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\title{\textcolor{yellow}{Vargaprak\d{r}ti}}
\author{Amba Kulkarni}
\institute{University of Hyderabad}
\date{21st March 2018}

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\frametitle{Vargaprak\d{r}ti: Some terminology}
Vargaprak\d{r}ti (equation of a \textbf{square} with some given \textbf{co-efficient}) is the name given to the equations of type $Nx^2+C=y^2$\\
which is a general form of Pell's equation $Nx^2+1=y^2$, where $N$ is is a positive nonsquare integer.\\

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\frametitle{Vargaprak\d{r}ti: Some terminology}

i\d{s}\d{t}a\.m h\d{r}sva\.m tasya varga\d{h} prak\d{r}ty{\=a}\d{h}\\
k\d{s}u\d{n}\d{n}au yukto varjit{\=a} v{\=a} sa\d{h} yena\\
m{\=u}la\.m dady{\=a}t k\d{s}epaka\.m tam dhanar\d{n}e\\
m{\=u}la\.m tat ca jye\d{s}\d{t}a m{\=u}la\.m vadanti\\

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Assume a number (x), which is called hrasva or kani\d{s}\d{t}a, its square should be multiplied by prak\d{r}ti and add or subtract a suitable number, called k\d{s}epaka, from it to make the remainder a perfect square. The square root of this number is called a larger root or jye\d{s}\d{t}ha m{\=u}lam.
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\frametitle{Vargaprak\d{r}ti: Some terminology}
In the equation $Nx^2+C=y^2$ \\
\begin{itemize}
\item N: Prak\d{r}ti
\item C: k\d{s}epaka
\item x: Kani\d{s}\d{t}ha root (smaller root)
\item y: Jye\d{s}\d{t}ha root (larger root)
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\frametitle{Bh{\=a}van{\=a}}
Bh{\=a}van{\=a} is the process of obtaining several solutions from a given solution of a vargaprak\d{r}ti equation. Assuming that there is one solution, below Bhaskaracharya describes various methods of obtaining other solutions derived from this basic solution.
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\frametitle{Bh{\=a}van{\=a}}
\textbf{hrasva jye\d{s}\d{t}a k\d{s}epak{\=a}n nyasya te\d{s}{\=a}m}\\
\vspace{0.2cm}
For the equation of the form $Nx^2+C=y^2$, write down suitable values of $x$, $y$ and $c$ in order.\\
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\textbf{t{\=a}n anya-anya-ada\d{h} nive\'{s}ya krame\d{n}a}\\
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Below them write the values of $x$,$y$ and $c$ of other similar(or the same) equation (with different $C$) in the same order.\\
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\textbf{s{\=a}dhya-anyebhya\d{h} bh{\=a}van{\=a}bhi\d{h} bah{\=u}ni}\\
\vspace{0.2cm}
By the process of bh{\=a}van{\=a}, we can obtain many values of $x$ and $y$.\\
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\textbf{m{\=u}l{\=a}ni e\d{s}{\=a}m bh{\=a}van{\=a} procchate ata\d{h}}\\
\vspace{0.2cm}
The process is called Bh{\=a}van{\=a}.
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\frametitle{Vargaprak\d{r}ti: Rule 1a}
\textbf{vajra-abhy{\=a}sau jye\d{s}\d{t}a laghvo\d{h} tadaikya\.m h\d{r}sva\.m}\\
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The sum of the cross-products of smaller and greater roots will be the smaller root of the new equation.\\
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\textbf{laghu {\=a}rohati\d{h} ca prak\d{r}ty{\=a} k\d{s}u\d{n}\d{n}{\=a}}\\
\textbf{jye\d{s}\d{t}a abhy{\=a}sayuk jye\d{s}\d{t}am{\=u}la\.m}\\
\vspace{0.2cm}
The products of smaller roots multiplied by the prak\d{r}ti added to the product of the greater roots will be the greater root of the new equation.\\
