where z and r may be taken at pleasure, except in so far as the values of x, y, z may be required to be all positive; for from such restriction the values of z and r may be confined within certain limits to be determined from the given equation. For more advanced treatment of linear indeterminate equations see [Combinatorial Analysis].
6. We proceed to indeterminate problems of the second degree: limiting ourselves to the consideration of the formula y² = a + bx + cx², where x is to be found, so that y may be a rational quantity. The possibility of rendering the proposed formula a square depends altogether upon the coefficients a, b, c; and there are four cases of the problem, the solution of each of which is connected with some peculiarity in its nature.
Case 1. Let a be a square number; then, putting g² for a, we have y² = g² + bx + cx². Suppose √(g² + bx + cx²) = g + mx; then g² + bx + cx² = g² + 2gmx + m²x², or bx + cx² = 2gmx + m²x², that is, b + cx = 2gm + m²x; hence
| x = | 2gm − b | , y = √(g² + bx + cx²)= | cg − bm + gm² | . |
| c − m² | c − m² |
Case 2. Let c be a square number = g²; then, putting √(a + bx + g²x²) = m + gx, we find a + bx + g²x² = m² + 2mgx + g²x², or a + bx = m² + 2mgx; hence we find
| x = | m² − a | , y = √(a + bx + g²x²) = | bm − gm² − ag | . |
| b − 2mg | b − 2mg |
Case 3. When neither a nor c is a square number, yet if the expression a + bx + cx² can be resolved into two simple factors, as f + gx and h + kx, the irrationality may be taken away as follows:—
Assume √(a + bx + cx²) = √{ (f + gx) (h + kx) } = m (f + gx), then (f + gx) (h + kx) = m² (f + gx)², or h + kx = m² (f + gx); hence we find
| x = | fm² − h | , y = √{ (f + gx) (h + kx) } = | (fk − gh) m | ; |
| k − gm² | k − gm² |
and in all these formulae m may be taken at pleasure.