lyukhson wrote: Determine whether there exists an infinite sequence of nonzero digits $a_1 , a_2 , a_3 , \cdots $ and a positive integer $N$ such that for every integer $k > N$, the number $\overline{a_k a_{k-1}\cdots a_1 }$ is a perfect square.

There seems to be a mistake in last two conclusions: $l_1=t_1=\frac{k}{2}$. Example $25,625,5625$ then $x=25$ and $y=75$ we have $l_1=2,t_1=1$ , $l=1,t=5$. $l_1+t_1=k$ * When $2|k$ we get $l_1=t_1$ and $1\le l-t \le 2$. In this case $x=10^{t_1}$ or $5*10^{t_1-1}$ where $t_1 \le 1$ or $k \le 2$ * When $k$ is odd, we have $l_1-t_1=1$ and $l=1$, $t \le 6$ Hence $x=\frac{10-t}{2} \cdot 10^{t_1}$ and as the number which represents $x^2$ can't contain zero's, we have $t_1 \le 1$ or $k \le 3$. Third mistake: $l*t=10^r*a$ doesn't mean $l*t=a$ as $l=8$,$t=25$ gives $l*t=200$ Hence this hint alone doesn't lead to a solution. Remark: $min(t_1,l_1)\le 1$ or else $10|x$ and we have a zero in $x^2$.

The answer is no. Just a sketch since the solution is a bit long: assume the contrary and take $N$ even, and let $x_{N+1}, x_{N+2}, \dots$ denote the square roots. Thusly for every $k$ we have \[ x_{k+1}^2 - x_k^2 = a_{k+1} \cdot 10^k. \] By consider the exponent of $5$, prove that $5 = a_{N+2} = a_{N+4} = \dots$ (this is the main portion). Then, take $x_{k+2}^2 - x_K^2$ modulo $9$ to show that $4 = a_{N+3} = a_{N+5} = \dots$. Then show some $x_i^2$ is a nonresidue modulo $11$.

Basically,I first showed that if you have $v_5(x_k) \le \frac{k}{2}$,then the highest power of five dividing all the $x_i$'s would be same and used this to show a contradiction.Then I used this fact to show that $y_{2k+2} > 10^{2k+2}$ which is again a contradiction.So we are through.

Since a full solution hasn't been posted, here's a link to the official one - https://www.imo-official.org/problems/IMO2013SL.pdf

Answer: no. Suppose otherwise and let $x_{N + 1}, x_{N + 2}, \cdots$ be positive integers such that $x_k^2 = \overline{a_ka_{k - 1} \cdots a_1}.$ Recall that \begin{align*} v_p(a + b) \ge \min\{v_p(a), v_p(b)\} \text{ with equality if } v_p(a) \ne v_p(b). \quad (1) \end{align*}______________________________________________________________________________________________________________________________________________________ Claim: $v_5\left(x_k^2\right) \ge k$ for all $k > N.$ Proof: Suppose for the sake of contradiction that there exists an $M > N$ satisfying $v_5\left(x_M^2\right) < M.$ Then for any $k > M$, (1) implies that \begin{align*} v_5\left(x_k^2\right) = v_5\left(x_M^2 + \sum_{i = M + 1}^{k} a_i \cdot 10^{i - 1}\right) = v_5(x_M^2) < M. \quad (2) \end{align*}Now take $k >> N$ and set $\alpha = x_{k + 1} + x_k$ and $\beta = x_{k + 1} - x_k.$ Combining (1) and (2), we obtain \begin{align*} \min\{v_5(\alpha), v_5(\beta)\} \le v_5\left(2x_{k + 1}\right) < M. \quad (3) \end{align*}Then from the difference of squares $a_{k + 1} \cdot 10^k = x_{k + 1}^2 - x_k^2 = \alpha\beta$, we deduce that \begin{align*} k \le v_5\left(a_{k + 1} \cdot 10^k\right) = v_5(\alpha) + v_5(\beta) < \max\{v_5(\alpha), v_5(\beta)\} + M, \quad (4) \end{align*}where the last step follow from (3). Exponentiating, we obtain \begin{align*} 5^k < 5^{\max\{v_5(\alpha), v_5(\beta)\}} \cdot 5^M \le \alpha \cdot 5^M. \end{align*}However, since $\alpha < 2x_{k + 1} < 2 \sqrt{10^{k + 2}}$, (4) is obviously false for sufficiently large $k.$ This contradiction proves the claim. $\blacksquare$ It follows from (1) that \begin{align*} v_5(\alpha) &\ge \min\{v_5(x_{k + 1}), v_5(x_k)\} \ge k / 2, \\ v_5(\beta) &\ge \min\{v_5(x_{k + 1}), v_5(x_k)\} \ge k / 2. \quad (5) \end{align*}______________________________________________________________________________________________________________________________________________________ Now, we repeat the same shenanigans as above, with $5$ replaced by $2.$ Choose some $M > N$ and note that if $v_2\left(x_M^2\right) \ge M$, then the claim implies that \begin{align*} 10^M > x_M^2 \ge 2^{v_2\left(x_M^2\right)} \cdot 5^{v_5\left(x_M^2\right)} \ge 2^M \cdot 5^M, \end{align*}which is clearly false. Therefore $v_2\left(x_M^2\right) < M.$ Replacing $5$ by $2$ and mimicking the proof of the claim, we arrive at a modified version of (4): \begin{align*} k < \max\{v_2(\alpha), v_2(\beta)\} + M + 1. \quad (6) \end{align*}Exponentiating, we deduce that \begin{align*} \frac{2^k}{2^{M + 1}} < 2^{\max\{v_2(\alpha), v_2(\beta)\}}. \quad (7) \end{align*}Combining (5) and (7), we find that \begin{align*} 2\sqrt{10^{k + 2}} > 2x_{k + 1} > \max\{\alpha, \beta\} \ge 2^{\max\{v_2(\alpha), v_2(\beta)\}} \cdot 5^{k / 2} > \frac{2^k \cdot 5^{k / 2}}{2^{M + 1}}. \quad (8) \end{align*}However, it is clear that (8) is false for sufficiently large $k.$ This final contradiction completes the proof. $\square$

The answer is no. Here's a much shorter solution than the one I wrote three years ago. Same as the one in post #7, I think. The answer is no. Assume for contradiction such a sequence exists, and let $x_k = \sqrt{\overline{a_k a_{k-1} \dots a_1}}$ for $k$ large enough. Difference of squares gives \[ A_k \cdot B_k \coloneqq (x_{k+1} - x_k)(x_{k+1} + x_k) = a_{k+1} \cdot 10^k \]with $\gcd(A_k, B_k) = 2\gcd(x_k, x_{k-1})$ since $x_k$ and $x_{k-1}$ have the same parity. We now split the proof in two cases: First assume $\nu_5(x_k^2) = 2e < k$ for some index $k$. Then we actually need to have \[ 2e = \nu_5(x_k^2) = \nu_5(x_{k+1}^2) = \dotsb. \]Thus in this situation, we need to have $\min(\nu_5(A_k), \nu_5(B_k)) = e$, and thus \[ \max(\nu_5(A_k), \nu_5(B_k)) = k-e. \]So \[ \max\left( A_k, B_k \right) \ge 5^{k-e}. \] Otherwise, assume $\nu_5(x_k^2) \ge k$ for all $k$. Note in particular that $a_0 = 5$, thus all $x_k$ are always odd. So one of $A_k$ and $B_k$ is divisible by $2^{k-1}$. Hence for each $k$, \[ \max(A_k, B_k) \ge 2^{k-1} \cdot 5^{k/2}. \] However, we also know that \[ \max(A_k, B_k) < 2x_{k+1} < 2 \cdot \sqrt{9 \cdot 10^k} \]which is incompatible with both cases above, for sufficiently large $k$.

