subsists between the coefficients p and q. necessary connection between p and q is the present question. To find the object of the Using the ordinary rule for square root, we have If therefore x2+px+q be a perfect square, the remainder, p2 g must be zero. 4' Hence the condition is determined by placing this remainder equal to zero and solving the resulting equation. 319. Symmetry. An expression is said to be symmetrical with respect to the letters it contains when its value is unaltered by the interchange of any pair of them; thus x+y+z, bc + ca+ab, 23 + y + z3 - xyz are symmetrical functions of the first, second, and third degrees respectively. It is worthy of notice that the only symmetrical expression of the first degree in x, y, z is of the form M(x+y+z), where M is independent of x, y, z. 320. It easily follows from the definition that the sum, difference, product, and quotient of any two symmetrical expressions must also be symmetrical expressions. The recognition of this principle is of great use in checking the accuracy of algebraic work, and in some cases enables us to dispense with much of the labor of calculation. In the following examples we shall assume as true a principle which will be demonstrated in Chap. XLII. Ex. 1. Find the expansion of (x + y + z)3. We know that the expansion must be a homogeneous expression of three dimensions, T and therefore of the form x + y3 + z3 + A(x2y + xy2 + y2z + yz2 + z2x + zx2)+Bxyz, where A and B are quantities independent of x, y, z. Put z = O, then A, the coefficient of x2y, is equal to 3, the coefficient of x2y in the expansion of (x + y)3. Put x=y=z = 1, and we get 27=3+ (3×6)+B; whence B = 6. Thus (x + y + z)8 =x8+ y3 + 23 + 3 x2y. + 3 xy2 + 3 y2z + 3 yz2 + 3 z2x + 3 zx2 + 6 xyz. Ex. 2. Find the factors of (b3 + c3) (b − c) + (c3 + a3) (c − a) + (a3 + b3) (a — b). Denote the expression by E; then E is an expression involving a, which vanishes when ab, and therefore contains ab as a factor [Art. 317]. Similarly it contains the factors bc and c E contains (b − c) (c − a) (a - b) as a factor. a; thus Also since E is of the fourth degree, the remaining factor must be of the first degree; and since it is a symmetrical expression involving a, b, c, it must be of the form m(a+b+c). [Art. 319.] ... E=m(b −c) (c − a) (a - b) (a + b + c). To obtain m we may give to a, b, c any values that we find most convenient; thus by putting a = 0, b = 1, c2, we find m = 1, and we have the required result. NOTE. For further information on the subject of Symmetry, the reader may consult Hall and Knight's Higher Algebra, Chap. xxxiv. EXAMPLES XXX. b. Without actual division find the remainder when 1. x5 - 5 x2 + 5 is divided by x 5. 2. 3x5 + 11 x2 + 90 x2 - 19 x + 53 is divided by x + 5. 3. x3- 7 x2a + 8 xa2 + 15 a3 is divided by x + 2 a. Without actual division show that 4. 32 x10 33x5 + 1 is divisible by x 1. 5. 3x+5x3- 13 x2 - 20x + 4 is divisibic by x2 - 4. 6. x44x3- 5 x2 - 36 x 36 is divisible by x2 Resolve into factors: X 6. 7. x3- 6x2 + 11 x − 6. Find the values of x which will make each of the following expressions a perfect square: 15. x+6x8 + 13 x2 + 13 x 1. 16. x+6x3 + 11 x2 + 3x + 31. Find the values of x which will make each of the following expres sions a perfect cube: 21. 8 x8-36 x2 + 56 x 39. 22. хо a2x4 27 3 +4a4x2-28 ao. 23. m3x6-9 m2nx4 + 39 mn2x2 - 51 n3. 24. If n be any positive integer, prove that 52n-1 is always divisible by 24. 26. a(b ~ c)2+b(c − a)2 + c(a − b)2 + 8 abc. CHAPTER XXXI. INDETERMINATE EQUATIONS OF THE FIRST DEGREE. 321. In Art. 167 we saw that if the number of unknown quantities is greater than the number of independent equations, there will be an unlimited number of solutions, and the equations will be indeterminate. By introducing conditions, however, we can limit the number of solutions. When positive integral values of the unknown quantities are required, the equations are called simple indeterminate equations. The introduction of this restriction enables us to express the solutions in a very simple form. Ex. 1. Solve 7x+12 y = 220 in positive integers. Since x and y are to be integers, we must have Now multiplying the numerator by such a number that the division of the coefficient of y may give a remainder of unity, in this case 3, we have 9-15y-integer; 7 Substituting this value in the original equation, we obtain 7x2484 m = 220; .. x 28+ 12 m. (2). Equation (1) shows that m may be 0 or have any negative integral value, but cannot have a positive integral value. Equation (2) shows in addition that m may be 0, but cannot have a negative integral value greater than 2. Thus the only positive integral values of x and y are obtained by placing m = 0, − 1, The complete solution may be exhibited as follows: m = 0, −1, −2, - 2. Ex. 2. Solve 5x-14 y 11 in positive integers and from (1), .. y = 5m - 4, X= 14 m - 9. This is called the general solution of the equation, and by giving to m any positive integral value, we obtain an unlimited number of values for x and y: thus we have |