Can you root Pi?

Claim: Given any string of digits $N$ , there always exists an integer $n$ such that $n^2$ has leading digits of $N$ .

Proof: There exists a large enough integer $m$ such that $\sqrt{ N + 1 } - \sqrt{ N} > 10 ^ { - m }$ . Then $\sqrt{ (N+1) \times 10 ^ {2m} } - \sqrt{ N \times 10 ^ {2m} } > 1$ . Hence, there exists an integer that is between these two numbers, or that

$\sqrt{ (N+1) \times 10 ^ { 2m } } > n \geq \sqrt{ N \times 10 ^ {2m } }$

Squaring both sides of the inequality, we conclude that $n^2$ does indeed have leading digits $N$ .

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As an explicit example, to find an $n$ such that $n^2$ has leading digits of 314159265359, we look at

$\sqrt{ 314159265359} = 560499.1216398$ and $\sqrt{314159265360} = 560499.1216407$ , with enough digits where they first disagree. This tells us that 56049912164 would satisfy the conditions, and indeed it does.

1772453850906^2=a number that begins with the digits 314159265359, so the answer is TRUE.