lyukhson wrote: Determine whether there exists an infinite sequence of nonzero digits $a_1 , a_2 , a_3 , \cdots $ and a positive integer $N$ such that for every integer $k > N$, the number $\overline{a_k a_{k-1}\cdots a_1 }$ is a perfect square. Suppose $a_1, a_2, \dots$ is such a sequence. Let $b_k \overset{\text{def}}{:=} \sqrt{\overline{a_k \dots a_1}}$ and $(b_k)_{k>N}$ be a sequence of positive integers. Observe that $$\underbrace{(b_{k+1}-b_k)}_{x_k} \cdot \underbrace{(b_{k+1}+b_k)}_{y_k}=10^{k}a_{k+1}$$for all $k>N$. Since $x_k<y_k<2 \cdot 10^{(k+1)/2}$ we see $5 \mid \gcd(x_k, y_k)$. (Otherwise, $5^k \mid x_k$ or $5^k \mid y_k$ leading to the *embarrassing* bound $5^k \le 2 \cdot 10^{(k+1)/2}$ for all $k>N$.) Consequently, $5 \mid \gcd(b_k, b_{k+1})$ and all terms $b_j$ for $j>N$ are odd (since $a_1 \ne 0$). Suppose $v_5(b_k^2) \ge k$ for all $k$ large enough. Then $\min{v_2(x_k), v_2(y_k)}=1$ so $2^{k-1} \cdot 5^{k/2}<\max \{x_k, y_k\}<2 \cdot 10^{(k+1)/2}$, a contradiction! So let $j>M$ be a large number, with $v_5(b_j^2)<j$. Suppose $v_5(b_j) \ne v_5(b_{j+1})$ then $v_5(x_j \cdot y_j) \le 2\min \{v_5(b_j), v_5(b_{j+1})\}<j$ a contradiction! Hence $v_5(b_{\ell})=t$ for some constant $t>0$ and all $\ell \ge j$. Then for large $\ell$ we have $5^{\ell-t} \mid x_{\ell}$ or $5^{\ell-t} \mid y_{\ell}$ forcing $5^{\ell-t} \le 2 \cdot 10^{(\ell+1)/2}$ again a contradiction! Note. The main idea is to let divisibility cause size issues. It's better to not focus on what the actual answer is; rather the heuristic is to "add necessary constraints" and see if we have any freedom at all.

Funny enough, this problem appeared in Crux Mathematicorum (p. 25). Interestingly, the solution also shows that the longest possible sequence of squares is actually length five: $25, 625, 5625, 75625,275625$.

We claim that there is no such sequence. Suppose there was, so let \[c_k=\overline{a_k\cdots a_1},\]and for sufficiently large $k$, let $c_k=b_k^2$. Note that \[b_{k+1}^2-b_k^2=10^k a_{k+1}.\]This equation will be the driver for most of our analysis. Lemma 1: For all $k\ge 1$, we have $\nu_5(c_k)\ge k$. Proof: Suppose there was some $k$ such that $\nu_5(c_k)=\ell<k$. Then, whenever we extend $c_k$ by any number of digits, it still has $\nu_5$ equal to $\ell$. Therefore, for all sufficiently large $k$, we have that $\nu_5(b_k)=\ell/2$. We see that $2b_k=(b_{k+1}+b_k)-(b_{k+1}-b_k)$, so \[\ell/2=\nu_5(b_k)\ge\min\{\nu_5(b_{k+1}-b_k),\nu_5(b_{k+1}+b_k)\}.\]However, since $(b_{k+1}-b_k)(b_{k+1}+b_k)=10^k a_{k+1}$, we see that one of $b_{k+1}-b_k$ or $b_{k+1}+b_k$ is at most $5^{\ell/2}2^k a_{k+1}$, so one of them is at least $5^{k-\ell/2}$. Thus, $b_{k+1}\ge 5^{k-\ell/2}/2$, so \[b_{k+1}^2\ge\frac{1}{4}\cdot 25^{k-\ell/2}>10^{k+1}.\]This is a contradiction as $b_{k+1}^2$ is a $k+1$ digit number. Thus, the lemma is proved. $\blacksquare$ Lemma 2: We have \[\min\{\nu_5(b_k),\nu_5(b_{k+1})\}\le\frac{k+\nu_5(a_{k+1})}{2}\]for all sufficiently large $k$. Proof: Suppose to the contrary we have that \[\nu_5(b_k),\nu_5(b_{k+1})>\frac{k+\nu_5(a_{k+1})}{2}.\]Then, \[\nu_5(b_{k+1}-b_k),\nu_5(b_{k+1}+b_k)>\frac{k+\nu_5(a_{k+1})}{2},\]so $\nu_5(b_{k+1}^2-b_k^2)>k+\nu_5(a_{k+1})$, which is clearly false as the left side is just $\nu_5(10^ka_{k+1})=k+\nu_5(a_{k+1})$. Thus, the lemma is proved. $\blacksquare$ Suppose $k$ is even. We see that $\nu_5(b_k),\nu_5(b_{k+1})\ge k/2$ from lemma 1, and in fact \[\min\{\nu_5(b_k),\nu_5(b_{k+1})\}\le k/2\]since the left side is an integer. Thus, $\min\{\nu_5(b_k),\nu_5(b_{k+1})\}=k/2$, so $\nu_5(b_k)=k/2$ since $\nu_5(b_{k+1})\ge\frac{k+1}{2}$ by lemma 1. Thus, for all even $k$, $\nu_5(b_k)=k/2$. Now, suppose $k$ is odd. By lemma 2 we have \[\min\left\{\nu_5(b_k),\frac{k+1}{2}\right\}\le\frac{k+\nu_5(a_{k+1})}{2}.\]By lemma 1, we have $\nu_5(b_k)\ge k/2$, but since $k$ is odd, this becomes $\nu_5(b_k)\ge\frac{k+1}{2}$. Thus, the left side of the above displayed equation is $\frac{k+1}{2}$, so $\nu_5(a_{k+1})=1$, so $a_{k+1}=5$. Thus, we have \[(b_{k+1}-b_k)(b_{k+1}+b_k)=5^{k+1}2^k.\]Since $\nu_5(b_{k+1})=\frac{k+1}{2}$ and $\nu_5(b_k)\ge\frac{k+1}{2}$, we must have \[\nu_5(b_{k+1}-b_k)=\nu_5(b_{k+1}+b_k)=\frac{k+1}{2}.\]Therefore, $b_{k+1}-b_k=5^{\frac{k+1}{2}}2^r$ and $b_{k+1}+b_k=5^{\frac{k+1}{2}}2^{k-r}$ for some $1\le r\le\frac{k-1}{2}$. From this, we can compute \[\frac{b_{k+1}}{b_k}=\frac{2^{k-2r}+1}{2^{k-2r}-1}.\]Note that \[\frac{b_{k+1}^2}{b_k^2}=\frac{5\cdot 10^k+b_k^2}{b_k^2}=1+5\frac{10^k}{b_k^2}>6,\]so we must have \[\frac{2^{k-2r}+1}{2^{k-2r}-1}>\sqrt{6}.\]This implies that $r=\frac{k-1}{2}$, so in fact \begin{align*} b_k &= 5^{\frac{k+1}{2}}2^{\frac{k-3}{2}} \\ b_{k+1} &= 3\cdot b_k \end{align*}for all odd $k$. However, this implies that $10\mid b_k$, or that $a_1=a_2=0$, which is the desired contradiction.

I see that most solution invoke $\nu_5(c_k) \ge k$ or $\nu_5(b_k^2) \ge k$, which involves bounding over $\nu_5$ over an object. I understand that the statement of the problem could rewrite to \[ x_{k + 1}^2 - x_k^2 = 10^{k + 1} a_{k + 1} \]But what motivates you to bound each $x_i$s? I mean the easiest thing I could relate to is \[ \nu_5(x_{k + 1}^2 - x_k^2) = k + 1 + \nu_5(a_{k + 1}) \]which doesn't work. Anyone?

