k = ±1

Using Brahmagupta's identity to compose the triple ${\displaystyle (a,b,k)}$ with itself:

${\displaystyle (a^{2}+Nb^{2})^{2}-N(2ab)^{2}=k^{2}}$ ${\displaystyle \Rightarrow }$ ${\displaystyle (2a^{2}-k)^{2}-N(2ab)^{2}=k^{2}}$

The new triple can be expressed as ${\displaystyle (2a^{2}-k,2ab,k^{2})}$.

Substituting ${\displaystyle k=-1}$ gives a solution:

${\displaystyle x=2a^{2}+1,y=2ab}$

For ${\displaystyle k=1}$, the original ${\displaystyle (a,b)}$ was already a solution. Substituting ${\displaystyle k=1}$ yields a second:

${\displaystyle x=2a^{2}-1,y=2ab}$

k = ±2

Again using the equation, ${\displaystyle (2a^{2}-k)^{2}-N(2ab)^{2}=k^{2}}$${\displaystyle \Rightarrow }$${\displaystyle \left({\frac {2a^{2}-k}{k}}\right)^{2}-N\left({\frac {2ab}{k}}\right)^{2}=1}$

Substituting ${\displaystyle k=2}$,

${\displaystyle x=a^{2}-1,y=ab}$

Substituting ${\displaystyle k=-2}$,

${\displaystyle x=a^{2}+1,y=ab}$

k = 4

Substituting ${\displaystyle k=4}$ into the equation ${\displaystyle ({\frac {2a^{2}-4}{4}})^{2}-N({\frac {2ab}{4}})^{2}=1}$ creates the triple ${\displaystyle ({\frac {a^{2}-2}{2}},{\frac {ab}{2}},1)}$.

Which is a solution if ${\displaystyle a}$ is even:

${\displaystyle x={\frac {a^{2}-2}{2}},y={\frac {ab}{2}}}$

If a is odd, start with the equations ${\displaystyle ({\frac {a}{2}})^{2}-N({\frac {b}{2}})^{2}=1}$ and ${\displaystyle ({\frac {2a^{2}-4}{4}})^{2}-N({\frac {2ab}{4}})^{2}=1}$.

Leading to the triples ${\displaystyle ({\frac {a}{2}},{\frac {b}{2}},1)}$ and ${\displaystyle ({\frac {a^{2}-2}{2}},{\frac {ab}{2}},1)}$. Composing the triples gives ${\displaystyle ({\frac {a}{2}}(a^{2}-3))^{2}-N({\frac {b}{2}}(a^{2}-1))^{2}=1}$

When ${\displaystyle a}$ is odd,

${\displaystyle x={\frac {a}{2}}(a^{2}-3)),y=({\frac {b}{2}}(a^{2}-1))}$

k = -4

When ${\displaystyle k=-4}$, then ${\displaystyle ({\frac {a}{2}})^{2}-N({\frac {b}{2}})^{2}=-1}$. Composing with itself yields ${\displaystyle ({\frac {a^{2}+Nb^{2}}{4}})^{2}-N({\frac {ab}{2}})^{2}=1}$${\displaystyle \Rightarrow }$${\displaystyle ({\frac {a^{2}+2}{2}})^{2}-N({\frac {ab}{2}})^{2}=1}$.

Again composing itself yields ${\displaystyle ({\frac {(a^{2}+2)^{2}+Na^{2}b^{2})}{4}})^{2}-N({\frac {ab(a^{2}+2)}{2}})^{2}=1}$${\displaystyle \Rightarrow }$ ${\displaystyle ({\frac {a^{4}+4a^{2}+2}{2}})^{2}-N({\frac {ab(a^{2}+2)}{2}})^{2}=1}$

Finally, from the earlier equations, compose the triples ${\displaystyle ({\frac {a^{2}+2}{2}},{\frac {ab}{2}},1)}$ and ${\displaystyle ({\frac {a^{4}+4a^{2}+2}{2}},{\frac {ab(a^{2}+2)}{2}},1)}$, to get

${\displaystyle ({\frac {(a^{2}+2)(a^{4}+4a^{2}+2)+Na^{2}b^{2}(a^{2}+2)}{4}})^{2}-N({\frac {ab(a^{4}+4a^{2}+3)}{2}})^{2}=1}$${\displaystyle \Rightarrow }$${\displaystyle ({\frac {(a^{2}+2)(a^{4}+4a^{2}+1)}{2}})^{2}-N({\frac {ab(a^{2}+3)(a^{2}+1)}{2}})^{2}=1}$${\displaystyle \Rightarrow }$${\displaystyle ({\frac {(a^{2}+2)[(a^{2}+1)(a^{2}+3)-2)]}{2}})^{2}-N({\frac {ab(a^{2}+3)(a^{2}+1)}{2}})^{2}=1}$.

