# Pell's equation

We will discuss below whether Pell's equation is properly named. By this we mean simply: did Pell contribute at all to the study of Pell's equation? There is no doubt that the equation had been studied in depth for hundreds of years before Pell was born. In fact the first contribution by Brahmagupta was made around 1000 years before Pell's time and it is with Brahmagupta's contribution that we begin our historical study.

First let us say what Pell's equation is. We are talking about the indeterminate quadratic equation

Now, although it is fair to say that Brahmagupta was the first to study this equation, it is equally possible to see that earlier authors had studied problems related to Pell's equation. To mention some briefly: Diophantus examines problems related to Pell's equation and we can reduce Archimedes' "cattle problem" to solving Pell's equation although there is no evidence that Archimedes made this connection.

Let us first note that

One property that he deduced was that if $(a, b)$ satisfies Pell's equation so does $(2ab, b^{2} + na^{2})$. This follows immediately applying the method of composition to $(a, b)$ and $(a, b)$. Now of course the method of composition can be applied again to $(a, b)$ and $(2ab, b^{2}+ na^{2})$ to get another solution and Brahmagupta immediately saw that from one solution of Pell's equation he could generate many solutions.

This was not the only way that Brahmagupta used the method of composition. He also noted that, using a similar argument to what we have just given, if $x = a, y = b$ is a solution of $nx^{2} + k = y^{2}$ then applying the method of composition to $(a, b)$ and $(a, b)$ gave $(2ab, b^{2} + na^{2})$ as a solution of $nx^{2} + k^{2} = y^{2}$ and so, dividing through by $k^{2}$, gives

How does this help? These values of $x$ and $y$ do not look like integers. Well if $k = 2$ then, since $(a, b)$ is a solution of $nx^{2} + k = y^{2}$ we have $na^{2} = b^{2} - 2$. Thus

For example, if we attempt to solve $23x^{2} + 1 = y^{2}$ we see that $a = 1, b = 5$ satisfies $23a^{2} + 2 = b^{2}$ so, by the above argument, $x = 5, y = 24$ satisfies Pell's equation. Applying the method of composition to the pair (5, 24) gives

Among the examples Brahmagupta gives himself is a solution of Pell's equation

and applies his method to find the solution

One can only marvel at this brilliant work done by Brahmagupta in 628 AD.

The next step forward was taken by Bhaskara II in 1150. He discovered the cyclic method, called

The method relies on a simple observation, namely that, for any $m, (1, m)$ satisfies the 'Pell type equation'

Next Bhaskara II knows that there are infinitely many $m$ such that $am + b$ is divisible by $k$. He chooses the one which makes $m^{2} - n$ as small as possible in absolute value. If $\large \frac {m^{2} - n}{k}$ is one of 1, -1, 2, -2, 4, -4 then we can apply Brahmagupta's method to find a solution to Pell's equation $nx^{2} + 1 = y^{2}$. If $\large \frac {m^{2} - n}{k}$ is not one of these values then repeat the process starting this time with the solution $x = \large \frac {am + b}{k} \normalsize, y = \large \frac {bm + na}{k}$ to the 'Pell type equation' $nx^{2} + \large \frac {m^{2} - n}{k} \normalsize = y^{2}$ in exactly the same way as we applied the process to $na^{2} + k = b^{2}$. Bhaskara II knows (almost certainly by experience rather than by having a proof) that the process will end after a finite number of steps. This happens when an equation of the form $nx^{2} + t = y^{2}$ is reached where $t$ is one of 1, -1, 2, -2, 4, -4.

Bhaskara II gives examples in

Why do we suspect that Bhaskara II had no proof of the method? Well there are at least two reasons. Firstly the proof is long and difficult and would appear to be well beyond 12th Century mathematics. Secondly the algorithm always reaches a solution of Pell's equation after a finite number of steps without stopping when an equation of the type $nx^{2} + k = y^{2}$ where $k$ = -1, 2, -2, 4, or -4 is reached and then applying Brahmagupta's method. If experience of the algorithm is only via examples then, knowing how to proceed when $k$ = -1, 2, -2, 4, or -4 is reached, it is natural to switch to Brahmagupta's method at that point. However, when one writes down a proof it should become clear that the algorithm switching to Brahmagupta's method is never necessary (although can reach the solution more quickly).