Define $n_k=\overline{a_k\cdots a_1}$, and $x_k^2=n_k\forall k>N$. Now, note that $x_k^2-x_{k-1}^2=10^{k-1}*d$ for some $d\in\{1,\ldots,9\}$. So, if we define the sequence $y_k=\nu_5(x_k)$, we must have $y_k=y_{k-1}\le \left\lfloor\frac{k}{2}\right\rfloor$ or $\min(y_k,y_{k-1})=\left\lfloor\frac{k}{2}\right\rfloor$, since if $y_k\neq y_{k-1}$, $\nu_5(x_k^2-x_{k-1}^2)=2\min(y_k,y_{k-1})=k-1, k$, and exactly one will work by parity. First, suppose there exists $i$ such that $y_i=y_{i-1}<\left\lfloor\frac{i}{2}\right\rfloor$. Then, $y_i$ makes $\min(y_{i+1},y_i)$ too small, so $y_{i+1}=y_i$ as well, and inducting upwards gives that the sequence is constant. If we let this constant value be $v$, note that at most one of $x_k-x_{k-1}$, $x_k,x_{k+1}$ for $k>i$ has $\nu_5>v$. So, we must have one of them has $\nu_5=v$, and the other one $\nu_5=k-v, k-1-v$. However, this means that $\max(x_k,x_{k+1})>\frac{5^{k-1-v}}{2}$, and for large enough $k$ we have $\left(\frac{5^{k-1-v}}{2}\right)^2\gg 10^k$, which is a contradiction, since $\max(x_k,x_{k+1})^2$ has at most $k$ digits. So, we must always have $\min(y_k,y_{k-1})=\left\lfloor\frac{k}{2}\right\rfloor$. If $\exists y_{2m}>m$, note that we need $y_{2m+1}=m$. However, then we get $\min(y_{2m+1},y_{2m+2})\le m < m+1$, which is a contradiction. So, $y_{2m}=m, y_{2m+1}\ge m+1\forall m$. Now, note that we can WLOG all $x_i$s are odd, since otherwise they end with $0$s, and we can remove 0s will preserving the goodness of the sequence. As exactly one of $x_{k}-x_{k-1}$ and $x_k+x_{k-1}$ is divisible by $4$, we must have that one of them is divisible by $2^{k-2}$. So, $\max(y_k,y_{k-1})\ge\frac{2^{k-2}\cdot 5^{\lceil k/2\rceil}}{2}=\Omega((2\sqrt{5})^k)>\Omega(\sqrt{10}^k)$. So, we eventually reach a contradiction by order of magnitude.

Let $x_k^2=\overline{a_k a_{k-1}\cdots a_1}$. Then $x_{k+1}^2-x_k^2=10^ka_{k+1}$; note that $1\le a_{k+1} \le 9$. We do casework on $v_5(x_k^2)=2\ell$. Case 1: $v_5(x_{k_0}^2)< k_0$ for some $k_0$. Then $x_{k+1} \equiv x_k^2 \pmod{10^{k_0}}$ for all $k\ge k_0$. If $v_5(x_{k+1}^2) \ge v_5(x_k^2)$, then $v_5(x_{k+1}^2-x_k^2)=v_5(x_k)^2<k$, contradiction. Hence $v_5(x_{k+1}^2)=2\ell$. Hence at least one of $v_5(x_{k+1}-x_k)$ or $v_5(x_{k+1}+x_k)$ is equal to $\ell$. We know $v_5((x_{k+1}-x_k)(x_{k+1}+x_k)) \in \{k,k+1\}$; hence the other is equal to $k-\ell$ or $k+1-\ell$. In particular, $x_{k+1}-x_k \ge 5^{k-\ell}$. On the other hand, $(x_{k+1}-x_k)^2\le x_{k+1}^2-x_k^2\le 10^{k+1}$, so $5^{k-\ell} \le x_{k+1}-x_k < 3.4^k$. This is a contradiction for large enough $k$. Case 2: $v_5(x_k^2)\ge k$ for all $k$. Then $v_5(x_1^2)=v_5(a_1^2)\ge 1$, so $5\mid a_1$. Since $a_i$ are nonzero, hence $a_1=5$. So all the $x_i$'s are odd. Now, $v_2((x_{k+1}-x_k)(x_{k+1}+x_k)) \ge k$, but the smaller of $v_2(x_{k+1}-x_k)$ and $v_2(x_{k+1}+x_k)$ is 1 since all the $x_i$'s are odd. Therefore, the larger of the two is at least $k-1$. By the assumption of this case, we have $v_5(x_k) \ge \lfloor k/2 \rfloor$ for all $k$. Therefore, $v_5(x_{k+1}-x_k),v_5(x_{k+1}-x_k)$ are both at least $k/2$. In particular, one of these two is at least $2^{k-1}5^{k/2}$, while it can be at most $\sqrt{10^{k+1}}$. The former is a constant times $(2\sqrt5)^k$, while the latter is $\sqrt{10}^k$; contradiction.

The answer is no. Firstly let $x_n=\overline{a_na_{n-1}...a_1}$ and $y_n=\sqrt{x_n}$ Notice that if $i<j$ then \begin{align} x_i&\equiv x_j\pmod{10^i}\\ 10^{n-1}\leq & x_n<10^n \end{align}The key observation is CLAIM. For all sufficiently large $n$ we have $$v_5(x_{2n-1})\geq 2n$$Proof. For all $2n-1>N$, $x_{2n-1}$ is a perfect square, so $y_{2n-1}$ is an integer and $v_5(x_{2n-1})$ is even. Suppose on the contrary that $v_5(x_{2n-1})<2n$, then $v_5(x_{2n-1})\leq 2n-2$. Therefore, for every $j>2n-1$, from $(1)$ we have $$5^{2n-1}|10^{2n-1}|x_j-x_{2n-1}$$Therefore $v_5(x_j)=v_5(x_{2n-1})$. Let this quantity be $2k$. For all $j>2n+1$ we may let $y_j=5^kz_j$, where $5\nmid z_j$. Now pick a sufficiently large $j$. From $(1)$ we have \begin{align*} y_{j+1}^2&\equiv y_j^2\pmod{10^j}\\ 5^{2k}z_{j+1}^2&\equiv 5^{2k}z_j^2\pmod{10^j}\\ \end{align*}Hence we have $$5^{j-2k}|z_{j+1}^2-z_j^2=(z_{j+1}+z_j)(z_{j+1}-z_j)$$Since $5\nmid z_j$, only one of $z_{j+1}+z_j$ and $z_{j+1}-z_j$ is divisible by $5$. Therefore either $5^{j-2k}|z_{j+1}+z_j$ or $5^{j-2k}|z_{j+1}-z_j$. In either case, we must have $z_{j+1}+z_j\geq 5^{j-2k}$. Hence $$y_{j+1}+y_j\geq 5{j-k}$$Since $y_j^2\geq 10^{j-1}$, we have \begin{align*} 10^{j+1}>y_{j+1}&\geq(5^{j-k}-10^{\frac{j-1}{2}})^2\\ 10^{\frac{j+1}{2}}+10^{\frac{j-1}{2}}&>5^{j-k}\\ 10^{\frac{j-1}{2}}\cdot 11&>5^{j-k}\\ 2^{\frac{j-1}{2}}\cdot 11&>5^{\frac{j+1}{2}-k}\\ \end{align*}which fails for sufficiently large $j$. Hence let $x$ be a sufficiently large integer. Let $y_{2x-1}=a\cdot 5^x$ and $y_{2x+1}=b\cdot 5^{x+1}$, where $a$ and $b$ are integers. Then from $(1)$ we have $$2^{2x-1}|10^{2x-1}|y_{2x+1}^2-y_{2x-1}^2=5^{2x+2}b^2-5^{2x}a^2$$Hence $$2^{2x-1}|(5b)^2-a^2=(5b+a)(5b-a)$$Since $v_5(x_{2n-1})\geq 2n$ and $a_1\neq 0$, $a_1=5$, hence $a$ and $b$ are odd. Therefore, either one of $5b+a$ and $5b-1$ is divisible by $4$. Therefore either $2^{2x-2}|5b+a$ or $2^{2x-2}|5b-a$ which implies $$5b+a\geq 2^{2x-2}$$However, from $(2)$, we have $$10^{2x-2}<a^2\cdot 5^{2x}<10^{2x-1}$$$$5^{2x+2}b^2<10^{2x+1}$$and hence $$5b<\sqrt{10}\cdot 2^x$$and $$a<\frac{2^{2x-2}}{25}<5b$$Therefore, $$2\sqrt{10} 2^x\geq 5b+a>2^{2x-2}$$which cannot hold for sufficiently large $x$, so we are done.

