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\title{\textcolor{yellow}{Cakrav{\=a}la}}
\author{Amba Kulkarni}
\institute{ University of Hyderabad}
\date{27th March 2018}

\begin{document}
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\titlepage

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\begin{frame}
\frametitle{Cakrav{\=a}la Method}
Let the Vargaprak\d{r}ti be $N x^2 + C = y^2$, \\
where $x$ is the smaller root (hrasva),\\
$y$ is the greater root (jye\d{s}\d{t}ha),\\
and $C$ is the constant (k\d{s}epaka).\\
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\begin{frame}
\frametitle{Cakrav{\=a}la Method}
\textbf{hrasva-jye\d{s}\d{t}ha-pada-k\d{s}ep{\=a}n} \\
\textbf{bh{\=a}jya-prak\d{s}epa-bh{\=a}jak{\=a}n}\\
\textbf{k\d{r}tvo kalpyo gu\d{n}a\d{h} tatra tath{\=a} prak\d{r}ti tat cyute}\\
\textbf{gu\d{n}avarge prak\d{r}tya\d{h} une athav{\=a} alpa\.m \'{s}e\d{s}aka\.m yath{\=a}}\\
\vspace{0.2cm}
Assume (integer) values for smaller root, greater root and the constant.\\
Take them to be the dividend, constant and divisor (in order of a ku\d{t}\d{t}aka).\\

Solve it and find the values of guna and labdhi. Subtract the square of guna from prak\d{r}ti or vice versa so that the remainder is the smallest.\\
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\begin{frame}
\frametitle{Cakrav{\=a}la Method}
\textbf{tattu k\d{s}epah\d{r}ta\.m k\d{s}epo vyasta\d{h} prak\d{r}ti tat cyute}\\
\textbf{gu\d{n}a labdhi\d{h} padam h\d{r}sva\.m tata\d{h} jye\d{s}\d{t}a\d{h} mata\d{h} asak\d{r}t}\\
\textbf{tyaktv{\=a} p{\=u}rvapada k\d{s}ep{\=a}n cakrav{\=a}la\.m idam jagu\d{h}}\\
\textbf{catur dva-ekayut{\=a} eva\.m abhinne bhavata\d{h} pade}\\
\vspace{0.5cm}
Divide this difference by the previous argument(constant) and obtain the new value of the constant.\\

If the square of the guna is subtracted from the prak\d{r}ti, its sign changes.\\

Assume the guna so obtained as the new value of the smaller root and find the new greater root with prak\d{r}ti and constant. 
%Again the new smaller root, greater root and the constant should be taken as the dividend, constant and the divisor respectively for the Kuttaka, and evaluate guna and labdhi as above. 

Repeat the process till the constant becomes 4,2 or 1. This is called 'cakrav{\=a}la'.\\

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\begin{frame}
\frametitle{Cakrav{\=a}la Method}
\textbf{caturdvi k\d{s}epa m{\=u}l{\=a}bhy{\=a}\.m r{\=u}pa k\d{s}ep{\=a}rtha bh{\=a}van{\=a}}\\

\vspace{0.5cm}
You get the constant unity from the value of the constant of 4 or 2 or 1 by Bh{\=a}van{\=a}. \\