The next contribution to Pell's equation was made by Narayana who, in the 14th Century, wrote a commentary on Bhaskara II's

In

Wallis published

We should note that by this time several mathematicians had claimed that Pell's equation $nx^{2} + 1 = y^{2}$ had solutions for any $n$. Wallis, describing Brouncker's method, had made that claim, as had Fermat when commenting on the solutions proposed to his challenge. In fact Fermat claimed, correctly of course, that for any $n$ Pell's equation had infinitely many solutions.

Euler gave Brahmagupta's lemma and its proof in a similar form to that which we have given above. He was, of course, aware of the work of Brouncker on Pell's equation as presented by Wallis, but he was totally unaware of the contributions of the Indian mathematicians. He gave the basis for the continued fractions approach to solving Pell's equation which was put into a polished form by Lagrange in 1766. The other major contribution of Euler was in naming the equation "Pell's equation" and it is generally believed that he gave it that name because he confused Brouncker and Pell, thinking that the major contributions which Wallis had reported on as due to Brouncker were in fact the work of Pell.

Lagrange published his

For example √19 has the continued fraction expansion

$\sqrt{19} = 4+ \frac{1}{2+} \frac{1}{1+} \frac{1}{3+} \frac{1}{1+} \frac{1}{2+} \frac{1}{8+} \frac{1}{2+} \frac{1}{1+} \frac{1}{3+} \frac{1}{1+} \frac{1}{2+} \frac{1}{8+} \frac{1}{2+} \frac{1}{1+} ...$

which stands for

$\sqrt{19} = 4+ \large\frac{1}{1+ \large\frac{1}{2+\large\frac{1}{1 + \large\frac{1}{3 + \large\frac{1}{2 + \large\frac{1}{8 + \large\frac{1}{2 + \large\frac{1}{1 + ...}}}}}}}}$

which recurs with length 6. The convergent immediately before the point from which it repeats is $\large\frac{170}{39}\normalsize$ and Lagrange's theory says that

First let us say what Pell's equation is. We are talking about the indeterminate quadratic equation

$nx^{2} + 1 = y^{2}$

which we can also write as
$y^{2} - nx^{2} = 1$

where $n$ is a given integer and we are looking for integer solutions $(x, y)$.
Now, although it is fair to say that Brahmagupta was the first to study this equation, it is equally possible to see that earlier authors had studied problems related to Pell's equation. To mention some briefly: Diophantus examines problems related to Pell's equation and we can reduce Archimedes' "cattle problem" to solving Pell's equation although there is no evidence that Archimedes made this connection.

Let us first note that

$(b^{2} - na^{2})(d^{2} - nc^{2}) = (bd + nac)^{2} - n(bc + ad)^{2}$

and
$(b^{2} - na^{2})(d^{2} - nc^{2}) = (bd - nac)^{2} - n(bc - ad)^{2}$

From this we see that if
$b^{2} - na^{2} = 1$ and $d^{2} - nc^{2} = 1$

then
$(bd + nac)^{2} - n(bc + ad)^{2} = 1$

and
$(bd - nac)^{2} - n(bc - ad)^{2} = 1$.

In other words, if $(a, b)$ and $(c, d)$ are solutions to Pell's equation then so are
$(bc + ad, bd + nac)$ and $(bc - ad, bd - nac)$.

This fundamentally important fact generalises easily to give Brahmagupta's lemma, namely that if $(a, b$) and $(c, d)$ are integer solutions of 'Pell type equations' of the form
$na^{2} + k = b^{2}$ and $nc^{2} + k' = d^{2}$

then
$(bc + ad, bd + nac)$ and $(bc - ad, bd - nac)$

are both integer solutions of the 'Pell type equation'
$nx^{2} + kk' = y^{2}$.

