Prove that for any given sequence of digits, there is a perfect square starting with that sequence Prove that for any given sequence of digits, there is a perfect square starting with that sequence. With more details, prove that for $\forall a\in \mathbb{N}$, such that $a=\overline{a_{1}a_{2}...a_{n}}$, $a_{i}\in \left \{ 0,..,9 \right \},\forall i=\overline{1..n}$ and $a_{1}>0$, there $\exists p\in \mathbb{N}$ such that $p^{2}=\overline{a_{1}a_{2}...a_{n}...}$
To make long story short, back in 1995 I was attending a maths summer school and this problem was a "homework" related to Dirichlet's Approximation Theorem and Kronecker's Approximation Theorem, so the solutions provided should include these tools (a bit of constraint to the problem).
 A: This is probably not what you want, but suppose you want a perfect square starting with $a_1a_2\dots a_n$ then for sufficiently large $2k$  we shall have 
$10^k\sqrt{a_1a_2\dots a_n+1}-10^k\sqrt{a_1a_2\dots a_n}= 10^k(\sqrt{a_1a_2\dots a_n+1}-\sqrt{a_1a_2\dots a_n})>1$. 
This means there is a perfect square between $10^{2k}(a_1a_2\dots a_n)$ and  $10^{2k}(a_1a_2\dots a_n+1)$ This number is a perfect square starting with $a_1a_2\dots a_n$
A: Probably not en-style with the rules, but after few days I came to this solution, which probably isn't the most elegant. Still looking for a better one...


*

*If $a$ is a perfect square itself, we're done. Now, let's assume ...

*$a$ is not a perfect square, then $\sqrt{a}\in \mathbb{R} \setminus  \mathbb{Q}$ and so is $\frac{1}{\sqrt{a}}\in \mathbb{R} \setminus  \mathbb{Q}$. Then, according to Kronecker's Approximation Theorem, $M=\left \{ \frac{m}{\sqrt{a}} + n \mid m,n\in \mathbb{Z} \right \}$ is dense in $\mathbb{R}$. This basically means that for $\forall x,y\in \mathbb{R},x<y$ there $\exists \gamma \in M$ such that $x< \gamma < y$. 
If we take $x=10^{k}$ and $y=10^{k} + \frac{1}{\sqrt{a}}$ then $\exists m,n\in \mathbb{Z}$ such that $$10^{k}\cdot \sqrt{a}< n\cdot \sqrt{a}+m< 10^{k}\cdot \sqrt{a}+1$$
Notating $$n\cdot \sqrt{a}+m=\alpha=\left \lfloor \alpha \right \rfloor+\left \{ \alpha \right \} $$
We have $$10^{k}\cdot \sqrt{a}\leq 10^{k}\cdot \sqrt{a}+1-\left \{ \alpha  \right \}< \left \lfloor \alpha  \right \rfloor+1< 10^{k}\cdot \sqrt{a}+1+1-\left \{ \alpha  \right \}\leq 10^{k}\cdot \sqrt{a}+2$$
Where $\left \lfloor \alpha  \right \rfloor+1=p$ we are looking for. So $$10^{k}\cdot \sqrt{a}< p < 10^{k}\cdot \sqrt{a}+2$$ or $$10^{2\cdot k}\cdot a< p^{2}<10^{2\cdot k}\cdot a+4\cdot 10^{k}\sqrt{a}+4$$ and obviously there $\exists t\in \mathbb{N}$ such that $10^{2\cdot k}\cdot a+t=p^{2}$. This imposes the restriction $$0<t<4\cdot 10^{k}\sqrt{a}+4$$
What we want now, taking into account arbitrary nature of $k$, is $$0<t<4\cdot 10^{k}\sqrt{a}+4<10^{2 \cdot k}$$ i.e. $$4\cdot \sqrt{a}+\frac{4}{10^{k}}<10^{k}$$ which is true for $\forall k> \log_{10}\left ( 4\cdot \sqrt{a}+1 \right )$. This way, we have $$p^{2}=10^{2\cdot k}\cdot a+t, 0<t<10^{2 \cdot k}$$