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\textbf{tatr{\=a}bhy{\=a}sa\d{h} k\d{s}epayo\d{h} k\d{s}epaka\d{h} sy{\=a}t}\\
The product of the constants will be the constant of the new equation.
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\frametitle{Vargaprak\d{r}ti: Rule 1a}
If $(x,y) = (\alpha,\beta)$ is a solution for the equation $Nx^2 + k = y^2$\\
and $(x,y) = (\alpha_{1},\beta_{1})$ is a solution for the equation $Nx^2 + k_1 = y^2$\\
then the solution of $Nx^2+kk_1=y^2$ is given by\\
$(x,y)=(\alpha\beta_{1}+\alpha_{1}\beta,N\alpha\alpha_{1}+\beta\beta_{1})$\\
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\frametitle{Vargaprak\d{r}ti: Rule 1b}
\textbf{hrasva\.m vajr{\=a}bhy{\=a}sayo\d{h} antaram v{\=a}}\\
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Otherwise, the difference of the cross-products of smaller and greater roots will be the new smaller root \\
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\textbf{laghvo\d{h} gh{\=a}ta\d{h} ya\d{h} prak\d{r}ty{\=a} vinighna\d{h}}\\
\vspace{0.2cm}
and the products of smaller roots multiplied with prakriti \\
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\textbf{gh{\=a}ta\d{h} ya\d{h} ca jye\d{s}\d{t}ayo\d{h} tad viyoga\d{h} jye\d{s}\d{t}a\.m} \\
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reduced by the product of greater roots is the new greater root. \\
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\textbf{k\d{s}epa\d{h} atra api ca k\d{s}epagh{\=a}ta\d{h}}\\
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Here also the product of the constants will be the constant of the new equation.
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\frametitle{Vargaprak\d{r}ti: Rule 1b}
If $(x,y) = (\alpha,\beta)$ is a solution for the equation $Nx^2 + k = y^2$\\
and $(x,y) = (\alpha_{1},\beta_{1})$ is a solution for the equation $Nx^2 + k_1 = y^2$\\
then the solution of $Nx^2+kk_1=y^2$ is given by\\
$(x,y)=(\alpha\beta_{1}-\alpha_{1}\beta,N\alpha\alpha_{1}-\beta\beta_{1})$\\
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\frametitle{Vargaprak\d{r}ti: Rule 1: Proof}
Let $Nx^2 + k = y^2$, and\\
$N_1x_1^2 + k_1 = y_1^2$.\\
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Hence, \\
$k k_1 = (y^2-Nx^2) (y_1^2-N_1x_1^2)$\\
$k k_1 = (y^2y_1^2+N^2x^2x_1^2-Nx^2y_1^2-Nx_1^2y^2)$\\
$Nx^2y_1^2+Nx_1^2y^2 + k k_1 = y^2y_1^2+Nx^2Nx_1^2$\\
$Nx^2y_1^2+Nx_1^2y^2\pm2Nxx_1yy_1 + k k_1 = y^2y_1^2+N^2x^2x_1^2\pm2Nxx_1yy_1$\\
$N(xy_1 \pm x_1y)^2 + k k_1 = (yy_1 \pm Nxx_1)^2$\\
\vspace{0.2cm}
which is of the form\\
\vspace{0.2cm}
$NX^2+K = Y^2$, where\\
$X = xy_1 \pm x_1y$, $Y = yy_1 \pm Nxx_1, and, K = kk_1$ \\
\vspace{0.2cm}
hence the solution of $Nx^2+kk_1=y^2$ is given by\\
$(x,y)=(\alpha\beta_{1}\pm\alpha_{1}\beta,N\alpha\alpha_{1}\pm\beta\beta_{1})$\\
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\frametitle{Vargaprak\d{r}ti: Rule 1:Special case}
If $(x,y) = (\alpha,\beta)$ is a solution for the equation $Nx^2 + k = y^2$\\
then $Nx^2 + k^2 = y^2$ will have the solution\\
$(x,y)=(2\alpha\beta,N\alpha^2+\beta^2)$\\
\vspace{0.5cm}
If $(x,y) = (\alpha,\beta)$ is a solution for the equation $Nx^2 + 1 = y^2$\\
then another solution is give by\\
$(x,y)=(2\alpha\beta,N\alpha^2+\beta^2)$\\
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\frametitle{Vargaprak\d{r}ti: Rule 1 Special Case :Proof}
$Nx^2 + 1 = y^2$\\
$1 = (y^2-Nx^2)^2$\\
$1 = y^4+N^2x^4-2Nx^2y^2$\\
$1+4Nx^2y^2 = y^4+N^2x^4+2Nx^2y^2$\\
$1+N(2xy)^2 = (y^2+Nx^2)^2$\\
Let $2xy$ = $X$ and $y^2+Nx^2$ = $Y$\\
Then $(X,Y)$ is a solution of $NX^2+1=Y^2$\\

