# 4: An Example

In this blog post I will show you how you can use the tools developed in parts 1, 2 and 3 to solve a problem in number theory.

## Determine all the integer solutions of the equations x2 − 3y2 = m for m = −1 and 13.

Noting that x2 − 3y2 = (x – √3y)(x + √3), first we find the fundamental unit, e = a + b√3, of the number field Q(√3):

For b = 1, 3b2 + 1 = 4 = 2and so a = 2. Hence the fundamental unit is e = 2 +√3.

First we note that the norm of an element in Q(√d) is:

N(a + b√d) = a2 − db2

We will use the following three facts:

1. N(x + √3y) = x2 − 3y2 = m.
2. If we have a unit u, then N(a) = N(ua) for some a in Q(√d), as N(u) = 1 and N(ua) = N(u)N(a) (i.e. norm is multiplicative).
3. In the last post, we saw how the units in the ring of algebraic integers of L is of the form: Thus, multiplying x + √3y by a unit is equal to multiplying x + √3y by a power of the fundamental unit e.

So how do we solve this equation? First, by fact 1 we have to do find the element of Q(√3) that has norm +m or -m (there will be one such element for each, up to multiplication by a unit). Using fact 1 and 2, we deduce all the solutions to the equation are given by (x, y) such that:

• N(e) = – 1, N(x + √3y) = -m: en(x + √3y) for n a natural number.
• N(e) = – 1, N(x + √3y) = m: e2n(x + √3y) for n a natural number.
• N(e) = 1, N(x + √3y) = m: en(x + √3y) for n a natural number.

## m = -1

Noting that for any element a,|N(a)| = 1 if and only if a is a unit. Hence as the solutions are given by (x, y) where N(x + √3y) = -1, we must have that x + √3y is a unit. But, so all units have norm 1, and thus there are NO integer solutions to this equation.

## m = 13

We need to find (x, y) such that N(x + √3y) = x2 − 3y2 = 13 (because N(e) = 1).

• When y = 1, x2 = 14, so x is not an integer.
• When y = 2, x2 = 25, so x = 5.

Hence we get that all solutions are of the form:

en(5 + 2√3), for n a natural number and e = 2 +√3, fundamental unit

This is the last post of this series! Next I’ll be delving into Galois Theory. M x