Someone claims that the first five decimal digits of the square root of 51 in a row match the first five significant digits of the square root of 2. Let's check this claim. What rational approximating fraction can we obtain from this observation for $\sqrt{2}$? How many significant digits does this match with the actual value of $\sqrt{2}$? What is the relationship between this approximating value and the 1061st and related earlier problem 1 number sequences?[^0]


[^0]:    ${ }^{1}$ See K. M. L. 23 (1961) 62. o.
$$
\sqrt{51} \approx 7+\frac{\sqrt{2}}{10}
$$

from which, by squaring and rearranging,

$$
\sqrt{2} \approx \frac{99}{70}=1.41428 \ldots,
$$

This also matches $\sqrt{2}$ to 5 significant digits.

Problem 945 ${ }^{2}$ gave the fractions $\frac{y_{k}}{x_{k}}$ as approximations to $\sqrt{2}$ from the sequences defined by $x_{1}=2, y_{1}=3$, and for $k=1,2,3, \ldots$, $x_{k+1}=3 x_{k}+2 y_{k}, y_{k+1}=4 x_{k}+3 y_{k}$. In these sequences, the third term is $x_{3}=70, y_{3}=99$.

Remark. Can we obtain from the mentioned sequence pairs further numbers $a$ above 51, for which the first few decimal places of their square roots match the consecutive digits of the decimal form of $\sqrt{2}$, that is, for which

$$
\sqrt{a} \approx b+\frac{\sqrt{2}}{10}, \quad \text { and from this } \quad \sqrt{2} \approx \frac{50\left(a-b^{2}\right)-1}{10 b} ?
$$

The denominator of the resulting approximation for $\sqrt{2}$ thus ends in 0, and the numerator ends in 49 or 99. We want to find such pairs $x_{k}, y_{k}$. The search for denominators ending in 0 is facilitated by the expressions from problem 1062. ${ }^{3}$

$$
x_{k}=6 x_{k-1}-x_{k-2}=34 x_{k-2}-x_{k-4}=198 x_{k-3}-x_{k-6}
$$

which allow us to advance without calculating the terms of the $y$ sequence. For now, it is enough to examine the last digit of the terms. From $x_{1}=2, x_{2}=12$, and $x_{3}=70$, the last digit of $x_{4}$ is 8, the same for $x_{5}$, and again 0 for $x_{6}$. Using the third expression above and the $x_{0}=0$ from problem 1061, we get $x_{6}=198 \cdot 70-0=13860$. We immediately see that every third term of the $x$ sequence ends in 0, among them, alternating terms ending in 2 and 8. Using the additional formulas from problem 1062, $y_{6}=19601$, which does not meet the condition in (3). However, it is easy to show that $y_{k}=198 y_{k-3}-y_{k-6}$, and from $y_{0}=1, y_{3}=99$, that the $y_{3k}$ terms end in 1 and 9 alternately, so $y_{9}$ ends in a digit that meets the condition. Indeed, $y_{9}=198 \cdot 19601-99=3880899=50 \cdot 77618-1$, and since $x_{9}=198 \cdot 13860-70=2744210=10 b$ from which $b=274421$, then from $50\left(a-b^{2}\right)-1=y_{9}$, we get $a=b^{2}+\left(1+y_{9}\right) / 50=75306962859$, a number with the desired property. Thus,

$$
\sqrt{a}=274421, \underbrace{1414213562373} 140 \ldots
$$

and here, after the decimal point, 13 correct digits of $\sqrt{2}$ appear. [^0]

[^0]:    ${ }^{1}$ See K. M. L. 23 (1961/9) 62. p.

    ${ }^{1}$ K. M. L. $19(1959 / 10)$ 56. p.

    ${ }^{3}$ K. M. L. $23(1961 / 10) 64$. p.

    ${ }^{4}$ K. M. L. $23(1961 / 10) 64$. p.