Thus if $(\alpha,\beta)$ is a solution of $1+Nx^2=y^2$, then\\
$(X,Y)=(2\alpha\beta,N\alpha^2+\beta^2)$ is also its another solution.\\
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\frametitle{Vargaprak\d{r}ti: Rule 2}
\textbf{i\d{s}\d{t}a varga h\d{r}ta\d{h} k\d{s}epa\d{h}}\\
\textbf{k\d{s}epasya {\=a}di\d{s}\d{t}a bh{\=a}jite}\\
\textbf{m{\=u}le te sta\d{h} athav{\=a} k\d{s}epa\d{h}}\\
\textbf{k\d{s}u\d{n}\d{n}a\d{h} k\d{s}u\d{n}\d{n}e tad{\=a} pade}\\

\vspace{0.5cm}
If the constant is divided by the square of a desired number to form new constant, the values of $x$ and $y$ should be divided by the same desired number to get the new values of $x$ and $y$. \\
\vspace{0.2cm}
Similarly, if the constant is multipied by the square of a desired number, the values of $x$ and $y$ are to be multipled by the same desired number to get the new values of $x$ and $y$.
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\frametitle{Vargaprak\d{r}ti: Rule 2}
If $(x,y) = (\alpha,\beta)$ is a solution for the equation $Nx^2 + c = y^2$\\
then the solution of $Nx^2+ck^2=y^2$ is given by\\
$(x,y)=(\alpha * k ,\beta * k)$\\
\vspace{0.5cm}
If $(x,y) = (\alpha,\beta)$ is a solution for the equation $Nx^2 + c = y^2$\\
then the solution of $Nx^2+\frac{c}{k^2}=y^2$ is given by\\
$(x,y)=(\frac{\alpha}{k} ,\frac{\beta}{k})$\\
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\frametitle{Vargaprak\d{r}ti: Rule 3}
How to get the initial solution?\\
\vspace{0.2cm}
\textbf{i\d{s}\d{t}a varga prak\d{r}tyo\d{h}} \\
\textbf{yad vivara\d{m} tena v{\=a} bhajed}\\
\textbf{dvighnami\d{s}\d{t}a\.m kani\d{s}\d{t}a\.m} \\
\textbf{tat padam sy{\=a}d eka sa\.myutau}\\
\textbf{tata\d{h} jye\d{s}\d{t}am} \\
\vspace{0.2cm}
If the constant (c) is equal to 1, the value of $x$ can be obtained by dividing twice a desired number by the difference between the prak\d{r}ti and the square of this desired number. From this the greater root (y) can be obtained.\\
\vspace{0.2cm}
So $x = \frac{2a}{N-a^2}$ is a possible value of the smaller root for an equation $Nx^2+1=y^2$ where $a$ be some desired number.\\
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\textbf{iha anantyam bh{\=a}van{\=a} tathe\d{s}ata\d{h}}\\
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By Bh{\=a}van{\=a}, a number of such solutions can be obtained.
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\frametitle{Vargaprak\d{r}ti: Rule 3 Proof}
Consider the equation $Nx^2+1=y^2$.\\
let $a$ be some desired number.\\
Then, $(N-a^2)^2 + 4Na^2 = (N+a^2)^2$\\
\vspace{0.5cm}
Dividing both sides by $(N-a^2)^2$, we get\\
$1 + \frac{4Na^2}{(N-a^2)^2} = \frac{(N+a^2)^2}{(N-a^2)^2}$\\
i.e., $1 + N (\frac{2a}{N-a^2})^2 = \frac{(N+a^2)^2}{(N-a^2)^2}$\\
\vspace{0.5cm}
So $x = \frac{2a}{N-a^2}$ is a possible value of the smaller root.
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\frametitle{An Example}
\textbf{ko varga a\d{s}\d{t}a hata\d{h} sa-eka\d{h}} \\
\textbf{k\d{r}ti\d{h} sy{\=a}d ga\d{n}aka ucyat{\=a}m}\\
\textbf{ek{\=a}da\'{s}a gu\d{n}a\d{h} ko v{\=a} }\\
\textbf{varga\d{h} sa-eka\d{h} k\d{r}tibhavet}\\
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O, Mathematician, tell me the number whose square multiplied by 8 and increased by one becomes a square number.\\
\vspace{0.2cm}
Also tell me the number whose square multiplied by 11 and increased by one gives a square number.\\
\vspace{0.2cm}
Solve the eqautions: $8x^2 + 1 = y^2$, and $11x^2 + 1 = y^2$.\\
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\frametitle{An Example: Solution}
Consider the eqaution: $8x^2 + 1 = y^2$\\
\vspace{0.2cm}
For some $a$, $x = \frac{2*a}{8-a^2}$.\\
For $a=1$, $x = \frac{2}{7}$\\
For $a=2$, $x = \frac{4}{4} = 1$\\

