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

Author(s):
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 $$x=a,$$ $$y=1.$$

Friedrich Katscher (Vienna University of Technology), "Extracting Square Roots Made Easy: A Little Known Medieval Method - A Close Connection Between Our Square Roots and the Pell Equation," Loci (November 2010), DOI:10.4169/loci003494