No such sequence exists. Assume such a sequence exists, and let \(x_k^2=\overline{a_ka_{k-1}\cdots a_1}\) for all \(k\). Then for all \(k\), \[x_{k+1}^2-x_k^2\in\{10^k,2\cdot10^k,\ldots,9\cdot10^k\}.\] Suppose there is a \(k>N\) with \(\nu_5(x_k^2)<k\). Since \[x_k^2\equiv x_{k+1}^2\equiv x_{k+2}^2\equiv\cdots\pmod{10^k},\]we have \(\nu:=\nu_5(x_k^2)=\nu_5(x_{k+1}^2)=\cdots\). Then take \(j\gg k\), and note that \(10^j\mid(x_{j+1}-x_j)(x_{j+1}+x_j)\). Evidently \[\nu=\nu_5(2x_{j+1})\ge\min\big\{\nu_5(x_{j+1}-x_j),\nu_5(x_{j+1}+x_j)\big\},\]so \(\nu_5(x_{j+1}\pm x_j)\ge j-\nu\) by choosing the sign appropriately. Then \[x_{j+1}^2\ge\left(\frac{x_{j+1}\pm x_j}2\right)^2\ge\left(\frac{5^{j-\nu}}2\right)^2\ge10^{j+1}\]for large enough \(j\), contradiction. $~$ Otherwise, suppose \(\nu_5(x_k^2)\ge k\) for all \(k>N\). Since 5 divides \(x_k\), \(a_1=5\), so all the \((x_k)\) are odd. Take \(j\gg k\), and note that \(10^j\mid(x_{j+1}-x_j)(x_{j+1}+x_j)\). Evidently \[1=\nu_2(2x_{j+1})\ge\min\big\{\nu_2(x_{j+1}-x_j),\nu_2(x_{j+1}+x_j)\},\]so \(\nu_2(x_{j+1}\pm x_j)\ge j-1\) by choosing the sign appropriately. Recall that \(\nu_5(x_{j+1}\pm x_j)\ge j/2\), so \[x_{j+1}^2\ge\left(\frac{x_{j+1}\pm x_j}2\right)^2\ge\left(2^{j-2}5^{j/2}\right)^2\ge10^{j+1}\]for large enough \(j\), contradiction.

The answer is no. Suppose otherwise, and denote $\overline{a_ka_{k-1}\ldots a_1}$ as $b_k^2$ (where $k>N$). We note that: $$b_k^2-b_{k-1}^2=10^{k-1}a_k \in \{10^k,2\cdot 10^k,\ldots,9\cdot 10^k\}$$which can also be written as: $$b_{k-1}^2+10^{k-1}a_k=b_{k}^2.$$We will break up the remainder of the proof into a number of key claims: Claim 1: We must have $v_5(b_k^2)\geq k$ for all sufficiently large $k$. Proof: Suppose that for some $k$ we have $v_5(b_k^2)<k$. Using the fact that $v_p(a+b)=\min\{v_p(a),v_p(b)\}$ when $v_p(a) \neq v_p(b)$, we obtain that $v_5(b_{k+1}^2)<k,v_5(b_{k+2}^2)<k$ and so on. Call this value $n$. Now pick some $j\gg k$. We require: $$b_j-b_{j-1}^2=10^{j-1}a_j \implies v_5(b_j+b_{j-1})+v_5(b_j-b_{j-1})\geq j-1.$$We can find that $v_5(\gcd(b_j+b_{j-1},b_j-b_{j-1}))\leq v_5(2b_{j-1})=n$, so at least one of $v_5(b_j+b_{j-1})$ and $v_5(b_j-b_{j-1})$ will be equal to $n$. Suppose the latter is equal to $n$. Then we require: $$v_5(b_j+b_{j-1})\geq j-n \implies b_j+b_{j-1} \geq 5^{j-1-n}$$since $b_j+b_{j-1}$ is nonzero (and positive). However, by definition we require $10^{(i-1)/2} \leq b_i<10^{i/2}$ for all $i$, so we have: $$b_j+b_{j-1}<10^{j/2}+10^{(j-1)/2}=\sqrt{10}^j+\sqrt{10}^{j-1}<5^{j-1-n}$$since $j$ is arbitrarily large and $\sqrt{10}<5$. The case where $v_5(b_j-b_{j-1})=n$ can be dealt with similarly. This generates a contradiction, so we cannot have $v_5(b_k^2)<k$ for any $k$. Observe that since the digits are nonzero, this claim implies that we require $a_1=5$, and that all the $b_i$ are odd. Claim 2: For all sufficiently large $k$, we require $a_{2k}=5$. Proof: Suppose otherwise. Then $v_5(10^{2k-1}a_{2k})=2k-1$. Now, by claim 1, we require $v_5(b_{2k-1}^2) \geq 2k-1$. But $b_{2k-1}^2$ is a perfect square, so we require $v_5(b_{2k-1}^2)$ even. This implies $v_5(b_{2k-1}^2 \geq 2k$. Then we have: $$v_5(b_{2k}^2)=v_5(b_{2k-1}^2+10^{2k-1}a_{2k})=\min\{v_5(b_{2k-1}^2),v_5(10^{2k-1}a_{2k})\}=2k-1$$which implies that $v_5(b_{2k}^2)<2k$: contradiction. Thus we require $a_{2k}=5$ for sufficiently large $k$. Claim 3: We require $v_5(b_{2k})=k$ for sufficiently large $k$. Proof: Suppose otherwise. Then $v_5(b_{2k}) \geq k+1$ and therefore $v_5(b_{2k}^2+10^{2k}a_{2k+1})\in \{2k,2k+1\}$. However it cannot be the second since it must be even, so we end up with $v_5(b_{2k+1}^2)=2k$ which is a clear contradiction to claim 1. Note that this claim combined with claim 1 is enough to imply that $v_5(b_{2k+1}-b_{2k})=v_5(b_{2k+1}-b_{2k})=k$, since $v_5(b_{2k+1})>v_5(b_{2k})$. Now we will finish the proof. Looking at $b_{2k+1}^2-b_{2k}^2=10^{2k}a_{2k+1}$ and taking the $v_2$ of both sides, we obtain: $$v_2(b_{2k+1}+b_{2k})+v_2(b_{2k+1}-b_{2k})\geq 2k.$$Since $b_{2k}$ and $b_{2k+1}$ are both odd, we must have one of $v_2(b_{2k+1}+b_{2k})$ and $v_2(b_{2k+1}-b_{2k})$ be equal to one. Suppose the latter is equal to one. Then we require: $$v_2(b_{2k+1}+b_{2k})\geq 2k-1.$$Combining this with $v_5(b_{2k+1}+b_{2k})=k$, we require (since $b_{2k+1}+b_{2k}$ is positive): $$b_{2k+1}+b_{2k}\geq 5^k2^{2k-1}.$$However, we also have: $$b_{2k+1}+b_{2k}<10^{k+1/2}+10^{k}.$$Since $5\cdot 2^2>10$, for sufficiently large $k$ we obtain $10^{k+1/2}+10^{k}<5^k2^{2k-1}$, which is a contradiction. The case where $v_2(b_{2k+1}+b_{2k})=1$ is similar and leads to the same size contradiction. Therefore we conclude that such a sequence does not exist. $\blacksquare$