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\begin{frame}
\frametitle{Procedure}
Let the equation to be solved be $Pq^2 + 1 = r^2$\\
\begin{itemize}
\item [1] Choose $s$ such that $Pq^2+s=r^2$, so that for $q=1$, $P+s$ is a perfect square. Get $r$ from $r^2 = Pq^2+s$.
\item [2] If $s$=1 or 2 or 4, $(q,r)$ is the solution for $Pq^2+s=r^2$\\
When $s$ = 1, you are done. When $s$ = 2 or 4, continue using Bh{\=a}van{\=a} until $s$ = 1.
\item [3] Otherwise, solve the ku\d{t}\d{t}aka $qx+r=sy$.
\item [4] Let $(X,Y) = (x+ms,y+mq)$, m=1,2,3,... be the possible solutions.
\item [5] Choose $X$ such that $P-X^2$ is minimum.
\item [6] Calculate new value of $s$ as\\
$s_1 = (-1). \frac{P-X^2}{s}$.
\item [7] Now find $r_1$ from $Pq_1^2+s_1=r_1^2$, where $q_1=Y$.
\item [8] Let $(q_1, r_1, s_1)$ be the new values of $(q,r,s)$. Go to step 2.
\end{itemize}
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\begin{frame}
\frametitle{Justification}
Let the equation to be solved be $Pq^2 + 1 = r^2$\\
Choose $s$ such that $Pq^2+s=r^2$, so that for $q=1$, $P+s$ is a perfect square. Get $r$ from $r^2 = Pq^2+s$.
\begin{itemize}
\item If $s$ = 1, we are done.
\item If $s$ = -1, use the rule R(1) viz.\\
\vspace{0.2cm}
\textbf{If $(x,y) = (\alpha,\beta)$ is a solution of $Nx^2 + k = y^2$}\\
\textbf{then $Nx^2 + k^2 = y^2$ will have the solution}\\
\textbf{$(x,y)=(2\alpha\beta,N\alpha^2+\beta^2)$}\\
\vspace{0.2cm}
to get the solution of $Pq^2 + 1 = r^2$\\
\end{itemize}
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\begin{frame}
\frametitle{Justification ...}
\begin{itemize}
\item If $s$ = $\pm$2, use the rule R(1) viz.\\
\vspace{0.2cm}
\textbf{If $(x,y) = (\alpha,\beta)$ is a solution of $Nx^2 + k = y^2$}\\
\textbf{then $Nx^2 + k^2 = y^2$ will have the solution}\\
\textbf{$(x,y)=(2\alpha\beta,N\alpha^2+\beta^2)$}\\
\vspace{0.2cm}
to get the solution of $Pq^2 + 4 = r^2$, and then use the rule R(2) viz.\\
\vspace{0.2cm}
\textbf{If $(x,y) = (\alpha,\beta)$ is a solution for the equation $Nx^2 + c = y^2$}\\
\textbf{then the solution of $Nx^2+\frac{c}{k^2}=y^2$ is given by}\\
\textbf{$(x,y)=(\frac{\alpha}{k} ,\frac{\beta}{k})$, with $k=2$}\\
\vspace{0.2cm}
to get the solution of $Pq^2 + 1 = r^2$\\
\end{itemize}
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\begin{frame}
\frametitle{Justification ...}
\begin{itemize}
\item If $s$ = 4, use the rule R(2) above viz.\\
\vspace{0.2cm}
\textbf{If $(x,y) = (\alpha,\beta)$ is a solution for the equation $Nx^2 + c = y^2$}\\
\textbf{then the solution of $Nx^2+\frac{c}{k^2}=y^2$ is given by}\\
\textbf{t  get the solution of $Pq^2 + 1 = r^2$}\\
\vspace{0.2cm}
\item If $s$ = -4, use the rule R(2) above with $k=2$\\
to get the solution of $Pq^2 - 1 = r^2$, and then use rule R(1) above viz.\\
\vspace{0.2cm}
\textbf{If $(x,y) = (\alpha,\beta)$ is a solution of $Nx^2 + k = y^2$}\\
\textbf{then $Nx^2 + k^2 = y^2$ will have the solution}\\
\textbf{$(x,y)=(2\alpha\beta,N\alpha^2+\beta^2)$}\\
\vspace{0.2cm}
to get the solution of $Pq^2 + 1 = r^2$\\
\end{itemize}
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\begin{frame}
\frametitle{Justification ... Contd}
\begin{itemize}
\item If $s$ is other than $\pm$ 1,2,4, then
we need to find some better approximation of $q$ and $s$ such that $s$ is close to 1.\\
\vspace{0.2cm}
Let the solution for $Pq^2+s=r^2$ be $(q,r)$ = $(1,A)$, where $s = A^2-P$\\
Let another solution for $Pq^2+k=r^2$ be $(q,r,k)$ = $(x,y,k)$\\
\vspace{0.2cm}
Then following Bh{\=a}van{\=a}, we have \\
$(q,r)$ = $(A.x+y, P.x+A.y)$ as the solution for $Pq^2+s.k=r^2$\\
Or,
$(q,r)$ = $(\frac{A.x+y}{s}, \frac{P.x + A.y}{s})$ as the solution for $Pq^2+\frac{k}{s}=r^2$\\
\vspace{0.2cm}
Thus we have two diophantine linear equations viz.\\
$q.s = y + A.x$, and \\
$r.s = P.x + A.y$.\\
\end{itemize}
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\begin{frame}
\frametitle{Justification ... Contd}
Let us now consider the first equation:\\
$q.s = y + A.x$  -- (1)\\
\vspace{0.2cm}
Now we solve this diophantine equation, for $(A,q)$, where $(x,y,s) = (1,a,a^2-P)$.\\
$q.(a^2-P) = a + A$ for the pair of values $(A,q)$.\\
\vspace{0.2cm}
We will have infinite solutions of the form $(A+m(a^2-P),q+m)$ for integer values of $m$.\\

