\section*{Problem 3B - 271043B}

A method for the approximate calculation of \(\sqrt{2}\) states:

From an approximate value \(\frac{a}{b}\), whose numerator \(a\) and denominator \(b\) are positive integers, a new approximate value \(\frac{a^{\prime}}{b^{\prime}}\) is obtained according to the rule

\[
a^{\prime}=a^{2}+2 b^{2} \quad(1) \quad ; \quad b^{\prime}=2 a b
\]

To assess whether \(\frac{a}{b}\) can be a suitable initial value for this method, the following problem is addressed (where the number \(\sqrt{2}\) is used as a known value):

Determine all those \(\frac{a}{b}\) ( \(a, b\) positive integers) for which the rule (1), (2) leads to a better approximate value \(\frac{a^{\prime}}{b^{\prime}}\), i.e.,

\[
\left|\frac{a^{\prime}}{b^{\prime}}-\sqrt{2}\right|<\left|\frac{a}{b}-\sqrt{2}\right|
\]

holds.
}

The inequality to be discussed is

\[
\left|\frac{a^{2}+2 b^{2}}{2 a b}-\sqrt{2}\right|0\) if and only if

\[
\left|a^{2}+2 b^{2}-2 a b \sqrt{2}\right|0\) holds, (3) is equivalent to

\[
|a-b \sqrt{2}|0\) is equivalent to \(b \sqrt{2}\frac{1}{3} \sqrt{2}\) to a better approximation value.

Adapted from \([5]\)