The proof that we have given is due to European mathematicians in the 17th Century (and we shall comment further on it later in this article), but Brahmagupta's lemma was discovered by Brahmagupta in 628 AD. The method was called *samasa*by the Indian mathematicians but we shall call it the 'method of composition'. In fact this method of composition allowed Brahmagupta to make a number of fundamental discoveries regarding Pell's equation.One property that he deduced was that if $(a, b)$ satisfies Pell's equation so does $(2ab, b^{2} + na^{2})$. This follows immediately applying the method of composition to $(a, b)$ and $(a, b)$. Now of course the method of composition can be applied again to $(a, b)$ and $(2ab, b^{2}+ na^{2})$ to get another solution and Brahmagupta immediately saw that from one solution of Pell's equation he could generate many solutions.

This was not the only way that Brahmagupta used the method of composition. He also noted that, using a similar argument to what we have just given, if $x = a, y = b$ is a solution of $nx^{2} + k = y^{2}$ then applying the method of composition to $(a, b)$ and $(a, b)$ gave $(2ab, b^{2} + na^{2})$ as a solution of $nx^{2} + k^{2} = y^{2}$ and so, dividing through by $k^{2}$, gives

$x = \large \frac {2ab}{k}\normalsize, y = \large \frac {b^{2} + na^{2}}{k}$

as a solution of Pell's equation $nx^{2} + 1 = y^{2}$.
How does this help? These values of $x$ and $y$ do not look like integers. Well if $k = 2$ then, since $(a, b)$ is a solution of $nx^{2} + k = y^{2}$ we have $na^{2} = b^{2} - 2$. Thus

$x = \large \frac{ 2ab}{2} \normalsize = ab$,

$y = \large \frac {b^{2} + na^{2}}{2} \normalsize = \large \frac {2b^{2} - 2}{2} \normalsize = b^{2} - 1$

and this is an integer solution to Pell's equation. If $k = -2$ then essentially the same argument works while if $k = 4$ or $k = -4$ then a more complicated method, still based on the method of composition, shows that integer solutions to Pell's equation can be found. So Brahmagupta was able to show that if he could find $(a, b)$ which "nearly" satisfied Pell's equation in the sense that $na^{2} + k = b^{2}$ where $k$= 1, -1, 2, -2, 4, or -4 then he could find one, and therefore many, integer solutions to Pell's equation. Often he could find trial solutions which worked for one of these values of $k$ and so in many cases he was able to give solutions.
$y = \large \frac {b^{2} + na^{2}}{2} \normalsize = \large \frac {2b^{2} - 2}{2} \normalsize = b^{2} - 1$

For example, if we attempt to solve $23x^{2} + 1 = y^{2}$ we see that $a = 1, b = 5$ satisfies $23a^{2} + 2 = b^{2}$ so, by the above argument, $x = 5, y = 24$ satisfies Pell's equation. Applying the method of composition to the pair (5, 24) gives

$x = 2\times 5\times 24 = 240, y = 24^{2} + 23\times 5^{2} = 1151$

as another solution. Applying the method of composition again yields
$x = 11515, y = 55224$

and yet again gives
$x = 552480, y = 2649601$

and so on.
Among the examples Brahmagupta gives himself is a solution of Pell's equation

$83x^{2} + 1 = y^{2}$

where he notes that the pair (1, 9) satisfies
83×12 - 2 = 92

and applies his method to find the solution

$x = 9, y = 82$.

We can now generate a sequence of solutions $(x,y)$:
(9, 82),

(1476, 13447),

(242055, 2205226),

(39695544, 361643617),

(6509827161, 59307347962),

(1067571958860, 9726043422151),

(175075291425879, 1595011813884802)

and so on.
(1476, 13447),

(242055, 2205226),

(39695544, 361643617),

(6509827161, 59307347962),

(1067571958860, 9726043422151),

(175075291425879, 1595011813884802)

One can only marvel at this brilliant work done by Brahmagupta in 628 AD.