lyukhson wrote: Determine whether there exists an infinite sequence of nonzero digits $a_1 , a_2 , a_3 , \cdots $ and a positive integer $N$ such that for every integer $k > N$, the number $\overline{a_k a_{k-1}\cdots a_1 }$ is a perfect square. The answer is negative. Define $b_k^2:\stackrel{\text{def}}{=}\overline{a_k a_{k-1}\cdots a_1 }$ ____________________________________________________________________________________________ Claim:- For all $k\ge 1$, we must have $\nu_5(b_k^2)\ge k$. Proof: FTSOC, assume that for some $k$ we have $\nu_5(b_k)=\ell<k$ However \[\ell/2=\nu_5(b_k)\ge\min\{\nu_5(b_{k+1}-b_k),\nu_5(b_{k+1}+b_k)\}.\]However, since $(b_{k+1}-b_k)(b_{k+1}+b_k)=10^k a_{k+1}$,so atleast one of $b_{k+1}-b_k$ or $b_{k+1}+b_k$ is at most $5^{\ell/2}2^k a_{k+1}$, so one of them is at least $5^{k-\ell/2}$. Thus, $b_{k+1}\ge 5^{k-\ell/2}/2$, so\[b_{k+1}^2\ge\frac{1}{4}\cdot 25^{k-\ell/2}>10^{k+1}.\], contradiction. ____________________________________________________________________________________________ However since by difference of squares $$(b_k+b_{k-1})(b_k-b_{k-1})=b_k^2-b_{k-1}^2=10^{k-1}a_{k+1} \in \{10^k,2\cdot 10^k,\ldots,9\cdot 10^k\}$$so $v_5(a_{k+1})>2k-1$,impossible if $k$ is large enough.

Suppose there is such a sequence. Let $a^2=\overline{a_k a_{k-1} \dots a_1 a_0}$ for some huge $k$ and $b^2=a^2+10^ka_{k+1}.$ Then $(b-a)(a+b)=10^ka_{k+1}$. Since the digits are non-zero, this means $(a,10)=(b,10)=1$. Anyways, write $$b-a=2^x\cdot 5^y\cdot d$$$$a+b=2^{k-x}\cdot 5^{k-y}\cdot \frac{a_{k+1}}{d}$$where $d\mid a_{k+1}$. Hence $$b=2^{x-1}5^yd+2^{k-x-1}5^{k-y}\frac{a_{k+1}}{d}$$Now, it is relatively easy to prove that $x<k-x$ and $y<k-y$ using that $b-a<<b+a$ and $a_{k+1}<10$. Case 1: $x>1$ Since $10\nmid b\Rightarrow y=0\Rightarrow b=2^{x-1}d+2^{k-x-1}5^{k}\frac{a_{k+1}}{d}$. Let $c^2=b^2+10^{k+1}a_{k+2}$. Work similarly to now and get $$2^{x-1}d+2^{k-x-1}5^{k}\frac{a_{k+1}}{d}=b=2^{k-x_1}5^{k+1-y_1}\frac{a_{k+2}}{d_1}-2^{x_1-1}5^{y_1}d_1$$where $x_1-1<k-x_1, y_1<k+1-y_1, d_1\mid a_{k+2}$. It is obvious why $d\neq 5\Rightarrow y_1=0\Rightarrow 2^{k-x_1}5^{k+1}\frac{a_{k+2}}{d_1}=2^{x_1-1}d_1+2^{x-1}d+2^{k-x-1}5^k\frac{a_{k+1}}{d}$. Since $v_2(LHS)=v_2(RHS)\Rightarrow x=x_1\Rightarrow 5^k\mid d+d_1$, which can't be true for big enough $k$'s. Case 2: $x=1$ $b=5^yd+2^{k-2}5^{k-y}\frac{a_{k+1}}{d}$. It is obvious why $2\nmid d$. Cow consider $c$ as in the previous case and get $$5^yd+2^{k-2}5^{k-y}\frac{a_{k+1}}{d}=b=2^{k-x_1}5^{k+1-y_1}\frac{a_{k+2}}{d_1}=2^{x_1-1}5^{y_1}d_1$$Since $10\nmid b\Rightarrow x_1=1\Rightarrow 5^yd+2^{k-2}5^{k-y}\frac{a_{k+1}}{d}=2^{k-1}5^{k+1-y}\frac{a_{k+2}}{d_1}-5^{y_1}d_1$. Since $v_5(LHS)=v_5(RHS)\Rightarrow y=y_1\Rightarrow 2^{k-1}\mid d+d_1$, which is false for big enough $k$'s. Case 3: $x=0$ In this case we have $a-b\equiv 1(\text{mod }2)$, which is false because $b^2=a^2+10^ka_{k+1}$. Finally, we got to a contradiction, hence there is not such a sequence.

I forgot to post this after writing it up, here is a solution with some motivation (integrated within the solution as my thought process). $\textbf{Pre-Solution:}$ The answer is naturally no. We have the following natural claim.

$\textbf{Lemma 1:}$ For all $k \in \mathbb{N}$, $5^k \mid \sum_{i=0}^{k-1} a_{i+1} \cdot 10^i$ $\textbf{Proof)}$ Before proving the lemma, we shall prove the following sub-lemma. $\textbf{Sub-Lemma 1:}$ For all $C \in \mathbb{N}$ there exists some $k \in \mathbb{N}$ such that $v_5(x_k^2) \geq C$ $\textbf{Proof)}$ Assume for the sake of contradiction that $\nu_5(x_k^2) < C$ for some $C \in \mathbb{N}$ and for all $k \in \mathbb{N}$. We have established that $$(x_k-x_{k-1})(x_k+x_{k-1}) = a_k \cdot 10^{k-1}$$Consequently, $x_k-x_{k-1} = a_k^{\varepsilon_1} \cdot 2^{\alpha_1} \cdot 5^{\beta_1}$ and $x_k+x_{k-1} = a_k^{\varepsilon_2} \cdot 2^{\alpha_2} \cdot 5^{\beta_2}$ where $\{\varepsilon_1, \varepsilon_2 \} = \{0,1\}$, $\alpha_1+\alpha_2 = k-1, \alpha_1, \alpha_2 \geq 1$ and $\beta_1 + \beta_2 = k-1$. Then, notice that $\min(\beta_1, \beta_2) < C$ and therefore $max(\beta_1, \beta_2) \geq k-C$ because otherwise $v_5(x_k^2) \geq C$. Then as $$5^{k-C} > 9 \cdot 2^{k-1} \geq a_k \cdot 2^{k-1}$$for all sufficiently large $k \in \mathbb{N}$ we must have that $\nu_5(x_k+x_{k-1}) \geq k-C$ we therefore get as $\alpha_1, \alpha_2 \geq 1$ that $$x_k \geq \frac{(x_k+x_{k-1})+(x_k-x_{k-1})}{2} \geq \frac{2 \cdot 5^{k-C}+2}{2} > 5^{k-C}$$Also, noticing that $10^k > x_k^2 > 5^{2(k-C)}$ we consequently get that for all sufficiently large $k \in \mathbb{N}$, $$(\frac{10}{25})^k > \frac{1}{5^C}$$which is clearly not the case and thus a contradiction. $\blacksquare$ Now that we have concluded the sub-lemma, we can quickly prove the initial lemma. Then for each $T \in \mathbb{N}$ there exists sufficiently large $k \in \mathbb{N}$ such that $$0 \equiv x_k^2 \equiv \sum_{i=0}^{k-1} a_{i+1} \cdot 10^i \equiv \sum_{i=0}^{T-1} a_{i+1} \cdot 10^i \pmod{5^T}$$which immediately proves the lemma. $\blacksquare$ $\textbf{Lemma 2:}$ For all sufficiently large even $k$, $a_k = 5$ $\textbf{Proof)}$ Assume for the sake of contradiction that $a_k \neq 5$. We can again recall the equations $x_k-x_{k-1} = a_k^{\varepsilon_1} \cdot 2^{\alpha_1} \cdot 5^{\beta_1}$ and $x_k+x_{k-1} = a_k^{\varepsilon_2} \cdot 2^{\alpha_2} \cdot 5^{\beta_2}$ where $\{\varepsilon_1, \varepsilon_2 \} = \{0,1\}$, $\alpha_1+\alpha_2 = k-1, \alpha_1, \alpha_2 \geq 1$ and $\beta_1 + \beta_2 = k-1$. We will split into the following cases. $\textbf{Case 1)}$ $k$ is even This means that $\beta_1 \neq \beta_2$ due to parity reasons. Then $\min(\beta_1, \beta_2) = v_5(x_k) \geq \frac{k}{2}$ which is clearly not possible as $$\min(\beta_1,\beta_2) \leq \frac{\beta_1+\beta_2}{2} = \frac{k-1}{2}$$which concludes the lemma. $\blacksquare$ Now, again working with the previous set-up, one can see that $$\min(v_5(\varepsilon_1+\beta_1, \varepsilon_2+\beta_2) \geq \frac{k}{2}$$where $\beta_1+\beta_2 = k-1$ ultimately meaning that $\min(\beta_1,\beta_2) \geq \frac{k-2}{2}$ one can also see that $4 \nmid gcd(x_k-x_{k-1},x_k+x_{k-1}))$. This immediately gives us that $max(\alpha_1,\alpha_2) \geq k-2$ We can therefore see that $$x_k \geq 2^{\alpha_1-1} \cdot 5^{\beta_1}+2^{\alpha_2-1} \cdot 5^{\beta_1} \geq 2^{k-3} \cdot 5^{\frac{k-2}{2}}$$Then $$10^k > x_k^2 > 5^{k-2} \cdot 2^{2k-6}$$This ultimately gives us that for all sufficiently large even $k \in \mathbb{N}$, in fact, the above inequality fails for any $k \geq 12$ as $2^{k-6} \geq 25$ This is ultimately a contradiction and hence no such sequence exists. $\blacksquare$