We choose that pair $(A,q)$ such that $s = A^2-P$ is the smallest.
\vspace{0.2cm}
For this value of $q$, now we find $r$ from $Pq^2+s=r^2$, or\\
from $r.s = P.x + A.y$, where ($x,y,s) = (1,a,a^2-P)$.\\
\vspace{0.2cm}
And new value of $s$ as $s = (-1) \frac{A^2-P}{s}$.\\
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\begin{frame}
\frametitle{Example}
This is illustrated by the following example.\\

\textbf{k{\=a} sapta \'{s}a\d{s}\d{t}i gu\d{n}it{\=a} k\d{r}ti\d{h} eka yukt{\=a}}\\
\textbf{k{\=a} ca eka\d{s}a\d{s}\d{t}i nihat{\=a} ca sakhe sar{\=u}p{\=a}}\\
\textbf{sy{\=a}t m{\=u}lad{\=a} yadi k\d{r}ti prak\d{r}ti\d{h} nit{\=a}ntam}\\
\textbf{tvat cetasi pravada t{\=a}ta tat{\=a}lat{\=a}vat}\\

\vspace{0.5cm}
If you are clear and firm in your mind, tell me the following. What is that number, square of which, when multiplied by 67 or 61 and increased by 1 yields a square root?\\
\vspace{0.5cm}
Find the solutions of $67q^2+1=r^2$ and $61q^2+1=r^2$.
\end{frame}
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\begin{frame}
\frametitle{Example1: Solution}
The equation to be solved is $67q^2 + 1 = r^2$\\
\begin{itemize}
\item [1] Choose $s$ such that $67q^2+s=r^2$, and for $q=1$, $67+s$ is a perfect square. \\
$s = -3$.\\
$r^2 = 67-3$. Hence $r = 8$.
\item First Iteration
\begin{itemize}
\item [3.1] Solve the ku\d{t}\d{t}aka $x+8=3y$. (Ignore the sign of $s$).
\item [4.1] $(X,Y) = (1,3), (4,4), (7,5), (10,6), ...,$ \\
$(1+4m, 3+m)$, m=0,1,2,3,... are the possible solutions.
\item [5.1] Choose $X$ such that $67-X^2$ is minimum.\\
$67-7^2$ is minimum.
\item [6.1] New value of $s$ is\\
$s_1 = (-1).\frac{67-7^2}{-3} = 6$. 
\item [7.1] Now from $67.5^2+6=r_1^2$, $r_1^2 = 41^2$. \\
Hence $r_1 = 41$.
\item [8.1] Let $(q,r,s) = (5,41,6)$.
\end{itemize}
\end{itemize}
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\begin{frame}
\frametitle{Example1: Solution ...}
\begin{itemize}
\item Second Iteration
\begin{itemize}
\item [3.1] Solve the ku\d{t}\d{t}aka $5x+41=6y$.
\item [4.1] $(X,Y) = (5,11), (11,16), ...,$ \\
$(5+6m, 11+5m)$, m=0,1,2,3,... are the possible solutions.
\item [5.1] Choose $X$ such that $67-X^2$ is minimum.\\
$67-5^2$ is minimum.
\item [6.1] New value of $s$ is\\
$s_1 = (-1).\frac{67-5^2}{6} = -7$. 
\item [7.1] Now from $67*11^2-7=r_1^2$, $r_1^2 = 90^2$. \\
Hence $r_1 = 90$.
\item [8.1] $(q,r,s) = (11,90,7)$.
\end{itemize}
\end{itemize}
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\begin{frame}
\frametitle{Example1: Solution ...}
\begin{itemize}
\item Third Iteration
\begin{itemize}
\item [3.1] Solve the ku\d{t}\d{t}aka $11x+90=7y$.
\item [4.1] $(X,Y) = (2,16),(9,27), ...,$ \\
$(2+7m, 9+11m)$, m=0,1,2,3,... are the possible solutions.
\item [5.1] Choose $X$ such that $67-X^2$ is minimum.\\
$67-9^2$ is minimum.
\item [6.1] New value of $s$ is\\
$s_1 = (-1).\frac{67-9^2}{-7} = -2$. 