This give us the solutions

${\displaystyle x={\frac {(a^{2}+2)[(a^{2}+1)(a^{2}+3)-2)]}{2}}y={\frac {ab(a^{2}+3)(a^{2}+1)}{2}}}$^[11]

(Note, ${\displaystyle k=-4}$ is useful to find a solution to Pell's Equation, but it is not always the smallest integer pair. e.g. ${\displaystyle 36^{2}-52*5^{2}=-4}$. The equation will give you ${\displaystyle x=1093436498,y=151632270}$, which when put into Pell's Equation yields ${\displaystyle 1195601955878350801-1195601955878350800=1}$, which works, but so does ${\displaystyle x=649,y=90}$ for ${\displaystyle N=52}$.

n = 61

The n = 61 case (determining an integer solution satisfying ${\displaystyle a^{2}-61b^{2}=1}$), issued as a challenge by Fermat many centuries later, was given by Bhaskara as an example.^[9]

We start with a solution ${\displaystyle a^{2}-61b^{2}=k}$ for any k found by any means. In this case we can let b be 1, thus, since ${\displaystyle 8^{2}-61\cdot 1^{2}=3}$, we have the triple ${\displaystyle (a,b,k)=(8,1,3)}$. Composing it with ${\displaystyle (m,1,m^{2}-61)}$ gives the triple ${\displaystyle (8m+61,8+m,3(m^{2}-61))}$, which is scaled down (or Bhaskara's lemma is directly used) to get:

${\displaystyle \left({\frac {8m+61}{3}},{\frac {8+m}{3}},{\frac {m^{2}-61}{3}}\right).}$

For 3 to divide ${\displaystyle 8+m}$ and ${\displaystyle |m^{2}-61|}$ to be minimal, we choose ${\displaystyle m=7}$, so that we have the triple ${\displaystyle (39,5,-4)}$. Now that k is −4, we can use Brahmagupta's idea: it can be scaled down to the rational solution ${\displaystyle (39/2,5/2,-1)\,}$, which composed with itself three times, with ${\displaystyle m={7,11,9}}$ respectively, when k becomes square and scaling can be applied, this gives ${\displaystyle (1523/2,195/2,1)\,}$. Finally, such procedure can be repeated until the solution is found (requiring 9 additional self-compositions and 4 additional square-scalings): ${\displaystyle (1766319049,\,226153980,\,1)}$. This is the minimal integer solution.

n = 67

Suppose we are to solve ${\displaystyle x^{2}-67y^{2}=1}$ for x and y.^[12]

We start with a solution ${\displaystyle a^{2}-67b^{2}=k}$ for any k found by any means; in this case we can let b be 1, thus producing ${\displaystyle 8^{2}-67\cdot 1^{2}=-3}$. At each step, we find an m > 0 such that k divides a + bm, and |m² − 67| is minimal. We then update a, b, and k to ${\displaystyle {\frac {am+Nb}{|k|}},{\frac {a+bm}{|k|}}}$ and ${\displaystyle {\frac {m^{2}-N}{k}}}$ respectively.

First iteration

We have ${\displaystyle (a,b,k)=(8,1,-3)}$. We want a positive integer m such that k divides a + bm, i.e. 3 divides 8 + m, and |m² − 67| is minimal. The first condition implies that m is of the form 3t + 1 (i.e. 1, 4, 7, 10,… etc.), and among such m, the minimal value is attained for m = 7. Replacing (a, b, k) with ${\displaystyle \left({\frac {am+Nb}{|k|}},{\frac {a+bm}{|k|}},{\frac {m^{2}-N}{k}}\right)}$, we get the new values ${\displaystyle a=(8\cdot 7+67\cdot 1)/3=41,b=(8+1\cdot 7)/3=5,k=(7^{2}-67)/(-3)=6}$. That is, we have the new solution:

${\displaystyle 41^{2}-67\cdot (5)^{2}=6.}$

At this point, one round of the cyclic algorithm is complete.

Second iteration

We now repeat the process. We have ${\displaystyle (a,b,k)=(41,5,6)}$. We want an m > 0 such that k divides a + bm, i.e. 6 divides 41 + 5m, and |m² − 67| is minimal. The first condition implies that m is of the form 6t + 5 (i.e. 5, 11, 17,… etc.), and among such m, |m² − 67| is minimal for m = 5. This leads to the new solution a = (41⋅5 + 67⋅5)/6, etc.:

${\displaystyle 90^{2}-67\cdot 11^{2}=-7.}$

Third iteration

For 7 to divide 90 + 11m, we must have m = 2 + 7t (i.e. 2, 9, 16,… etc.) and among such m, we pick m = 9.

${\displaystyle 221^{2}-67\cdot 27^{2}=-2.}$

Final solution

At this point, we could continue with the cyclic method (and it would end, after seven iterations), but since the right-hand side is among ±1, ±2, ±4, we can also use Brahmagupta's observation directly. Composing the triple (221, 27, −2) with itself, we get

${\displaystyle \left({\frac {221^{2}+67\cdot 27^{2}}{2}}\right)^{2}-67\cdot (221\cdot 27)^{2}=1,}$

that is, we have the integer solution:

${\displaystyle 48842^{2}-67\cdot 5967^{2}=1.}$

This equation approximates ${\displaystyle {\sqrt {67}}}$ as ${\displaystyle {\frac {48842}{5967}}}$ to within a margin of about ${\displaystyle 2\times 10^{-9}}$.