1911 Encyclopædia Britannica/Equation/Indeterminate Equations

II. Indeterminate Equations.

1. When the number of unknown quantities exceeds the number of equations, the equations will admit of innumerable solutions, and are therefore said to be indeterminate. Thus if it be required to find two numbers such that their sum be 10, we have two unknown quantities x and y, and only one equation, viz. x + y = 10, which may evidently be satisfied by innumerable different values of x and y, if fractional solutions be admitted. It is, however, usual, in such questions as this, to restrict values of the numbers sought to positive integers, and therefore, in this case, we can have only these nine solutions,

x = 1, 2, 3, 4, 5, 6, 7, 8, 9;
y = 9, 8, 7, 6, 5, 4, 3, 2, 1;

which indeed may be reduced to five; for the first four become the same as the last four, by simply changing x into y, and the contrary. This branch of analysis was extensively studied by Diophantus, and is sometimes termed the Diophantine Analysis.

2. Indeterminate problems are of different orders, according to the dimensions of the equation which is obtained after all the unknown quantities but two have been eliminated by means of the given equations. Those of the first order lead always to equations of the form

ax ± by = ±c,

where a, b, c denote given whole numbers, and x, y two numbers to be found, so that both may be integers. That this condition may be fulfilled, it is necessary that the coefficients a, b have no common divisor which is not also a divisor of c; for if a = md and b = me, then ax + by = mdx + mey = c, and dx + ey = c/m; but d, e, x, y are supposed to be whole numbers, therefore c/m is a whole number; hence m must be a divisor of c.

Of the four forms expressed by the equation ax ± by = ±c, it is obvious that ax + by = −c can have no positive integral solutions. Also ax − by = −c is equivalent to by − ax = c, and so we have only to consider the forms ax ± by = c. Before proceeding to the general solution of these equations we will give a numerical example.

To solve 2x + 3y = 25 in positive integers. From the given equation 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

ax − by = c (aq − bp);

∴ 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

(x − cq) / b = (y − cq) / a = t.

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,

x = cq − bt, y = at − cp.

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,

x = mr + nc′, y = mr + n′c′,

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, (1/2 r)² − 1, (1/2 r)² + 1. If r be an odd number, they become (dividing by 2) r, 1/2 (r² − 1), 1/2 (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.