2

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).

rtybase
  • 17,398
  • The distance between 2 number that skip over that particular starting digit is exponential in term of number of digit. The distance between consecutive square is exponential with lower base in term of number of digit. – Gina Jul 16 '14 at 23:17

2 Answers2

3

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 $a_1a_2\dots a_n$

Asinomás
  • 107,565
  • Why does this mean there is a perfect square between those numbers? – DavidButlerUofA Jul 16 '14 at 22:37
  • @DavidButlerUofA: Because if $x$ and $y$ are real numbers with $x-y>1$, there is an integer between them. Here $x=10^k\sqrt{a_1a_2\ldots a_n+1}$ and $y=10^k\sqrt{a_1a_2\ldots a_n}$. – TonyK Jul 16 '14 at 23:10
  • Oh I see. Since there is an integer m between x and y (ie $x > m > y$), there is a square of an integer between $x^2$ and $y^2$ (ie $x^2 > m^2 > y^2$). – DavidButlerUofA Jul 16 '14 at 23:28
0

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...

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

  2. $a$ 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 $$p^{2}=10^{2\cdot k}\cdot a+t, 0<t<10^{2 \cdot k}$$

rtybase
  • 17,398
  • Why bother with all this, when Bananarama already posted a five-line proof? – TonyK Jul 16 '14 at 23:12
  • From the educational perspective "so the solutions provided should include these tools (a bit of constraint to the problem)." ... I am sure there are other ways to prove it. – rtybase Jul 16 '14 at 23:14