Solved with BarisKoyuncu and sevket12. The answer is no. FTSOC, assume for all $k>N$, $\overline{a_k a_{k-1}\cdots a_1} = b_k^2$. Note that $b_k^2 \leq 10^k - 1 \implies b_k \leq 10^{k/2}$ for all $k > N$. Claim: We cannot have $v_5(b_i)$ bounded. Proof: Assume that $v_5(b_i)$ is bounded, then since $b_k^2 + 10^ka_{k+1} = b_{k+1}^2$ for all $k>N$, $v_5(b_i)$ is eventually constant, call that constant $c$. Now take $k$ large enough, we have $(b_{k+1}-b_k)(b_{k+1}+b_k)=a_{k+1}10^k$. Note also that both $b_k$ and $b_{k+1}$ have the same parity and $\min({v_5(b_{k+1}-b_k),v_5(b_{k+1}+b_k)}) \leq c$. If $b_{k+1}-b_k \geq 2\cdot 5^{k-c}$, then $b_k^2+10^ka_{k+1} \geq b_k^2+4\cdot 5^{2k-2c}+4b_k\cdot 5^{k-c} \implies 9 \cdot 10^k \geq a_{k+1}10^k \geq 4 \cdot 5^{2k-2c} + 4b_k \cdot 5^{k-c} > 4 \cdot 5^{2k-2c}$ which fails for large enough $k$. If $b_{k+1}+b_k \geq 2\cdot 5^{k-c}$, then $b_k^2 + 10^ka_{k+1} \geq b_k^2 + 4 \cdot 5^{2k-2c} - 4b_k \cdot 5^{k-c} \implies 9 \cdot 10^k + 4 \cdot 5^{k-c}10^{k/2} \geq 4 \cdot 5^{2k-2c}$ which again fails for large enough $k$, so contradiction. $\blacksquare$ Since all $a_i$ are non-zero, this implies that $a_1=5$, as otherwise $v_5(b_i)=0$ for all $i$. So all $b_i$'s are odd. Now if $v_5(b_{2k}) < k$ for some $k$, then from $b_{2k}^2 + 10^{2k}a_{2k+1} = b_{2k+1}^2$, we get that $v_5(b_i) < k$ for all $i > 2k$, contradiction. If $v_5(b_{2k}) > k$, then $v_5(b_{2k+1})=k$ and from $b_{2k+1}^2 + 10^{2k+1}a_{2k+2} = b_{2k+2}^2$ we get $v_5(b_i) = k$ for all $i>2k$, contradiction. Thus $v_5(b_{2k}) = k$ for all $k>N/2$. Also a similar argument shows $v_5(b_{2k+1}) \geq k+1$. From $b_{2k+1}^2 + 10^{2k+1}a_{2k+2} = b_{2k+2}^2$ we get that $a_{2k+2} = 5$ for all $k$. Thus, $(b_{2k+2}-b_{2k+1})(b_{2k+2}+b_{2k+1})=5\cdot 10^{2k+1}$. Since both $v_5(b_{2k+2}-b_{2k+1})$ and $v_5(b_{2k+2}+b_{2k+1})$ are at least $k+1$, we must have equalities, in particular using that $b_i$'s are odd, we get $b_{2k+2}-b_{2k+1}=2 \cdot 5^{k+1} , b_{2k+2} + b_{2k+1}= 2^{2k} 5^{k+1} \implies b_{2k+2}=5^{k+1}(2^{2k-1}+1)$ and $b_{2k+1}=5^{k+1}(2^{2k-1}-1)$ for all $k$. But, using these in $10^{2k+2}a_{2k+3}=(b_{2k+3}-b_{2k+2})(b_{2k+3}+b_{2k+2})$, we reach a contradiction since $v_2(RHS)=3$ for large $k$, contradiction. $\square$

Consider two cases: A: $v_5 \sqrt{\overline{a_ka_{k-1} \cdots a_1}}$ is bounded above, then for sufficiently large $k$ we can look at the problem only mod $5^s$ instead of $10^s$ where $s$ is the number of digits of the number, then it is clear that for some amount of rounds after the $v_5$ stabilizes, the chain ends because asymptotically $5^{2s} > 10^{s+c}$ occurs (either in the congruent case or the second instance of the opposite sign mod $5^s$ case) for any fixed $c$ since each new number is congruent to the last mod $5^s$ so their square roots are $\pm$ each other mod $5^s$. B: $v_5 \sqrt{\overline{a_ka_{k-1} \cdots a_1}}$ is unbounded and must be at least $\frac{s}{2}$ for an $s$-digit number, so looking at $v_5$ from any number to the next gives that $5^{\frac{s-1}{2}}$ divides their difference, but since the number is also odd, we can get also $2^{s-1}$ divides their difference or sum, so since $20 > 10$, the chain ends after a few turns asymptotically similarly to the previous case (again, either in the congruent case or the second instance of the opposite mod $2^s$ by bounding). e.g. 3 -> 7 is a small jump since it is $\pm$ mod $5$ but the next time it needs to lift with opposite signs, it is big since $\pm$ mod 25 gives at least 43.

