308 = 9. positive integral solution of the equations x2 - Ny2 =±a, when a is not one of the above mentioned denominators; thus we easily find that the equation x2 - 7y3 = 53 is satisfied by y = 2, x= When one solution in integers has been found, any number of solutions may be obtained as explained in the next article. *376. Suppose that x=f, y=g is a solution of the equation solution of the equation x2 - Ny2 = a; and let x= = h, y = k be any -Ny1; then By putting x3 – Ny2 = (ƒ2 – Ng3) (h3 — Nk2) =(fhNgk) - N (fkgh). x=fhNgk, y=fk=gh, and ascribing to h, k their values found as explained in Art. 371, we may obtain any number of solutions. *377. Hitherto it has been supposed that N is not a perfect square; if, however, N is a perfect square the equation takes the form x2 - n'y2 = a, which may be readily solved as follows. Suppose that a = bc, where b and c are two positive integers, of which b is the greater; then (x + ny) (x − ny) = bc. if the values of x and y found Put x+ny=b, x— ny = c; from these equations are integers we have obtained one solution of the equation; the remaining solutions may be obtained by ascribing to b and c all their possible values. Example. Find two positive integers the difference of whose squares is equal to 60. Let x, be the two integers; then x2 y2=60; that is, (x+y) (x − y) = 60. Now 60 is the product of any of the pair of factors 1, 60; 2, 30; 3, 20; 4, 15; 5, 12; 6, 10; and the values required are obtained from the equations x+y=30, x+y=10, the other equations giving fractional values of x and y. Thus the numbers are 16, 14; or 8, 2. COR. In like manner we may obtain the solution in positive integers of ax2 + 2hxy + by3 + 2gx + 2fy + c = k, if the left-hand member can be resolved into two rational linear factors. *378. If in the general equation a, or b, or both, are zero, instead of employing the method explained in Art. 367 it is simpler to proceed as in the following example. must be equal to ±1, or ±2, or ±3, or ±6. The cases ±2, +6 may clearly be rejected; hence the admissible values of x are obtained from 2x-5=±1, 2x-5=±3; whence the values of x are 3, 2, 4, 1. Taking these values in succession we obtain the solutions x=3, y=11; x=2, y = −3; x=4, y=9; x=1, y= −1; and therefore the admissible solutions are x=3, y=11; x=4, y=9. *379. The principles already explained enable us to discover for what values of the variables given linear or quadratic functions of x and y become perfect squares. Problems of this kind are sometimes called Diophantine Problems because they were first investigated by the Greek mathematician Diophantus about the middle of the fourth century. Example 1. Find the general expressions for two positive integers which are such that if their product is taken from the sum of their squares the difference is a perfect square. This equation is satisfied by the suppositions mx=n(z+y), n(x − y) = m (z − y), where m and n are positive integers. Hence mx - ny - nz=0, nx + (m - n) y − z=0. From these equations we obtain by cross multiplication and since the given equation is homogeneous we may take for the general solution x=2mn-n2, y=m2-n2, z=m2 — mn+n2. Here m and n are any two positive integers, m being the greater; thus if m= =7, n=4, we have x= =40, y=33, z= =37. Example 2. Find the general expression for three positive integers in arithmetic progression, and such that the sum of every two is a perfect From the value of x it is clear that m and n are either both even or both odd; also their values must be such that x is greater than y, that is, or (m2 + n2)2 > 8mn (m2 — n2), m3 (m - 8n) +2m2n2 + 8mn3 +n*>0; which condition is satisfied if m>8n. If m=9, n=1, then x=3362, y = 2880, and the numbers are 482, 3362, 6242. The sums of these taken in pairs are 3844, 6724, 9604, which are the squares of 62, 82, 98 respectively. *EXAMPLES. XXVIII. Solve in positive integers: 1. 5x2 - 10xy + 7y2=77. 2. 7x2-2xy+3y2=27. 3. y2-4xy+5x2 – 10x=4. 4. xy-2xy=8. Find the general values of x and y which make each of the following expressions a perfect square: 15. x2 - 3xy+3y2. 16. x2+2xy +2y2. 17. 5x2+y2. 18. Find two positive integers such that the square of one exceeds the square of the other by 105. 19. Find a general formula for three integers which may be taken to represent the lengths of the sides of a right-angled triangle. 20. Find a general formula to express two positive integers which are such that the result obtained by adding their product to the sum of their squares is a perfect square. 21. "There came three Dutchmen of my acquaintance to see me, being lately married; they brought their wives with them. The men's names were Hendriek, Claas, and Cornelius; the women's Geertruij, Catriin, and Anna: but I forgot the name of each man's wife. They told me they had been at market to buy hogs; each person bought as many hogs as they gave shillings for one hog; Hendriek bought 23 hogs more than Catriin; and Claas bought 11 more than Geertruij; likewise, each man laid out 3 guineas more than his wife. I desire to know the name of each man's wife." (Miscellany of Mathematical Problems, 1743.) 22. Shew that the sum of the first n natural numbers is a perfect square, if n is equal to k2 or k'2 – 1, where k is the numerator of an odd, and the numerator of an even convergent to √2. CHAPTER XXIX. SUMMATION OF SERIES. 380. Examples of summation of certain series have occurred in previous chapters; it will be convenient here to give a synopsis of the methods of summation which have already been explained. (i) Arithmetical Progression, Chap. IV. (ii) Geometrical Progression, Chap. V. (iii) Series which are partly arithmetical and partly geometrical, Art. 60. (iv) Sums of the powers of the Natural Numbers and allied Series, Arts. 68 to 75. (v) Summation by means of Undetermined Coefficients, Art. 312. (vi) Recurring Series, Chap. XXIV. We now proceed to discuss methods of greater generality; but in the course of the present chapter it will be seen that some of the foregoing methods may still be usefully employed. 381. If the 7th term of a series can be expressed as the difference of two quantities one of which is the same function of r that the other is of r1, the sum of the series may be readily found. and its sum by S, and suppose that any term u can be put in the form vv,,; then S1 = (v,-v)+(v2-v,) + (v2 − v2) + ... + (v-1 — V-2)+(V_V-1) น |