we can clearly see that $p_0=5$ works with $k=1$ for $k\in\mathbb{N}-{1}$ let $n= \overline{p_{k-1}p_{k-2}\cdots p_0}\in\mathbb{N}$ which $n\equiv1(mod~2^k)$ and $n\equiv0(mod~5^k)$ thus $n^2\equiv n(mod~10^k)$ we can choose $p_{k}\in\{0,1,2,..,9\}$ such that $2^k(p_{k})+\frac{n}{5^k}\equiv 0(mod~5)$ $\Rightarrow 10^kp_{k}+n\equiv 0(mod~5^{k+1})~~~(*)$ and $5^k(p_{k})+\frac{n-1}{2^n}\equiv 0(mod~2)$ $\Rightarrow 10^kp_{k}+n\equiv1(mod~2^{k+1})~~~(**)$ from $(*),~(**)$ we'll get that $(10^kp_{k}+n)^2\equiv 10^kp_{k}+n(mod~10^{k+1})$ which is $(\overline{p_{k}p_{k-1}\cdots p_0})^2\equiv \overline{p_{k}p_{k-1}\cdots p_0}(mod~10^{k+1})$ $\therefore$ by induction we can conclude that for any $k\in\mathbb{N}$ we can choose $p_{k}\in\{0,1,2,..,9\}$ which square of any positive interger ending with $\overline{p_{k}p_{k-1}\cdots p_0}$ ends with the same digits $\square$ lmk if there is any mistakes