Starting Decimal Digits of Integer Square Roots

 Here is a cute result.

Given a finite string of digits $S = a_1 \dots a_k$, show that there is a positive integer $N$ such that the fractional part of $\sqrt{N}$ is of the form $0.S... = 0.a_1 \dots a_k \dots$.

For eg, is $S = "414"$, we have that $\sqrt{2} = 1.414\dots$ and fractional part is $0.414\dots$.

Scroll down for a solution.

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Given $S = a_1 \dots ...a_k$, treat it as the number $M = a_1 a_2\dots a_k$.

We can also assume $a_k < 9$ (otherwise, we can just extend $S$ by the digit $0$).

Now consider $X = 10^{2k} + M$.

There are $2X+1$ numbers between $X^2$ and $(X+1)^2$ (not including $(X+1)^2$). Since $X \ge 10^{2k}$, some number among those is divisible by $10^{2k}$. Say it is $P = 10^{2k} Q$.

We have that

$$X^2 \leq 10^{2k}Q < (X+1)^2$$

which implies

$$ X \leq 10^k \sqrt{Q} < (X + 1)$$

i.e

$$ 10^k + \frac{M}{10^k} \leq \sqrt{Q} < 10^{k} + \frac{M + 1}{10^k}$$

And thus $Q$ is the required number.

For eg for $S = 2026$, we get $Q = 100004053$ and $\sqrt{Q} = 10000.2026479\dots$.