Exercise 3.4.4 (Pell-Fermat equation $x^2 - 2 y^2 = 1$)

Question

Use the binomial theorem to give a formula for positive integers $x_k$ and $y_k$ such that $(3 + 2\sqrt{2})^k = x_k + y_k \sqrt{2}$. Show that $(3-2\sqrt{2})^k = x_k - y_k\sqrt{2}$. Deduce that $x_k^2 - 2 y_k^2 = 1$ for k = 1,2,3,\ldots. Show that $(x_k , y_k) = 1$ for each $k$. Show that $x_{k+1} = 3x_k + 5 y_k$ and $y_{k+1} = 2x_k + 3 y_k$ for $k = 1, 2, 3,\ldots$. Show that $\{x_k\}$ and $\{y_k\}$ are strictly increasing sequences. Conclude that the number $1$ has infinitely many proper representations by the quadratic form $x^2 - 2 y^2$.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} Note first that, for all integers $a,b,c,d$,
\begin{align}
 a + b \sqrt{2} = c + d \sqrt{2} \Rightarrow a =c, b = d.
 \end{align}
 Indeed, if $b 
e d$ then $\sqrt{2} = \frac{a-c}{d-b}$ would be a rational number, which is false, so $b = d$, and then $a = c + (d - b)\sqrt{2} = c$.
 
 \bigskip
 By the binomial formula
\begin{align*}
x_k + y_k\sqrt{2} &= (3 + 2\sqrt{2})^k \
&= \sum_{i = 0}^k \binom{k}{i} (2\sqrt{2})^i 3^{k-i}\
&= \sum_{j = 0}^{\lfloor k/2floor} \binom{k}{2j} (2\sqrt{2})^{2j} 3^{k-2j}+ \sum_{j = 0}^{\lfloor (k-1)/2floor}  \binom{k}{2j+1}(2\sqrt{2})^{2j+1} 3^{k-2j-1}\
&=  \sum_{j = 0}^{\lfloor k/2floor}  \binom{k}{2j} 2^{3j} 3^{k-2j} + \sqrt{2} \sum_{j = 0}^{\lfloor (k-1)/2floor}  \binom{k}{2j+1}2^{3j+1} 3^{k-2j-1}.
\end{align*}
By (1), since that these two sums are integers, we obtain
\begin{align*}
x_k &= \sum_{j = 0}^{\lfloor k/2floor} \binom{k}{2j} 2^{3j} 3^{k-2j},\
y_k &=\sum_{j = 0}^{\lfloor (k-1)/2floor}\binom{k}{2j+1} 2^{3j+1} 3^{k-2j-1}. 
\end{align*}
By the same calculation, we obtain
\begin{align*}
(3- 2\sqrt{2})^k &= \sum_{i = 0}^k \binom{k}{i} (-2\sqrt{2})^i 3^{k-i}\
&= \sum_{j = 0}^{\lfloor k/2floor}   \binom{k}{2j}(-2\sqrt{2})^{2j} 3^{k-2j}+ \sum_{j = 0}^{\lfloor (k-1)/2floor}  \binom{k}{2j+1}(-2\sqrt{2})^{2j+1} 3^{k-2j-1}\
&=\sum_{j = 0}^{\lfloor k/2floor} \binom{k}{2j} 2^{3j} 3^{k-2j} - \sqrt{2} \sum_{j = 0}^{\lfloor (k-1)/2floor}  \binom{k}{2j+1} 2^{3j+1} 3^{k-2j-1}\
&=x_k - \sqrt{2} y_k.
\end{align*}
(Alternatively, with some field theory, we can use the fact that $a + b \sqrt{2} \mapsto a - b \sqrt{2}$ is a $\Q$-automorphism of the field $\Q(\sqrt{2})$.)

\bigskip
For all integers $k \in \N = \{0,1,2,\ldots\}$,
\begin{align*}
x_k + y_k\sqrt{2} &= (3 + 2\sqrt{2})^k,\
x_k - y_k\sqrt{2} &= (3 - 2\sqrt{2})^k.
\end{align*}
By multiplying these two equalities,
\begin{align*}
x_k^2 - 2 y_k^2 &= [(3 + 2\sqrt{2})(3 - 2\sqrt{2})]^k\
&= 1.
\end{align*}
For any integer $d$, if $d \mid x_k$ and $d \mid y_k$, then $d \mid x_k^2 - 2 y_k^2 = 1$.This shows that $x_k \wedge y_k = 1$.

Now, for any $k\in \N$,
\begin{align*}
x_{k+1} + y_{k+1}\sqrt{2} &= (3+2\sqrt{2})^{k+1}\
&=(3+2\sqrt{2})^{k}(3+2\sqrt{2})\
&=(x_k + y_k\sqrt{2})(3+2\sqrt{2})\
&= (3x_k + 4 y_k) + (2x_k + 3y_k)\sqrt{2}.
\end{align*}
Using (1) anew, we obtain for every $k \in \N$,
\begin{align*}
x_{k+1} &= 3x_k + 4 y_k,\
y_{k+1} &= 2x_k + 3y_k.
\end{align*}
First,  $x_1 = 3>0, y_1 = 2>0$. If we suppose that $x_k>0, y_k>0$, then $x_{k+1} = 3x_k + 4 y_k >0$, and $y_{k+1} = 2x_k + 3y_k>0$. This proves by induction that $x_n>0, y_n>0$ for every index $n>0$. Therefore
\begin{align*}
x_{k+1} - x_k &= 2x_k + 4y_k>0,\
y_{k+1} - y_k&= 2x_k + 2y_k >0.
\end{align*}
This shows that the two sequences $(x_k)_{k\in\N^*}$ and $(y_k)_{k \in\N^*}$ are strictly increasing sequences.

Therefore the solutions $(x_k, y_k)$ of the equation $x^2 - 2y^2 = 1$ are distinct for all $k\in \N^*$, and $x_k \wedge y_k = 1$. So the number $1$ has infinitely many proper representations by the quadratic form $x^2 - 2 y^2$.
\end{proof}

Exercise 3.4.4 (Pell-Fermat equation $x^2 - 2 y^2 = 1$)

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Exercise 3.4.4 (Pell-Fermat equation $x^2 - 2 y^2 = 1$)

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