we have x = (25 − 3y) / 2 = 12 − y − (y − 1) / 2. Now, since x must be a
whole number, it follows that (y − 1)/2 must be a whole number.
Let us assume (y − 1) / 2 = z, then y = 1 + 2z; and x = 11 − 3z, where
z might be any whole number whatever, if there were no limitation
as to the signs of x and y. But since these quantities are required
to be positive, it is evident, from the value of y, that z must be
either 0 or positive, and from the value of x, that it must be less than
4; hence z may have these four values, 0, 1, 2, 3.
| If | z = 0, | z = 1, | z = 2, | z = 3; |
| Then | x = 11, | x = 8, | x = 5, | x = 2, |
| y = 1, | y = 3, | y = 5, | y = 7. |
3. We shall now give the solution of the equation ax − by = c in positive integers.
Convert a/b into a continued fraction, and let p/q be the convergent immediately preceding a/b, then aq − bp = ±1 (see Continued Fraction).
(α) If aq − bp = 1, the given equation may be written
∴ a (x − cq) = b (y − cp).
Since a and b are prime to one another, then x − cq must be divisible by b and y − cp by a; hence
That is, x = bt + cq and y = at + cp.
Positive integral solutions, unlimited in number, are obtained by giving t any positive integral value, and any negative integral value, so long as it is numerically less than the smaller of the quantities cq/b, cp/a; t may also be zero.
(β) If aq − bp = −1, we obtain x = bt − cq, y = at − cp, from which positive integral solutions, again unlimited in number, are obtained by giving t any positive integral value which exceeds the greater of the two quantities cq/b, cp/a.
If a or b is unity, a/b cannot be converted into a continued fraction with unit numerators, and the above method fails. In this case the solutions can be derived directly, for if b is unity, the equation may be written y = ax − c, and solutions are obtained by giving x positive integral values greater than c/a.
4. To solve ax + by = c in positive integers. Converting a/b into a continued fraction and proceeding as before, we obtain, in the case of aq − bp = 1,
Positive integral solutions are obtained by giving t positive integral values not less than cp/a and not greater than cq/b.
In this case the number of solutions is limited. If aq − bp = −1 we obtain the general solution x = bt − cq, y = cp − at, which is of the same form as in the preceding case. For the determination of the number of solutions the reader is referred to H. S. Hall and S. R. Knight’s Higher Algebra, G. Chrystal’s Algebra, and other text-books.
5. If an equation were proposed involving three unknown quantities, as ax + by + cz = d, by transposition we have ax + by = d − cz, and, putting d − cz = c′, ax + by = c′. From this last equation we may find values of x and y of this form,
or x = mr + n (d − cz), y = m′r + n′ (d − cz);
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.
Case 4. The expression a + bx + cx2 may be transformed into a square as often as it can be resolved into two parts, one of which is a complete square, and the other a product of two simple factors; for then it has this form, p2 + qr, where p, q and r are quantities which contain no power of x higher than the first. Let us assume √(p2 + qr) = p + mq; thus we have p2 + qr = p2 + 2mpq + m2q2 and r = 2mp + m2q, and as this equation involves only the first power of x, we may by proper reduction obtain from it rational values of x and y, as in the three foregoing cases.
The application of the preceding general methods of resolution to any particular case is very easy; we shall therefore conclude with a single example.
Ex. It is required to find two square numbers whose sum is a given square number.
Let a2 be the given square number, and x2, y2 the numbers required; then, by the question, x2 + y2 = a2, and y = √(a2 − x2). This equation is evidently of such a form as to be resolvable by the method employed in case 1. Accordingly, by comparing √(a2 − x2) with the general expression √(g2 + bx + cx2), we have g = a, b = 0, c = −1, and substituting these values in the formulae, and also −n for +m, we find
| x = | 2an | , y = | a (n2 − 1) | . |
| n2 + 1 | n2 + 1 |
If a = n2 + 1, there results x = 2n, y = n2 − 1, a = n2 + 1. Hence if r be an even number, the three sides of a rational right-angled triangle are r, (12 r)2 − 1, (12 r)2 + 1. If r be an odd number, they become (dividing by 2) r, 12 (r2 − 1), 12 (r2 + 1).
For example, if r = 4, 4, 4 − 1, 4 + 1, or 4, 3, 5, are the sides of a right-angled triangle; if r = 7, 7, 24, 25 are the sides of a right-angled triangle.
1. Cubic equations, like all equations above the first degree, are divided into two classes: they are said to be pure when they contain only one power of the unknown quantity; and adfected when they contain two or more powers of that quantity.
Pure cubic equations are therefore of the form x3 = r; and hence it appears that a value of the simple power of the unknown quantity may always be found without difficulty, by extracting the cube root of each side of the equation. Let us consider the equation x3 − c3 = 0 more fully. This is decomposable into the factors x − c = 0 and x2 + cx + c2 = 0. The roots of this quadratic equation are 12 (−1 ± √−3) c, and we see that the equation x3 = c3 has three roots, namely, one real root c, and two imaginary roots 12 (−1 ± √−3) c. By making c equal to unity, we observe that 12 (−1 ± √−3) are the imaginary cube roots of unity, which are generally denoted by ω and ω2, for it is easy to show that (12 (−1 − √−3))2 = 12 (−1 + √−3).
2. Let us now consider such cubic equations as have all their terms, and which are therefore of this form,
where A, B and C denote known quantities, either positive or negative.
This equation may be transformed into another in which the second term is wanting by the substitution x = y − A/3. This transformation is a particular case of a general theorem. Let xn + Axn−1 + Bxn−2 ... = 0. Substitute x = y + h; then (y + h)n + A (y + h)n−1 ... = 0. Expand each term by the binomial theorem, and let us fix our attention on the coefficient of yn−1. By this process we obtain 0 = yn + yn−1(A + nh) + terms involving lower powers of y.
Now h can have any value, and if we choose it so that A + nh = 0, then the second term of our derived equation vanishes.
Resuming, therefore, the equation y3 + qy + r = 0, let us suppose y = v + z; we then have y3 = v3 + z3 + 3vz (v + z) = v3 + z3 + 3vzy, and the original equation becomes v3 + z3 + (3vz + q) y + r = 0. Now v and z are any two quantities subject to the relation y = v + z, and if we suppose 3vz + q = 0, they are completely determined. This leads to v3 + z3 + r = 0 and 3vz + q = 0. Therefore v3 and z3 are the roots of the quadratic t2 + rt − q2/27 = 0. Therefore
| v3 = | −12 r + √(127 q3 + 14 r2); z3 = −12 r − √(127 q3 + 14r2); |
| v = | ∛{−12 r + √(127 q3 + 14 r2) }; z = ∛{ (−12 r − √(127 q3 + 14 r2) }; |
| and y = | v + z = ∛{−12 r + √(127q3 + 14 r2) } + ∛{−12 r − √(127 q3 + 14 r2) }. |
Thus we have obtained a value of the unknown quantity y, in terms of the known quantities q and r; therefore the equation is resolved.
3. But this is only one of three values which y may have. Let us, for the sake of brevity, put
| and put | α = 12 (−1 + √−3), |
| β = 12 (−1 − √−3). |
Then, from what has been shown (§ 1), it is evident that v and z have each these three values,
z = ∛B, z = α∛B, z = β∛B.
To determine the corresponding values of v and z, we must consider that vz = −13 q = ∛(AB). Now if we observe that αβ = 1, it will immediately appear that v + z has these three values,
v + z = α∛A + β∛B,
which are therefore the three values of y.