\item [7.1] Now from $67*27^2-2=r_1^2$, $r_1^2 = 221^2$. \\
Hence $r_1 = 221$.
\item [8.1] $(q,r,s) = (27,221,2)$.
\end{itemize}
\end{itemize}
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\begin{frame}
\frametitle{Example1: Solution ...}
$s$=2, and hence\\
$(27,221)$ is the solution for $67q^2-2=r^2$.\\
\vspace{0.2cm}
Following the rule in Bh{\=a}van{\=a} viz.\\
If $(x,y) = (\alpha,\beta)$ is a solution of $Nx^2 + k = y^2$\\
then $Nx^2 + k^2 = y^2$ will have the solution\\
$(x,y)=(2\alpha\beta,N\alpha^2+\beta^2)$\\
\vspace{0.2cm}
$67q^2+4=r^2$ will have a solution given by $(2*27*221,67*27^2+221^2)$, or\\
$(11934,97684)$.\\
\vspace{0.2cm}
Following the rule in Bh{\=a}van{\=a} viz.\\
If $(x,y) = (\alpha,\beta)$ is a solution of $Nx^2 + k^2 = y^2$\\
then $Nx^2 + 1 = y^2$ will have the solution\\
$(x,y)=(\frac{\alpha}{k},N\frac{\alpha^2}{k})$\\
\vspace{0.2cm}
$67q^2+1=r^2$ will have a solution given by $(\frac{11934}{2},\frac{97684}{2})$ = $(5967,48842)$.\\
\end{frame}
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\begin{frame}
\frametitle{Example2: Solution}
The equation to be solved is $61q^2 + 1 = r^2$\\
\begin{itemize}
\item [1] Choose $s$ such that $61q^2+s=r^2$, ans for $q=1$, $61+s$ is a perfect square. \\
$s = 3$.\\
$r^2 = 61+3$. Hence $r = 8$.
\item First Iteration
\begin{itemize}
\item [3.1] Solve the ku\d{t}\d{t}aka $x+8=3y$.
\item [4.1] $(X,Y) = (1,3), (4,4), (7,5), (10,6), ...,$ \\
$(1+4m, 3+m)$, m=0,1,2,3,... are the possible solutions.
\item [5.1] Choose $X$ such that $61-X^2$ is minimum.\\
$61-7^2$ is minimum.
\item [6.1] New value of $s$ is\\
$s_1 = (-1).\frac{61-7^2}{3} = -4$. 
\item [7.1] Now from $61*5^2-4=r_1^2$, $r_1^2 = 39^2$. \\
Hence $r_1 = 39$.
\item [8.1] Let $(q,r,s) = (5,39,4)$.
\end{itemize}
\end{itemize}
\end{frame}
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\begin{frame}
\frametitle{Solution ... Contd}
$s$=4, and \\
$(5,39)$ is the solution for $61q^2-4=r^2$.\\
\vspace{0.5cm}
Following the rule in Bh{\=a}van{\=a} viz.\\
If $(x,y) = (\alpha,\beta)$ is a solution of $Nx^2 - k^2 = y^2$\\
then $Nx^2 - 1 = y^2$ will have the solution\\
$(x,y)=(\frac{\alpha}{k},\frac{\beta}{k})$\\
\vspace{0.5cm}
$61q^2-1=r^2$ will have a solution given by $(\frac{5}{2},\frac{39}{2})$\\
\vspace{0.5cm}
Following the rule in Bh{\=a}van{\=a} viz.\\
If $(x,y) = (\alpha,\beta)$ is a solution of $Nx^2 + k = y^2$\\
then $Nx^2 + k^2 = y^2$ will have the solution\\
$(x,y)=(2\alpha\beta,N\alpha^2+\beta^2)$\\
\vspace{0.5cm}
$61q^2+1=r^2$ will have a solution given by $(2*\frac{5}{2}*\frac{39}{2},61*\frac{5}{2}^2+\frac{39}{2}^2)$\\
$=(\frac{195}{2},\frac{1523}{2})$.
\end{frame}
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\begin{frame}
\frametitle{Solution ... Contd}
Following Bh{\=a}van{\=a} again,\\
$(q,r) = (2*\frac{195}{2}*\frac{1523}{2}, 61*\frac{195}{2}^2+\frac{1523}{2}^2)$\\
$=(\frac{296985}{2},\frac{2319527}{2})$.\\

