make a solution of the equation. We can also see that some values of x, y and z
are not solutions of the equation. For example x = 3, y = 7 and z = 9 is not a
solution because these values do not make the left-hand side x + y equal the
right hand side z.
The equation x^2 + y^2 = z^2
We’ll now think about squares. The square of a number is that number
multiplied by itself, the number which we write as x^2. If x = 3 then x^2 = 3 × 3 =
- The equation we are thinking of now is not x + y = z, but
x^2 + y^2 = z^2
Can we solve this as before, by choosing values for x and y and computing z?
With the values x = 3 and y = 7, for example, the left-hand side of the equation
is 3^2 + 7^2 which is 9 + 49 = 58. For this z would have to be the square root of
58 (z = √58) which is approximately 7.6158. We are certainly entitled to claim
that x = 3, y = 7 and z = √58 is a solution of x^2 + y^2 = z^2 but unfortunately
Diophantine equations are primarily concerned with whole number solutions. As
√58 is not a whole number, the solution x = 3, y = 7 and z = √58 will not do.
The equation x^2 + y^2 = z^2 is connected with triangles. If x, y and z represent
the lengths of the three sides of a right-angled triangle they satisfy this equation.
Conversely, if x, y and z satisfy the equation then the angle between x and y is a
right angle. Because of the connections with Pythagoras’s theorem solutions for
x, y and z are called Pythagorean triples.
How can we find Pythagorean triples? This is where the local builder comes to
the rescue. Part of the builder’s equipment is the ubiquitous 3–4–5 triangle. The