_{n-1} + 4y_{n-1} \\ y_n &= 2x_{n-1} + 3y_{n-1} \end{aligned}$ (an elementary proof of this fact can be found here on the Brilliant wiki page). Combining these equations gives $\begin{aligned} \dfrac{1}{4}\left(x_{n+1}-3x_n\right) &= 2x_{n-1}+\dfrac{3}{4}\left(x_n-3x_{n-1}\right) \\ \Longrightarrow x_{n+1} &= 6x_n - x_{n-1} \\ \Longrightarrow x_n &= 6x_{n-1}-x_{n-2},\end{aligned}$ where $x_1=3$ and $x_2=17$ . But notice that $a_1=3$ and $a_2=17$ , and since their recurrence relations are the same, we have $x_n=a_n\;\; \forall \;\; n\geq 1$ .

Hence, letting $x^2=(r^2)^2=r^4,\;\; r\in \mathbb{Z}^+$ , it suffices to find all positive, integral solutions for $r$ in the equation $r^4-2y^2=1$ (where $y$ is also an integer, since the corresponding relation for $y$ produces all integers by consequence). Notice this implies $r$ is odd, and hence let $r=2m+1, \;\; m\in \mathbb{Z}^+\cup \{0\}$ . Factorising yields $\begin{aligned} 2y^2 &= 2m(2m+2)(4m^2+4m+2) \\ \Longrightarrow y^2 &= 4m(m+1)(2m^2+2m+1). \end{aligned}$ So $m(m+1)(2m^2+2m+1)=m(m+1)(2m(m+1)+1)$ must itself be a perfect square. But $m$ , $m+1$ , and $2m(m+1)+1$ are pairwise coprime, so they must all be perfect squares themselves. The only two perfect squares that are $1$ apart are $0$ and $1$ , implying that $m=0$ , and hence $r=1$ . So the only solution to the equation is $(r,y)=(1,0)$ , which corresponds to $x=a_0=1^2$ as the only perfect square in the sequence. Hence, the answer is $\boxed{1}$ .

Note: Alternative 1 is simpler, but does not show the inspiration behind the construction of $a_n$ .