554 ALGEBRA [INDETERMINATE PROBLEMS. SECT. XV. INDETERMINATE PROBLEMS. 117. When the conditions of a question are such that the number of unknown quantities exceeds the number of equations, that question will admit of innumerable solu tions, and is therefore said to be indeterminate. Thus, if it be required to find two numbers subject to no other limi tation than that their sum be 10, we have two unknown quantities x and y, and only one equation, viz. x + y = 1 0, 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 solu tions, x = l, 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 be come the same as the last four, by simply changing x into y, and the contrary. 118. Indeterminate problems are of different orders, according to the dimensions of the equation which is obtained after all the iinknown quantities but two have been eliminated by means of the given equations. Those of the first order lead always to equations of this 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= ; but d, e, x, y, are supposed to be whole numbers, therefore is a whole number ; hence m m must be a divisor of c. We proceed to illustrate the manner of resolving inde terminate equations of the first order, by some numerical examples. Ex. 1. Given 2x + 3y = 25, to determine x and y in whole positive numbers. 25-3)/ From the given equation we have x = - = 1 2 - y -
^r Now, since x must be a whole number, it follows a ft I 1 n I "1 that ^- must be a whole number. Let us assume - = 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 or positive, and from the value of x, that it must be less than 4; hence z may have these three values, 0, 1, 2, 3. If 2 = 0, z=, z=2, z=3; nri ( X=, X=8, X=5, X=2, Jill (. II s -f n f >-f y i> y = 3, y = 5, y = l. Ex. 2. It is required to find all the possible ways in which 60 can be paid in guineas and moidores only. Let x be the number of guineas, and y the number of moidores. Then the value of the guineas, expressed in shillings, is 2lx, and that of the moidores 27y; therefore, from the nature of the question, 2lx + 27 y = 1200, or, dividing the equation by 3, 7^ + 9^ = 400; hence, proceed ing as before, we obtain
- =61 -9v.
From the value of x, it appears that v cannot exceed 6, and from the value of y, that it cannot be less than 1. Hence if v= 1, 2, 3, 4, 5, 6, we have . = 52, 43, 34, 25, 16, 7. y= 4, 11, 18, 25, 32, 39. 119. In the foregoing examples the unknown quantities x and y have each a determinate number of positive values; and this will evidently be the case as often as the proposed equation is of this form, ax + by c. If, however, b be negative, that is, if the equation be of this form, ax - by c, or ax = by + c, we shall have questions of a different kind, admitting each of an infinite number of solutions; these, however, may be resolved in the same manner as the preceding. 120. 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 = m r + 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 con fined within certain limits to be determined from the given equation. 121. We proceed to indeterminate problems of the second degree : limiting ourselves to the consideration of the for mula y 2 = a + bx + ex 2 , 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 pecu liarity in its nature. Case 1. Let a be a square number; then, putting y 2 for a, we have y 2 = g 2 + bx + ex-. Suppose /g 2 + bx + ex 2 = g + mx; then g 2 + bx + ex 1 = g- + 2gmx +. m 2 x 2 , or bx + ex 2 = 2ymx + m 2 x 2 , that is, b + ex 2gm + m"x; hence 2r/m b / ; - ; - ~ cq bm + qtn^ x=~ 5-,^= V+ bx + cx 2 =^ ~ c-m- c-m 2 Case 2. Let c be a square number = g 2 ; then, putting /s/a + bx + g 2 x 2 = m + gx, we find a + bx + f/ 2 x 2 m 2 + 2 mgx + g 2 x 2 , or a + bx = m 2 + 2mgx; hence we find m 2 a i - ; - ir-n bm am* 2 art
, , b - Zmg b- 2mg Case 3. When neither a nor c is a square number, yet if the expression a + bx + ex 1 can be resolved into two simple factors, usf+gx and h + kx, the irrationality may be taken away as follows : Assume Ja + bx + ex 2 = */(/ + gx) (h + kx) = r>i(f+ gx), then (f+gx)(h + kx) = m 2 (f+gx) 2 , or h + kx = m 2 (f+gx) ; hence we find (fk - gh)m and in all these formulae m may be taken at pleasure. Case 4. The expression a + bx + ex 2 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, p 2 + qr, where p, q, and r are quantities which contain no power of x higher than the first. Let us assume /p 2 + qr=p + mq; thus we have p- + qr = p 2 + 2mpq + m 2 q 2 and r 2mp + m 2 q, 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 reso lution to any particular case is very easy; we shall there
fore conclude with a single example.
