Brahmagupta’s equations

Section 8.1 Brahmagupta’s equations

Let us consider the following equation

\begin{equation} x^2-dy^2=1\quad\text{[Brahmagupta's equation]}\tag{8.1.1} \end{equation}

where \(d\) is a square-free integer. Problem is to find integers \(x,y\) satisfying the above equation. Indian mathematician Brahmagupta discovered, among other things, the following identity: for given integers \(x,y,z\text{,}\) and \(w\) there exists integers \(M,N\) such that

\begin{equation} (x^2-dy^2)(z^2-dw^2)=M^2-dN^2\quad \text{[Bramhagupta's identity]}\tag{8.1.2} \end{equation}

Using this identity Bramhagupta was able to produce infinitely many solutions of \(x^2-dy^2=1\text{,}\) and also able to solve \(x^2-dy^2=c\) for certain values of \(d\) and \(c\text{.}\)

In this section we give a connection between the ring theory and equations of type (8.1.1). We begin with the following.

Lemma 8.1.1.

Let \(d\) be a square-free integer. The set \(\Z[\sqrt{d}]=\{a+b\sqrt{d}:a,b\in\Z\}\) is an integral domain, and the map

\begin{equation*} N\colon\Z[\sqrt{d}]\to\Z\quad\text{given by}\quad N(a+b\sqrt{-2})=a^2-db^2. \end{equation*}

is multiplicative, i.e.,

\begin{equation*} N\left((a+b\sqrt{d})(c+d\sqrt{d})\right)=N(a+b\sqrt{d})\,N(c+d\sqrt{d})\text{.} \end{equation*}

Remark 8.1.2.

The map \(N\) defined above is a square of the absolute value in \(\C\text{.}\)

Lemma 8.1.3.

We keep notations of Lemma 8.1.1.

\(\alpha\in\Z[\sqrt{d}]\) is a unit in \(\Z[\sqrt{d}]\) if and only if \(N(\alpha)=\pm 1\text{.}\)
If \(\pm N(\alpha)\) is a prime number in \(\Z\) then \(\alpha\) is irreducible.

Checkpoint 8.1.4.

Show that \(\alpha\in\Z[\sqrt{3}]\) is a unit if and only if \(N(\alpha)=1\text{.}\)

Remark 8.1.5. (Connection with Brahmagupta’s equation).

The integer solutions of the equation \(x^2-dy^2=1\) or \(x^2-dy^2=-1\) are precisely units in the ring \(\Z[\sqrt{d}]\text{.}\)

Lemma 8.1.6. (Units in \(\Z[\sqrt{-2}]\)).

By the above lemma an element \(a+b\sqrt{-2}\in\Z[\sqrt{-2}]\) is a unit if and only if \(a^2+2b^2=1\text{.}\) Thus we must have \(a=\pm 1\) and \(b=0\text{.}\)

Suppose that \(\alpha,\beta\in\Z[\sqrt{-2}]\) with \(\beta\neq 0\text{.}\) Then we can consider \(\alpha\beta^{-1}\in\C\text{.}\) Note that we can always choose \(q\in\Z[\sqrt{-2}]\) with

\begin{equation*} N\left(\alpha\beta^{-1}-q\right)\leq N\left(\frac{1+\sqrt{-2}}{2}\right)\lt 1. \end{equation*}

If we plot integers on \(X=\R\)-axis and integer multiples of \(\sqrt{-2}\) on the \(Y=i\R\)-axis then the choice of \(q\) will be clear. Note that it is similar to the result proved for Gaussian integers (ref Section 5.3).

Proposition 8.1.7. (\(\Z[\sqrt{-2}]\) is ED).

For nonzero \(\alpha,\beta\in\Z[\sqrt{-2}]\) there exists \(q,r\in\Z[\sqrt{-2}]\) such that

\begin{equation*} \alpha=q\beta+r\quad\text{with}\;N(r)\lt N(\beta). \end{equation*}

Remark 8.1.8.

Consider \(\Z[\sqrt{-5}]=\{a+b\sqrt{-5}:a,b\in\Z\}\) with the map

\begin{equation*} N\colon\Z[\sqrt{-5}]\to\Z_+\quad\text{given by}\quad a+b\sqrt{-5}\mapsto a^2+5b^2. \end{equation*}

Suppose that \(\alpha,\beta\in\Z[\sqrt{-5}]\) are nonzero elements such that

\begin{equation*} \alpha\beta^{-1}=\frac{1}{2}+\frac{\sqrt{-5}}{2}\in\C. \end{equation*}

Suppose that we want to find \(q=a+b\sqrt{-5}\in\Z[\sqrt{-5}]\) such that \(\alpha\beta^{-1}-q\in\Z[\sqrt{-5}]\text{.}\) Thus

\begin{equation*} N\left(\alpha\beta^{-1}-q\right)=\frac{1}{4}(1-2a)^2+\frac{5}{4}(1-2b)^2 \end{equation*}

In order to have \(N\left(\alpha\beta^{-1}-q\right)\lt 1\) we must have

\begin{equation*} (1-2a)^2+5(1-2b)^2\lt 4\quad\text{for integers}\, a, b. \end{equation*}

Since for any \(n\in\Z\) we have \(1-2n\neq 0\) and hence \((1-2a)^2+5(1-2b)^2\) is always bigger than \(5\text{.}\) Thus with the map \(N\) there does not exists \(q\in\Z[\sqrt{-5}]\) with \(N\left(\alpha\beta^{-1}-q\right)\lt 1\text{.}\)

Algebra-II: Companion notes

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