The next step forward was taken by Bhaskara II in 1150. He discovered the cyclic method, called

*chakravala*by the Indians, which was an algorithm to produce a solution to Pell's equation $nx^{2} + 1 = y^{2}$ starting off from any "close" pair $(a, b)$ with $na^{2} + k = b^{2}$. We can assume that $a$ and $b$ are coprime, for otherwise we could divide each by their gcd and get a "closer" solution with smaller $k$. Clearly $a, k$ are then also coprime.The method relies on a simple observation, namely that, for any $m, (1, m)$ satisfies the 'Pell type equation'

$n.1^{2} + (m^{2} - n) = m^{2}$.

Bhaskara II now applies the method of composition to the pairs $(a, b)$ and $(1, m)$ to get
$n(am + b)^{2} + (m^{2} - n)k = (bm + na)^{2}$.

Dividing by $k$ we note that
$x = \large \frac {am + b}{k} \normalsize, y = \large \frac {bm + na}{k}$

is a solution to
$nx^{2} + \large \frac {m^{2} - n}{k}\normalsize = y^{2}$.

Since $a, k$ are coprime we can chose $m$ such that $am + b$ is divisible by $k$. Bhaskara II now knows (but he gives no proof) that when $m$ is chosen so that $am + b$ is divisible by $k$ then $m^{2} - n$ and $bm + na$ are also divisible by $k$. With such a choice of $m$ he therefore has integer solutions
$x = \large \frac {am + b}{k} \normalsize, y = \large \frac {bm + na}{k}$

to the 'Pell type equation' $nx^{2} + \large \frac {m^{2} - n}{k} \normalsize = y^{2}$ where $\large \frac {m^{2} - n}{k}$ is also an integer.
Next Bhaskara II knows that there are infinitely many $m$ such that $am + b$ is divisible by $k$. He chooses the one which makes $m^{2} - n$ as small as possible in absolute value. If $\large \frac {m^{2} - n}{k}$ is one of 1, -1, 2, -2, 4, -4 then we can apply Brahmagupta's method to find a solution to Pell's equation $nx^{2} + 1 = y^{2}$. If $\large \frac {m^{2} - n}{k}$ is not one of these values then repeat the process starting this time with the solution $x = \large \frac {am + b}{k} \normalsize, y = \large \frac {bm + na}{k}$ to the 'Pell type equation' $nx^{2} + \large \frac {m^{2} - n}{k} \normalsize = y^{2}$ in exactly the same way as we applied the process to $na^{2} + k = b^{2}$. Bhaskara II knows (almost certainly by experience rather than by having a proof) that the process will end after a finite number of steps. This happens when an equation of the form $nx^{2} + t = y^{2}$ is reached where $t$ is one of 1, -1, 2, -2, 4, -4.

Bhaskara II gives examples in

*Bijaganita*and the first we look at is$61x^{2} + 1 = y^{2}$.

Using the above method he chooses $m$ so the $(m + 8)/3$ is an integer, making sure that $m^{2} - 61$ is as small as possible. Taking $m = 7$ he gets
$x = 5, y = 39$

as a solution to the 'Pell type equation' $nx^{2} - 4 = y^{2}$. But this is an equation which Brahmagupta's method solves giving
$x = 226153980, y = 1766319049$

as the smallest solution to $61x^{2} + 1 = y^{2}$.
Why do we suspect that Bhaskara II had no proof of the method? Well there are at least two reasons. Firstly the proof is long and difficult and would appear to be well beyond 12th Century mathematics. Secondly the algorithm always reaches a solution of Pell's equation after a finite number of steps without stopping when an equation of the type $nx^{2} + k = y^{2}$ where $k$ = -1, 2, -2, 4, or -4 is reached and then applying Brahmagupta's method. If experience of the algorithm is only via examples then, knowing how to proceed when $k$ = -1, 2, -2, 4, or -4 is reached, it is natural to switch to Brahmagupta's method at that point. However, when one writes down a proof it should become clear that the algorithm switching to Brahmagupta's method is never necessary (although can reach the solution more quickly).

