in Mathematics, certain questions relating to square and cube numbers, and right-angled triangles, &c. the nature of which was determined by Diophantus, a mathematician of Alexandria, who is believed to have lived about the third century. We have his works, which were published with notes at Paris, in 1621, by Bachet de Meziriac; and another edition in 1679, with observations on every question by M. Fermat.

In these questions it is endeavoured to find commensurable numbers to answer indeterminate problems; which bring out an infinite number of incommensurable quantities. For example, it is proposed to find a right-angled triangle, whose sides, \( x, y, z \), are expressed by commensurable numbers; it is known that \( x^2 + y^2 = z^2 \), \( z \) being the supposed hypotenuse. But it is possible to assume \( x \) and \( y \) so, that \( z \) will be incommensurable; for if \( x = 1 \), and \( y = 2 \), \( z = \sqrt{5} \).

The art of resolving such problems consists in so managing the unknown quantity or quantities in such a manner, that the square or higher power may vanish out of the equation, and then by means of the unknown quantity in its first dimension, the equation may be resolved without having recourse to incommensurables; e.g. let it be supposed to find \( x, y, z \), the sides of a right-angled triangle, such as will give \( x^2 + y^2 = z^2 \). Suppose \( z = x + u \), then \( x^2 + y^2 = x^2 + 2xu + u^2 \); out of which equation \( x^2 \) vanishes, and \( x = \frac{y^2 - u^2}{2u} \): then assuming \( y \) and \( u \) equal to any numbers at pleasure, the sides of the triangle will be \( y, \frac{y^2 \times u^3}{2u}, \) and the hypothenuse \( x + u = \frac{y^2 \times u^3}{2u} \); if \( y = 3 \), and \( u = 1 \), then \( \frac{y^2 - u^2}{2u} = 4 \), and \( x + u = 5 \). It is evident that this problem admits of an infinite number of solutions.

For the resolution of such kind of problems, see Saunderton's Algebra, vol. ii. book 6.