Answer: No. Assume the contrary, there exists an infinite sequence of nonzero digits $a_0, a_1, \dots$ such that $\overline{a_k a_{k-1} \dots a_1 a_0}$ is a perfect square for all sufficiently large $k$. Let $x_k = \overline{a_k a_{k-1} \dots a_1 a_0}$ and let $y_k = \frac{x_k}{5^{\nu_5(x_k)}}$. Claim 1: For every $k$, we have $\nu_5(x_k) \ge k+1$. Proof. Assume the contrary, there exists some $k$ such that $\nu_5(x_k) \le k$. Then $x_{k+1} = 10^{k+1} \cdot a_{k+1} + x_k$, so $\nu_5(x_{k+1}) = k$. Thus inducting on $k$ gives $\nu_5(x_l) = k$ for all $l \ge k$. Now let $N$ be a large integer and let $z_k = \sqrt{y_k}$. Then $y_N = \frac{x_N}{5^k} < \frac{10^{N+1}}{5^k} = 2^{N+1}5^{N+1-k}$, so $z_N < \sqrt{5}^{N+1-k}\sqrt{2}^{N+1}$. On the other hand, $10^{N+1} \cdot a_{N+1} = x_{N+1} - x_{N} = 5^{k}(z_{N+1} - z_{N})(z_{N+1} + z_{N})$ and since $\gcd(5, z_N) = 1$, so exactly one of $z_{N+1} + z_N, z_{N+1} - z_{N}$ is divisible by $5$, so $z_{N+1} + z_N \ge 5^{N+1-k}$. Let $s = N+1-k$, then $\sqrt{5}^{s+1}\sqrt{2}^{s+k+1} > z_{N+1} > 5^s - z_N > 5^s - \sqrt{5}^s\sqrt{2}^{s+k}$, so $\sqrt{5}^s\sqrt{2}^{s+k}(1 + \sqrt{10}) > 5^s$, which is false for large $s$, a contradiction. $\blacksquare$ Claim 2: $\nu_5(x_{2k+1}) = 2k+2$ and $a_{2k+1} = 5$ for large $k$. Proof. Note that $\nu_5(x_{2k}) \ge 2k+2$, so $x_{2k+1} = 10^{2k+1}\cdot a_{2k+1} + x_{2k}$ so if $a_{2k+1} \neq 5$, then $\nu_5(x_{2k+1}) = 2k+1$, contradicting the claim. So $a_{2k+1} = 5$ for all large $k$. Now assume $\nu_5(x_{2k+1}) \ge 2k+4$ for some $k$. Then $x_{2k+2} = 10^{2k+2} \cdot a_{2k+2} + x_{2k+1}$ and since $\nu_5(10^{2k+2} \cdot a_{2k+2}) \le 2k+3 < 2k+4 \le \nu_5(x_{2k+1})$, so $2k+3 \le \nu_5(x_{2k+2}) \le 2k+3$. Therefore $\nu_5(x_{2k+2}) = 2k+3$, contradicting the fact that $x_{2k+2}$ is a perfect square. $\blacksquare$ Now taking modulo 9 gives $a_{2k} = 4$ for all sufficiently large $k$. Let $M$ be a large odd integer and let $N$ be sufficiently large integer. Then $x_{M + 2N} = 5454\dots54a_{M}a_{M-1}\dots a_1a_0 = 54 \cdot \frac{10^{2N} - 1}{99} + x_M \equiv 0 (5^{2N})$, so $5^{2N} \mid 99x_M - 54$, contradicting the fact that $N$ is sufficiently large. This completes proof. $\blacksquare$

There's a nice $\nu_5$ solution to this. Assume for the sake of contradiction that such a sequence exists; in particular, for each $k > N$, let $b_k = \sqrt{\overline{a_ka_{k-1} \cdots a_1}}$. Then, $$(b_k-b_{k-1})(b_k+b_{k-1}) = b_k^2 - b_{k-1}^2 = d \cdot 10^k$$for some $1 \leq d \leq 9$. Because all digits are nonzero, notice that $$\{b_k-b_{k-1}, b_k+b_{k-1}\} = \{2x \cdot 5^a, 2^{k-1} \cdot y \cdot 5^b\} \text{ or } \{2^a \cdot 5^k \cdot x, 2^b \cdot y\}$$where $a+b=k$ and $xy=d$. The second case is impossible because it implies $b_k^2 > (2^{a-1} \cdot 5^k \cdot x)^2 > 5^{2k} > 10^{k+1}$, so we must be in the first case. Then, we have $$10^{k-1} < b_{k-1}^2 = x^2 \cdot 25^a + 4^{k-2} \cdot 25^ b - 10^k \cdot \frac y2 < 10^k.$$Notice now that $a > \frac k2 + 1$; if otherwise, then $4^{k-2} \cdot 25^b > 4^{k-2} \cdot 5^{k-2} > 10^{k+3}$ which makes the LHS too large. Thus, $$\nu_5(x^2 \cdot 25^a) \geq 2a > k+2 > k+1 \geq \nu_5\left(10^k \cdot \frac y2\right)$$as $y \leq 9$, which implies that $\nu_5(b_{k-1}) = a+\nu_2(x)$. But by similar logic, $\nu_5(b_k) = a+\nu_2(x)$ too, so it follows that $\nu_5(b_k)$ is constant for all sufficiently large $k$. This means that the $x^2 \cdot 25^a$ term has constant $\nu_5$ in the expansion; but as $a+b = k$, by picking sufficiently large $k$, we can make $b > a$, which yields a contradiction by above discussion.

The answer is no. Suppose otherwise. Consider some sequence $a_1, a_2, \ldots, $ that satisfied the problem conditions. For all $k>N$, let $f(k)$ be the square root of $\overline{a_k a_{k-1} \cdots a_1} $. We have\[f(k+1) ^2 - f(k)^2 = (f(k+1) - f(k))(f(k+1) + f(k)) = 10^k \cdot a_{k+1}\]Let $b_k = \nu_5(f(k))$. We have that\[\nu_5(f(k+1) - f(k)) + \nu_5(f(k+1) + f(k)) \in \{k, k + 1\},\]then if $b_k \ne b_{k+1}$, $\min (b_k , b_{k+1} )$ equals $\frac k2$ if $k$ is even, and $\frac{k+1}{2}$ if $k$ is odd. Additionally, let $c_k = \nu_2(f(k))$. If $c_k \ne c_{k+1}$, then $c_k \in \left \{ \frac k2, \frac{k+2}{2} \right\}$ if $k$ is even, and $c_k = \frac{k+1}{2}$ if $k$ is odd. Claim: There exists some $k > N$ with $b_k = b_{k+1}$. Proof: Suppose not. This implies that for each $k>N$, $\min(b_k, b_{k+1}) \in \left \{ \frac k2, \frac{k+1}{2} \right \} $, implying that $b_k \ge \frac k2 $ for each $k$. Consider some even $k$. Then at least one of $b_k, b_{k+1}$ is equal to $\frac k2$. But since $b_{k+1} \ge \frac{k+1}{2} > \frac k2$, we must have $b_k = \frac k2$. For odd $t$, we have $\min(b_t, b_{t+1}) = \frac{t+1}{2}$, so $\nu_5(f(t+1)^2 - f(t)^2 ) = t + 1$, meaning $a_{t+1} = 5$. Now for each $k$, we have $f(k)$ is a multiple of $5$, meaning that $a_1 = 5$, so $f(k)$ is odd. For each odd $t$, $\nu_2(f(t+1) - f(t)) + \nu_2(f(t+1) + f(t)) = t$. Since $f(t), f(t+1)$ are odd, one of $f(t+1) - f(t), f(t+1) + f(t)$ has a $\nu_2$ of $1$ and the other has a $\nu_2$ of $t-1$. Now, since $(f(t+1) - f(t)) (f(t+1) + f(t)) = 5\cdot 10^t$, each of $f(t+1) - f(t)$ and $f(t+1) + f(t)$ are a power of $2$ times a power of $5$, hence they must be equal to $5^{ \frac{t+1}{2} } \cdot 2$ and $5^{ \frac{t+1}{2} } \cdot 2^{t-1}$ in some order, meaning that adding them up gives $2\cdot f(t+1)$, so $f(t+1) = 5^{ \frac{t+1}{2}} \cdot (2^{t-2} + 1) $ and $f(t) = 5^{ \frac{t+1}{2} } \cdot (2^{t-2} - 1)$. However, we have $f(t+1) < 10^{\frac{t+1}{2}} $ , which is absurd since $5^{\frac{t+1}{2} } \cdot (2^{t-2} + 1) > 10^{ \frac{t+1}{2}}$ for large enough $t$. $\square$ Claim: If $k>N$ satisfies $b_k = b_{k+1}$, then we have $b_k = b_{k+1} =b_{k+2} = \cdots $. Proof: We prove that for each $k > N$ satisfying $b_k = b_{k+1}$, we have $b_{k+1} = b_{k + 2}$. Suppose some $k$ satisfying this had $b_{k+1} \ne b_{k+2}$. Let $f(k+1) = 5^{b_k}\cdot m$ and $f(k) = 5^{b_k} \cdot n$. We have $(m-n)(m+n) = a_{k+1} \cdot 2^k \cdot 5^{k - 2b_k}$ for $5\nmid m,n$, which in fact implies that $b_k \le \frac{k}{2}$. But since $b_{k+1} \ne b_{k+2}$, we have $\min (b_{k+1}, b_{k+2}) \in \left \{ \frac{k+1}{2} , \frac{k+2}{2} \right \}$, which is absurd since $b_{k+1} = b_k \le \frac k2$. $\square$ Now let $c_k = \frac{f(k)}{5^{b_k}}$ for each $k > N$. Then, we have $(c_{k+1} - c_k) (c_{k+1} + c_k) = a_{k+1} \cdot 2^k \cdot 5^{k - 2 b_k}$. Hence one of $c_{k+1} - c_k, c_{k+1} + c_k$ has $\nu_5$ at least $5^{k - 2b_k}$, meaning that one of $f(k+1) - f(k), f(k+1) + f(k)$ has $\nu_5$ at least $5^{k - b_k}$. But then, this implies that $f(k+1) + f(k) \ge 5^{k - b_k}\implies f(k+1) \ge \frac{5^{k - b_k}}{2} $. For sufficiently large $k$ (as $(b_i)$ is eventually constant), we have $5^{k - b_k} > 10 ^{\frac{k+1}{2}} > f(k+1)$, contradiction. Therefore, no such sequence $(a_i)$ exists.