The next contribution to Pell's equation was made by Narayana who, in the 14th Century, wrote a commentary on Bhaskara II's

*Bijaganita.*Narayana gave some new examples of the cyclic method. Here are two of his examples:$103x^{2} + 1 = y^{2}$.

Choosing $a = 1, b = 10$ Narayana obtains
$103\times 1^{2} - 3 = 10^{2}$.

Choose $m$ so that $m + 10$ is divisible by -3 with $m^{2} - 103$ as small as possible leads to $m = 11$ and we obtain
$103\times 7^{2} - 6 = 71^{2}$.

Next we must choose $m$ so that $7m + 71$ is divisible by -6 and $m^{2} - 103$ as small as possible. Take $m = 7$ to get the equation
$103\times 20^{2} + 9 = 203^{2}$.

Continuing, choose $m$ so that $20m+203$ is divisible by 9 and $m^{2} - 103$ as small as possible. Take $m = 11$ to get the equation
$103\times 47^{2} + 2 = 477^{2}$.

Now Narayana applies Brahmagupta's method, in the form we gave above for equations with $k = 2$, to obtain the solutions
$x = 22419, y = 227528$.

His next example is a solution of Pell's equation
$97x^{2} + 1 = y^{2}$

which leads successively, by applying the cyclic method, to the equations
$97\times 1^{2} + 3 = 10^{2}$

$97\times 7^{2} + 8 = 69^{2}$

$97\times 20^{2} + 9 = 197^{2}$

$97\times 53^{2} + 11 = 522^{2}$

$97\times 86^{2} - 3 = 847^{2}$

$97\times 569^{2} - 1 = 5604^{2}$

Finally Narayana applies Brahmagupta's method to this last equation to get the solution
$97\times 7^{2} + 8 = 69^{2}$

$97\times 20^{2} + 9 = 197^{2}$

$97\times 53^{2} + 11 = 522^{2}$

$97\times 86^{2} - 3 = 847^{2}$

$97\times 569^{2} - 1 = 5604^{2}$

$x = 6377352, y = 62809633$

Now the brilliant ideas of Brahmagupta, Bhaskara II and Narayana were completely unknown to the European mathematicians in the 17th Century. The European interest began in 1657 when Fermat issued a challenge to the mathematicians of Europe and England. Fermat wrote:-
We await these solutions, which, if England or Belgic or Celtic Gaul do not produce, then Narbonese Gaul will.Narbonese Gaul, of course, was the area around Toulouse where Fermat lived! One of Fermat's challenge problems was the same example of Pell's equation which had been studied by Bhaskara II 500 years earlier, namely to find solutions to

$61x^{2} + 1 = y^{2}$.

Several mathematicians participated in Fermat's challenge, in particular Frenicle de Bessy, Brouncker and Wallis. There followed an exchange of letters between these mathematicians during 1657-58 which Wallis published in *Commercium epistolicum*in 1658. Brouncker discovered a method of solution which is essentially the same as the method of continued fractions which was later developed rigorously by Lagrange. Frenicle de Bessy tabulated the solutions of Pell's equation for all $n$ up to 150, although this was never published and his efforts have been lost. He challenged Brouncker who was claiming to be able to solve any example of Pell's equation to solve$313x^{2} + 1 = y^{2}$.

Brouncker found the smallest solutions, using his method, which is
$x = 1819380158564160, y = 32188120829134849$

and he sent these to Frenicle de Bessy claiming that they had only taken him "an hour or two" to find.
In

*Commercium epistolicum*Wallis gave two methods of proving Brahmagupta's lemma which are both essentially equivalent to the argument we gave at the beginning of this article based on the result$(b^{2} - na^{2})(d^{2} - nc^{2}) = (bd + nac)^{2} - n(bc + ad)^{2}$.

In 1658 Rahn published an algebra book which contained an example of Pell's equation. This book was written with help from Pell and it is the only known connection between Pell and the equation which has been named after him.
Wallis published

*Treatise on Algebra*in 1685 and Chapter 98 of that work is devoted to giving methods to solve Pell's equation based on the exchange of letters he had published in*Commercium epistolicum*in 1658. However, in his algebra text Wallis put all the methods into a standard form.We should note that by this time several mathematicians had claimed that Pell's equation $nx^{2} + 1 = y^{2}$ had solutions for any $n$. Wallis, describing Brouncker's method, had made that claim, as had Fermat when commenting on the solutions proposed to his challenge. In fact Fermat claimed, correctly of course, that for any $n$ Pell's equation had infinitely many solutions.