\vspace{0.2cm}
For $x=1$, $y = 3$.
If $(1,3)$ and $(1,3)$ are two roots, then using bh{\=a}van{\=a}, we get \\
$(x,y) = (1*3+1*3, 8*1*1+3*3) = (6,17)$ as the second solution.\\
\vspace{0.2cm}
If $(1,3)$ and $(6,17)$ are two roots, then using bh{\=a}van{\=a}, we get \\
$(x,y) = (1*17+3*6, 8*1*6+3*17) = (35,99)$ as the third solution.\\
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\frametitle{An Example: Solution}
Consider the eqaution: $11x^2 + 1 = y^2$\\
\vspace{0.2cm}
For some $a$, $x = \frac{2*a}{11-a^2}$.\\
For $a=1$, $x = \frac{2}{10}$\\
For $a=2$, $x = \frac{4}{7}$\\
For $a=3$, $x = \frac{6}{2} = 3$\\

\vspace{0.2cm}
For $x=3$, $y = 10$.
If $(3,10)$ and $(3,10)$ are two roots, then using bh{\=a}van{\=a}, we get \\
$(x,y) = (2*3*10, 11*3*3+10*10) = (60,199)$ as the second solution.\\
\vspace{0.2cm}
If $(3,10)$ and $(60,199)$ are two roots, then using bh{\=a}van{\=a}, we get \\
$(x,y) = (3*199+10*60, 11*3*60+10*199) = (1197,3970)$ as the third solution.\\
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\frametitle{An Example: Alternative method}
Consider the eqaution: $11x^2 + 1 = y^2$\\
\vspace{0.2cm}
Find some $x$ such that for some $k$, $11x^2+k$ is a perfect square.\\
Let $x = 1$ and $k = -2$ is such pair,  since $11 * 1^2 - 2 = 9 =3^2$.\\
Thus $(1,3)$ is a solution to $11x^2 - 2 = y^2$.\\
\vspace{0.2cm}
Using Bh{\=a}van{\=a},\\
If $(x,y) = (\alpha,\beta)$ is a solution for the equation $Nx^2 + k = y^2$\\
then $Nx^2 + k^2 = y^2$ will have the solution\\
$(x,y)=(2\alpha\beta,N\alpha^2+\beta^2)$\\
\vspace{0.2cm}
Hence $(2*1*3,11*1^2+3^2)$ = $(6,20)$ is a solution of $11x^2+4=y^2$.\\
\vspace{0.2cm}
Using Bh{\=a}van{\=a} again,\\
If $(x,y) = (\alpha,\beta)$ is a solution for the equation $Nx^2 + c = y^2$\\
then the solution of $Nx^2+\frac{c}{k^2}=y^2$ is given by\\
$(x,y)=(\frac{\alpha}{k} ,\frac{\beta}{k})$\\
\vspace{0.2cm}
$(\frac{6}{2}, \frac{20}{2}) = (3,10)$ is the solution of $11x^2+1=y^2$.
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\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}
$6x^2+3 = y^2$ \\
$6x^2+12 = y^2$ \\
$6x^2+75 = y^2$ \\
$6x^2+300 = y^2$ \\

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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 $(10,30)$\\

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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., $(\frac{2}{2},3) = (1,3)$.
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\begin{frame}
\frametitle{Example}
\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\d{r}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\\
$(2*\frac{2}{4}*3,32*\frac{2}{4}\frac{2}{4}*3*3)$ = $(3,17)$.
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\begin{frame}
\frametitle{Example}
\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 $y = \frac{7}{2}$.

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\begin{frame}
\frametitle{General Observations}
While, both the methods provide a way to get a solution, it is not clear whether we will get the solution at all, and if yes, how many such interations are needed. Or in other words, how fast the solution will converge?\\
\vspace{0.5cm}
Bh{\=a}skar{\=a}ch{\=a}rya uses a combination of ku\d{t}\d{t}aka and bh{\=a}van{\=a} for fast convergence.
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