IRL_ABooklet_2023

Let $p$ be the given number and suppose we extend it by $n$ digits. The value of $n$ is still to be determined. The smallest such extension is $p\cdot 10^n$ and the largest is $(p+1)\cdot 10^n-1$. It is possible to complete the number $p$ to a perfect square by appending $n$ digits if there exists an integer $m$ such that $$p\cdot 10^n\le m^2<(p+1)\cdot 10^n,\text{ i.e. }$$ $$\sqrt{p}\cdot 10^{\frac{n}{2}}\le m<\sqrt{p+1}\cdot 10^{\frac{n}{2}}.$$ Because $\sqrt{p+1}-\sqrt{p}>0$, for $n$ large enough we have $(\sqrt{p+1}-\sqrt{p})\cdot 10^{\frac{n}{2}}>1$. Hence, there is an integer between $\sqrt{p}\cdot 10^{\frac{n}{2}}$ and $\sqrt{p+1}\cdot 10^{\frac{n}{2}}$.

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Alternative solution.

Let the given number $p$ have less than $k$ digits, i.e. $p<10^k$. Let $N$ be the largest integer for which $N^2\le p\cdot 10^{k+1}$, i.e. $$N^2\le p\cdot 10^{k+1}<(N+1{)}^2.$$ Combining $N^2\le p\cdot 10^{k+1}$ with $p<10^k$ we get $N^2<10^{2k+1}$, and so $N<\sqrt{10}\cdot 10^k$. Using $1+2\sqrt{10}<10$, which follows from $40=(2\sqrt{10}{)}^2<9^2$, we obtain $$2N+1<2\sqrt{10}\cdot 10^k+1<(1+2\sqrt{10})10^k<10^{k+1}.$$ We then get $$0<p\cdot 10^{k+1}<(N+1{)}^2=N^2+(2N+1)<p\cdot 10^{k+1}+10^{k+1}=(p+1)\cdot 10^{k+1}.$$ This means that the number $(N+1{)}^2$ is obtained from the number $p$ by appending $k+1$ digits.