Find a number by the decimal part of its square root

Question

I have a math problem consisting of two questions:

• can we find a number N knowing only the decimal part of its square root up to a precision (only an approximation of the decimal part because the decimal part never ends)
• is the answer unique? which mean that we won't find two integer whose square root decimal values are equal (the first 50 for example).

Example:

if we have $0,4142135623730950488016887242097$, can we find that it's the decimal part of square root of 2 or $0,418286444621616658231167581$ for $1234567890$

The answer for the second question is pretty easy because, let's say we have 50 decimals, the number of possible integer's square root is much more than the 10^50-1 possible values of the decimals parts, so there will be more than one answer. but is there a pattern or example getting one from another

I am very grateful for your help or any research track.

Answer 1

The first try would be to compute the continued fraction of the given decimals as far as precision can carry us and see if we can "guess" a repeating pattern. For example $$0.414213\ldots = \frac1{2+\frac1{2+\frac1{2+\ddots}}}$$ so we assume it is the solution of $x=\frac1{2+x}$, i.e. $x^2+2x-1=0$, $x=-1\pm\color{red}{\sqrt{2}}$. Similarly, $$0.6457513110645905905\ldots=\frac1{1+\frac1{1+\frac1{1+\frac1{4+\frac1{1+\frac1{1+\frac1{1+\frac1{4+\ddots}}}}}}}} $$ so we suspect a solution of $x=\frac1{1+\frac1{1+\frac1{1+\frac1{4+x}}}}=\frac{2x+9}{3x+14}$, hence $x^2+4x-3=0$, $x=-2\pm\color{red}{\sqrt 7}$. However, with finte precision given, we only get finitly many continued fractions correct and as in simple IQ tests, we may continue a finite sequence arbitrarily and assume other repetitive patterns. This is expecially true for large radicands. Thus we can find many "valid" continued fractions, among these many valid repeating ones, and each of these leads to a quadratic equation in $x$. Strictly speaking, only quadratic equations $Ax^2+Bx+C$ with $2A\mid B$ and $A\mid C$ match your question, i.e. that $x$ is of the form $\sqrt n-k$.

Example: If we are given $0.414$, this may stand for the rounding of any number between $0.4135$ and $0.4145$. We have $0.4135 = [0;2,2,2,1,1,3,2,8]$ and $0.4145=[0;2,2,2,2,1,3,1,2,1,2]$. Hence we are sure only of the first three $2$'s. An example of an alternate solutions is $x=[0;2,2,2,1+x]$, i.e. $x^2+x-\frac7{12}=0$, $x=-\frac12\pm\sqrt{\frac{31}{12}}$, but that is not the fractional part of a square root.

Answer 2

A related question: "can we have integers $a$ and $b$ that are not perfect squares such that the decimal expansions of $\sqrt{a}$ and $\sqrt{b}$ are eventually (if you go far enough into the expansion) the same?"

That is to say, can it be the case that the expansion for $\sqrt{b} - \sqrt{a} = q$, where $q$ is a decimal that eventually ends (and hence a rational number)?

The answer is no, and we can prove it as follows: Suppose $a$ is not a perfect square, and let $q$ be a non-zero rational number such that $$ \sqrt{b} - \sqrt{a} = q \implies \sqrt{b} = \sqrt{a} + q $$ Note that $$ (\sqrt{b})^2 = b = (\sqrt{a} + q)^2 = q^2 + a + 2q\sqrt{a} $$ $b$ is an integer, and so it is a rational. It is clear then that both sides of the equation are rational numbers.

Noting that $q^2$ and $a$ are both rational, the sum can only be rational if $2q\sqrt{a}$ is rational. In order for $2r\sqrt{a}$ to be rational, $\sqrt{a}$ would have to be rational. However, since $a$ is not a perfect square, $\sqrt{a}$ must be irrational, which gives us a contradiction.

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So, the moral of the story is as follows: for any two integers $a$ and $b$ that are not perfect squares: if you go far enough into the decimal parts, you'll find that the expansions are eventually different.

More precisely, you can never find a point in the expansions after which the two are identical.