\vspace{0.5cm}
Following Bh{\=a}van{\=a} once again,\\
$(q,r) = (226153980,11766319049)$.
\end{frame}
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\begin{frame}
\frametitle{More rules}
\textbf{r{\=u}pa \'{s}uddhau khiloddi\d{s}\d{t}a\.m}\\
\textbf{varga yoge gu\d{n}e na cet}\\
\vspace{0.5cm}
If the argument is negative, the equation can not be solved unless the dividend is the sum of two squares.\\
i.e., for $Nx^2-1=y^2$ has no integer solution unless $N=a^2+b^2$\\
\end{frame}
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\begin{frame}
\frametitle{more}
\textbf{akhile k\d{r}ti mul{\=a}bhy{\=a}m}\\
\textbf{dvidh{\=a} r{\=u}pa\.m vibh{\=a}jita\.m}\\
\textbf{dvidh{\=a} hrasva pada\.m jye\d{s}\d{t}a\.m}\\
\textbf{tato r{\=u}pa vi\'{s}odhane}\\
\textbf{p{\=u}rvavad v{\=a} pras{\=a}dhyete}\\
\textbf{pade r{\=u}pa vi\'{s}odhane}\\

\vspace{0.5cm}

In the case of argument being -1, the roots of the sqaures whose sum is the prak\d{r}ti, should be obtained. There will be two such values. The reciprocals of these values will be the two smaller roots. From these smaller vaues, the larger values can be estimated.
\vspace{0.5cm}
In other words, for the equation $Nx^2-1=y^2$, and $N = a^2+b^2$\\
the two smaller roots are $x = \frac{1}{a}$ and $x = \frac{a}{b}$.\\
The larger roots can then be obtained from the given equation.
\end{frame}
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\begin{frame}
\frametitle{Example}
\textbf{trayoda\'{s}a gu\d{n}o vargo nireka\d{h} ka\d{h} k\d{r}tir bhavet}\\
\textbf{ko v{\=a}\d{s}\d{t}a gu\d{n}ito vargo nnireko m{\=u}lad{\=a} vada}\\

\vspace{0.5cm}
Which is the number, whose square multiplied by 13 and reduced by one becomes a square?\\

Also tell me the number whose square multipled by 8 and reduced by one gives a suqare root.
\end{frame}
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\begin{frame}
\frametitle{Solution}
There are two equations given\\
$13x^2-1=y^2$, and\\
$8x^2-1=y^2$\\

Example 1: $13x^2-1 = y^2$\\
The argument is -1. The prak\d{r}ti is 13.\\
Since $13 = 3^2 + 2^2$\\
we have $x = \frac{1}{3}$, and $x = \frac{1}{2}$, as two possible solutions.\\
Substituting these values, we get $y = \frac{2}{3}$ and $y = \frac{3}{2}$.\\

With $x = \frac{1}{2}$ and $y = \frac{2}{3}$, using bh{\=a}van{\=a},
the roots of $13x^2+1=y^2$ are\\
$x_1 = 2 * \frac{1}{2} * \frac{3}{2} = \frac{3}{2}$\\
$y_1 = 13 * \frac{1}{4} + \frac{9}{4} = \frac{11}{2}$\\
\vspace{0.5cm}
Using bh{\=a}van{\=a} again, with the solutions $(x,y)= (\frac{1}{2},\frac{3}{2})$ and $(x,y) = (\frac{3}{2},\frac{11}{2})$ for $13x^2-1=y^2$ and $13x^2+1=y^2$, we get,\\
$x_1 = \frac{1}{2} * \frac{11}{2} + \frac{3}{2} * \frac{3}{2} = \frac{20}{4} = 5$\\
$y_1 = 13 * \frac{3}{4} + \frac{33}{4} = \frac{72}{4} = 18$\\
\end{frame}
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\begin{frame}
\frametitle{Solution}
Example 1: $8x^2-1 = y^2$\\
The argument is -1. The prak\d{r}ti is 8.\\
Since $8= 2^2 + 2^2$\\
we have $x = \frac{1}{2}$ for both possible solutions.\\
Substituting this value in the given equation, we get $y = 1$.\\
So the possible solution is $(\frac{1}{2},1)$.\\
\vspace{0.5cm}
Now, With $x = \frac{1}{2}$ and $y = 1$, using bh{\=a}van{\=a},
we get the solutions for $13x^2+1=y^2$.\\
$x_1 = 2 * \frac{1}{2} * 1 = 1$\\
$y_1 = 8 * \frac{1}{4} + 1 = 3$\\
\vspace{0.5cm}
Now, With $x = \frac{1}{2}$ and $y = 1$, for $8x^2-1=y^2$ and $(1,3)$ for $8x^2+1=y^2$, using bh{\=a}van{\=a},
we get the solutions for $13x^2-1=y^2$ as.\\
$x_2 = \frac{1}{2}*3 + 1*1 = \frac{5}{2}$\\
$y_2 = 8 * \frac{1}{2}*1 + 1*3 = 7$\\

