Extracting Square Roots Made Easy: A Little Known Medieval Method - A Close Connection Between Our Square Roots and the Pell Equation

Friedrich Katscher (Vienna University of Technology)

There is a very intimate relation between the approximations of square roots and the Pell equation. Thus the ancient and medieval Indian approximations for \(\sqrt{2},\) namely \(\frac{17}{12}\) and \(\frac{577}{408},\) and the approximations \(\frac{26}{15}\) and \(\frac{1351}{780}\) for \(\sqrt{3},\) all satisfy the Pell equation, when we take the radicand for \(d,\) the numerator for \(x,\) and the denominator for \(y.\) For example, \[ 1351^2-3 \times 780^2=1825201 - 3 \times 608400=1825201-1825200=1.\]

We detect that \(p\) and \(q\) of our approximations \(\frac{p}{q},\) but only those with unit fraction excesses, are solutions of the Pell equation if we set the radicand \(N=d,\: p=x, \) and \(q=y,\) which yields the Pell equation \(p^2-Nq^2=1.\) With the formula \[\frac{2p^2-1}{2pq},\] it is easy to obtain new and larger solutions \(x\) and \(y\) for Pell equations with a given \(d.\)

It is not difficult to prove why in the first approximation \(p\) and \(q\) fulfill the Pell equation: You can apply our formula only if \(N = a^2\pm \frac{2a}{n}.\) Then the first approximation is \[ \frac{p}{q}=a\pm \frac{1}{n}=\frac{an\pm 1}{n},\] with \(p=an\pm 1\) and \( q=n.\) Their squares are \(p^2=a^2n^2\pm 2an+1\) and \(q^2=n^2.\) Then the Pell equation is \[ p^2-Nq^2=a^2n^2\pm 2an+1-\left(a^2\pm \frac{2a}{n}\right)n^2=1.\]

It is important to note that the numerator and the denominator of square roots whose \(\frac{r}{2a}\) does not have the form \(\frac{1}{n},\) where \(n\) is an integer, are not solutions to Pell equations with the result \(1.\) This can be easily shown: If we reduce \(\frac{r}{2a}\) to the lowest numerator \(m\) and denominator \(n,\) the first approximation is \[a\pm \frac{m}{n}=\frac{an\pm m}{n}.\] Its square is \[\frac{a^2n^2\pm 2amn+m^2}{n^2},\] where \[N=\frac{a^2n^2\pm 2amn}{n^2}\] and the excess is \(\frac{m^2}{n^2}.\) The Pell equation is \[ a^2n^2\pm 2amn+m^2-Nn^2=a^2n^2\pm 2amn+m^2-\frac{a^2n^2\pm 2amn}{n^2}n^2=m^2.\] Hence \(m^2=1\) only if \(m=1.\)

From this we can draw the conclusion that \(x^2-dy^2 = 1\) only if the square of the calculated \(\sqrt{d}\:\) has an excess in the form of a unit fraction \(\frac{1}{n}.\)

For example, for \[ \sqrt{22}=\sqrt{5^2-3}\approx 5 - \frac{3}{10} = 4\frac{7}{10}=\frac{47}{10},\] where \(a=5,\: m=3,\: n=10,\) and \(\frac{3}{10}\) is not a unit fraction, we get the wrong Pell equation: \[47^2-22\times 10^2=2209-2200=9,\] not \(1,\) because \(m=3\) and \(m^2=9.\)

For \(d=22,\) the smallest, or the so-called fundamental, Pell solution, calculated with the help of continued fractions, is \(x=197\) and \(y=42,\) yielding \(197^2=38809,\) \(22\times 42^2=38808,\) and \(\left(\frac{197}{42}\right)^2=22\frac{1}{1764}.\) Here we have the expected unit fraction. Therefore, we can now apply our formula for the next approximation, \[\frac{2p^2-1}{2pq},\] with \(p=197\) and \(q=42.\) We get \(\frac{77617}{16548},\) and the Pell equation \[ 77617^2-22\times 16548^2=6024398689-6024398688=1.\]

It is especially simple to find the fundamental solution of the Pell equation for the numbers \(d=a^2+1,\, a^2+2,\, \) and \(a^2-2.\) The first approximations for the square roots of these numbers are respectively \[ a+\frac{1}{2a}=\underline{\frac{2a^2+1}{2a}},\, a+\frac{2}{2a}=a+\frac{1}{a}=\underline{\frac{a^2+1}{a}},\,\,{\rm and} \,\,a -\frac{2}{2a}=a-\frac{1}{a}=\underline{\frac{a^2-1}{a}}.\] The numerators and denominators of the underlined fractions are \(x\) and \(y\) for \(d.\) For \(d=a^2-1,\) the fundamental solution is always \(x=a,\) \(y=1.\)