Euler gave Brahmagupta's lemma and its proof in a similar form to that which we have given above. He was, of course, aware of the work of Brouncker on Pell's equation as presented by Wallis, but he was totally unaware of the contributions of the Indian mathematicians. He gave the basis for the continued fractions approach to solving Pell's equation which was put into a polished form by Lagrange in 1766. The other major contribution of Euler was in naming the equation "Pell's equation" and it is generally believed that he gave it that name because he confused Brouncker and Pell, thinking that the major contributions which Wallis had reported on as due to Brouncker were in fact the work of Pell.

Lagrange published his

*Additions to Euler's Elements of algebra*in 1771 and this contains his rigorous version of Euler's continued fraction approach to Pell's equation. This established rigorously the fact that for every $n$ Pell's equation had infinitely many solutions. The solution depends on the continued fraction expansion of $√n$. In the continued fraction of the square root of an integer the same denominators recur periodically. Moreover, the pattern in most of the recurring sequence is "palindromic". i.e. up to the last element, the second half of the periodic sequence is the first half in reverse. The last number in the repeating sequence is double the integer part of the square root.For example √19 has the continued fraction expansion

$\sqrt{19} = 4+ \frac{1}{2+} \frac{1}{1+} \frac{1}{3+} \frac{1}{1+} \frac{1}{2+} \frac{1}{8+} \frac{1}{2+} \frac{1}{1+} \frac{1}{3+} \frac{1}{1+} \frac{1}{2+} \frac{1}{8+} \frac{1}{2+} \frac{1}{1+} ...$

which stands for

$\sqrt{19} = 4+ \large\frac{1}{1+ \large\frac{1}{2+\large\frac{1}{1 + \large\frac{1}{3 + \large\frac{1}{2 + \large\frac{1}{8 + \large\frac{1}{2 + \large\frac{1}{1 + ...}}}}}}}}$

which recurs with length 6. The convergent immediately before the point from which it repeats is $\large\frac{170}{39}\normalsize$ and Lagrange's theory says that

$x = 39, y = 170$

will be the smallest solution to Pell's equation
$19x^{2} + 1 = y^{2}$.

To find the infinite series of solutions take the powers of 170 + 39√19. For example
$(170 + 39√19)^{2} = 57799 + 13260√19$

and
$x = 13260, y = 57799$

will give a second solution to the equation. Again
$(170 + 39√19)^{3} = 19651490 + 4508361√19$

giving
$x = 4508361, y = 19651490$

as the next solution. Here are the first few powers of (170 + 39√19), starting with its square, which gives the first few solutions to the equation 19x2 + 1 = y2
57799 + 13260√19

19651490 + 4508361√19

6681448801 + 1532829480√19

2271672940850 + 521157514839√19

772362118440199 + 177192022215780√19

262600848596726810 + 60244766395850361√19

89283516160768675201 + 20483043382566906960√19

Although the continued fraction approach to solving Pell's equation is a very nice one for small values on $n$, the difficulty of the method has been analysed to see if it is the most efficient for large $n$. A polynomial time method in the length of the input $n$ would be an algorithm which took time bounded by a fixed power of log $n$ (the length of the input). The continued fraction method is not a polynomial time algorithm, and indeed it is now known that no polynomial time algorithm exists for solving Pell's equation.
19651490 + 4508361√19

6681448801 + 1532829480√19

2271672940850 + 521157514839√19

772362118440199 + 177192022215780√19

262600848596726810 + 60244766395850361√19

89283516160768675201 + 20483043382566906960√19

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### Additional Resources (show)

Other websites about Pell's equation:

Written by J J O'Connor and E F Robertson

Last Update February 2002

Last Update February 2002