So we do not get an integer solution.
\end{frame}
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%\begin{frame}
%\frametitle{example}
%\textbf{ko varga\d{h} \'{s}adgu\d{n}a\d{h} try{\=a}dhyo}\\
%\textbf{dv{\=a}da\'{s}{\=a}dhyo athav{\=a} k\d{r}ti\d{h}}\\
%\textbf{yuto v{\=a} pa\~nca saptaty{\=a}}\\
%\textbf{tri\'{s}aty{\=a} v{\=a} k\d{r}tir bhavet}\\
%
%\vspace{0.5cm}
%
%Which is that number whose square multiplied by 6 and increased by 3, 12, 75 and 300 respectively becomes square numbers.
%
%\vspace{0.5cm}
%$6x^2+3 = y^2$ ; Solution $(1,3)$\\
%$6x^2+12 = y^2$ ; Solution $(2,6)$\\
%$6x^2+75 = y^2$ ; Solution $(5,15)$\\
%$6x^2+300 = y^2$ ; Solution $(10,30)$\\
%
%\end{frame}
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%\begin{frame}
%\frametitle{Rule}
%\textbf{varga cchinne gu\d{n}e hrasvam tat padena vibh{\=a}jayet.}\\
%\vspace{0.5cm}
%If the gu\d{n}a (prak\d{r}ti) is divided by the square of any number, then the smaller root of the reduced equation should be divided by the square root of the square used for division, to get the correct root of the original euqation.\\
%
%\vspace{0.5cm}
%For example, in the equation $8x^2+1=y^2$, the prak\d{r}ti 8 can be divided by $2^2$ to simplify the given equation as $2x^2+1=y^2$.\\
%Root of this reduced euqation are $(2,3)$.\\
%\vspace{0.5cm}
%So the roots of the original equation are obtained by dividing the smaller root by the square root of 4, i.e., $(\frac{2}{2},3) = (1,3)$.
%\end{frame}
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%\begin{frame}
%\frametitle{Rule}
%\textbf{dv{\=a}tri\.m\'{s}at gu\d{n}ito varga\d{h} ka\d{h} sa-eko m{\=u}lado vada}\\
%
%\vspace{0.5cm}
%Tell the number, the square of which when multiplied by 32 and increased by 1 gives a perfect square.
%\vspace{0.5cm}
%The equation is $32x^2+1=y^2$.\\
%Dividing the prak\r{d}ti by $16$, we get $2x^2+1=y^2$.\\
%\vspace{0.5cm}
%This has a solution $(2,3)$.\\
%Hence the solution of the original equation is $(\frac{2}{4},3)$.\\
%Using bh{\=a}van{\=a}, we get\\
%$(2*\frac{2}{4}*3,32*\frac{2}{4}\frac{2}{4}*3*3)$ = $(3,17)$.
%\end{frame}
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%\begin{frame}
%\frametitle{Rule}
%\textbf{i\d{s}\d{t}a bhakto dvidh{\=a} k\d{s}epa i\d{s}\d{t}on{\=a}\d{d}hyo dal{\=\i}k\d{r}t{\=a}\d{h}}\\
%\textbf{gu\d{n}a m{\=u}la h\d{r}ta\d{h} c{\=a}dyo h\d{r}asva jye\d{s}\d{t}e kram{\=a}lpade}\\
%
%\vspace{0.5cm}
%If the prak\d{r}ti(gu\d{n}a) is a square number, divide the constant argument by any optional number and write the quotient at 2 places. From one subtract the optional number and add it to the second. Divide both by 2. Dividing the first one by the square root of the gu\d{n}a gives the smaller root $x$ and the second one gives the larger root.\\
%
%For example, consider $16x^2+10=y^2$. The gu\d{n}a is a perfect square.\\
%Divide the constant say by $2$. We get $\frac{10}{2} = 5$. Subtracting and adding $2$ from $5$ and then dividing it by $2$, we get $\frac{5-2}{2}$ and $\frac{5+2}{2}$. Dividing the first one by square root of 16, we get $\frac{3}{2*4}$ as value of $x$, and $y = \frac{7}{2}$.
%
%\end{frame}
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\end{document}