The answer is no. Assume otherwise. Define $s_k^2=\overline{a_k a_{k-1}\cdots a_1}$ for all $k\ge N$. Then, \[(s_k-s_{k-1})(s_k+s_{k-1})=s_k^2-s_{k-1}^2=a_k 10^{k-1}\]Note that since $a_1\neq 0$, $10\nmid s_k$ for all $k$. Let $j$ be any positive integer and let $k$ be large enough. Clearly, if $5^j\mid s_{k-1}$ then $5^j\mid s_{k}$. Thus, either all $s_k$ are divisible by $5^j$ or all are not. Since $5^j$ divides either $s_{k}+s_{k-1}$ or $s_k-s_{k-1}$, if $5^j\nmid s_{k-1}$ then $s_k$ is either $-s_{k-1}$ or $s_{k-1}\pmod {5^j}$ but not both because $5^j$ is odd. Therefore, one of the factors $s_k+s_{k-1}$ or $s_k-s_{k-1}$ has a power of $5$ of less than $j$. Hence, $5^{k-j}\mid s_k+s_{k-1}$ or $5^{k-j}\mid s_k-s_{k-1}$ which implies \[5^{k-j}<s_k+s_{k-1}<\sqrt{10^k}+\sqrt{10^{k-1}}\]Seeing as $5>\sqrt{10}$, this is absurd. Therefore, $5^j\mid s_k$ for all large enough $k$. As a corollary, $5^j\mid \overline{a_k a_{k-1}\cdots a_1}$ for all large enough $k$. Since only the last $j$ digits matter, $5^j\mid \overline{a_ja_{j-1}\dots a_1}$ for all $j$. However, for all large enough $j$, we also need this to be a perfect square. Let $\overline{a_{2j}a_{2j-1}\dots a_1}\equiv i_j\cdot 5^{2j}\pmod {5^{2j+2}}$ for some $0\le i_j\le 24$, where $i_j\neq 5,10,15,20$ due to perfect square. Then $5^{2j+1}\mid \overline{a_{2j+1}a_{2j}\dots a_1}$. Since this is also a perfect square, we have $5^{2j+2}$ divides that. Thus, \[a_{2j+1}\cdot 10^{2j}\equiv -i_j\cdot 5^{2j}\pmod {5^{2j+2}}\]Dividing by $5^{2j}$ we get $a_{2j+1} \cdot 2^{2j}\equiv -i_j\pmod {25}$. Then, we have $5^{2j+2}$ divides both $\overline{a_{2j+1}a_{2j}\dots a_1}$ and $\overline{a_{2j+2}a_{2j+1}\dots a_1}$ so $a_{2j+2}\cdot 10^{2j+1}$ is divisible by $5^{2j+2}$ which implies that $a_{2j+2}=5$. Now, we are ready to finish the problem. Note that \[(s_{2j+2}+s_{2j+1})(s_{2j+2}-s_{2j+1})=5^{2j+2}\cdot 2^{2j+1}\]and since $\nu_5(s_{2j+2}+s_{2j+1})=\nu_5(s_{2j+2}-s_{2j+1})$, both are equal to $j+1$. Let $s_{2j+2}+s_{2j+1}=5^{j+1} 2^a$ and $s_{2j+2}-s_{2j+1}=5^{j+1} 2^b$ where $a+b=2j+1$ then let $a-b=d$ where $d$ is odd. We have \[\frac{s_{2j+2}+s_{2j+1}}{s_{2j+2}-s_{2j+1}}=2^d\]for some odd $d$. For each $d$, this leads to \begin{align*} s_{2j+2} &= s_{2j+1} \cdot \frac{2^d+1}{2^d-1} \\ s_{2j+2}^2 &= s_{2j+1}^2\cdot \left(\frac{2^d+1}{2^d-1}\right)^2 \\ 5\cdot 10^{2j+1} &= s_{2j+1}^2\cdot \left(\left(\frac{2^d+1}{2^d-1}\right)^2-1\right) \\ &<10^{2j+1}\left(\left(\frac{2^d+1}{2^d-1}\right)^2-1\right) \end{align*}This is straight-up false for $d\ge 3$, so we can only have $d=1$. This, however, leads to $625\cdot 10^{2j-2} = s_{2j+1}^2$. This means that $s_{2j+1}^2$ has nonzero digits, contradiction.

I claim the answer is no. For the sake of contradiction, suppose one exists. Let $p_k$ denote the perfect square with $\overline{a_ka_{k-1} \dots a_0}$. For sufficiently large $k$, we have \[(p_{k+1}-p_k)(p_{k+1}+p_k) = a_{k+1} \cdot 10^{k+1}\]\[\implies v_5(p_{k+1}-p_k)+v_5(p_{k+1}+p_k) \ge k+1\]However, notice that \[p_{k+1} \pm p_k < 2p_{k+1} < 2 \cdot \sqrt{10^{k+2}} <2 \cdot 5^{\frac{k+2}{2}} \cdot 5^{\frac{k+2}{4}}\]\[\implies v_5(p_{k+1} \pm p_k) <\frac{3k+6}{4}\]Since $p_k^2$ and $p_{k+1}^2$ both end in the same units digit, the units digit of $p_k$ and $p_{k+1}$ are either the same or add up the $10$. For now, assume that their units digits are not $5$. Therefore, they must be $(1,9),(2,8), \ldots$. However, each of these pairs means that one of $p_{k+1}+p_k$ or $p_{k+1}-p_k$ do not end in $0$ or $5$. Therefore, one of them is not divisible by $5$. So \[\frac{3k+6}{4} > v_5(p_{k+1} \pm p_k) \ge k+1\]which is a clear contradiction. Thus we must have $p_k$ and $p_{k+1}$ end in $5$. By similar logic, all $p_k$ must end in $5$, so they're all odd. To finish, take $2$ cases: If there exists a $k$ such that $v_5(p_k^2) \le k$. By LTE and the first equation, we must have \[2e \coloneq v_5(p_k^2) = v_5(p_{k+1}^2) = \cdots\]Since $\frac{p_k}{5^e}$ and $\frac{p_{k+1}}{5^e}$ don't end in $0$ or $5,$ we must have that \[\min(v_5(p_{k+1}-p_k),v_5(p_{k+1}+p_k)) = e \implies \max(v_5(p_{k+1}-p_k),v_5(p_{k+1}+p_k)) = k+1-e\]\[\implies 5^{k+1-e} \le p_{k+1}+p_k <2p_{k+1}<2 \cdot 10^{\frac{k+2}{2}}\]This gives a contradiction for large enough $k$. If $v_5(p_k^2) >k$ for all $k$. Since all $p_k$ are odd, one of $p_{k+1}-p_k$ and $p_{k+1}+p_k$ is $0$ modulo $4$ while the other one is $2$ modulo $4$. Therefore, \[2^{k} \mid p_{k+1}-p_k \text{ or } p_{k+1}+p_k\]\[\implies 2^k \cdot 5^{\frac{k}{2}} \le p_{k+1}+p_k <2 \cdot 10^{\frac{k+2}{2}}\]This also gives a contradiction for large enough $k$.