4. Find the sum of the infinite series x+4x+9x+16x+ 3. Required the sum of the series x +3x2+6x2+10x2+&c. ad infinitum. 4. Required the sum of n terms of the series 186. Diophantine analysis is that part of algebra which relates to the finding particular rational values for general expressions under a surd form; the principal methods of effecting which are comprehended in the following problems: Problem I. To find such values of x as will render rational the expression (ax+bx+c). Before we can give any direct investigation of this problem, it will be necessary to consider the nature of the known quantities a, b, c, because there are several cases in which the thing here proposed to be done becomes impossible, and that solely on account of these known quantities. = CASE I. When a 0, or when the expression is of the form ✔(bx+c). Put ✔(bx + c) = p, or bx + c = p2, then p2 C x= ; consequently, whatever value be given to p, there b must necessarily result a corresponding value of x that will render the proposed equation rational, and equal to p. 1. Find a number such, that if it be multiplied by 5, and the product increased by 2, the result shall be a square. p2-2 Put 5x+2=p2, then x= -; if p=2, then x = ; and by assuming other values for p, different values of x may be had. 2. Find two numbers, whose difference shall be equal to a giv en number a, and the difference of whose squares shall be also a square. Let x be one number, then a +x is the other, and we have to make (a+x)2x2, or a2+2ax, a square. Put a2+2ax-p3, then x= CASE II. 2 p2-a2 2a where the value of p may be any number. When a = 0, or when the expression is of the form (ax+bx). Put ✅(ax2+bx)=px, or ax2+bx=p2x2, then ax + x=; b b=px; whence = p2 and whatever value may be given to a p in this expression, there will result a value of x that will make the proposed expression rational. 1. Find a number such, that if its half be added to double its square, the result shall be a square. = a square, Let x be the number, then we must have 2x2+x which denote by px, then 2x+-p'x, or 2x-pa={ ; ··· x==== 2' p being any number whatever: If p be taken = 2, then 2. Find two numbers, whose sum shall be equal to a given number a, and whose product shall be a square. Let x be one number, then a- x is the other, and we have to make ax-x2 a square: Put ax-x-p2x2, then a-x=p2x, whence x= a 71 ? being any number whatever. , p Otherwise. Assume ar+bry. Now we may take y=r(ax+b)}, where r and s must be so assumed that their pro y=sx duct may be 1; which is the case generally when Equating the values of x, we get rax+rb=sx; bm2 S -ar n2 — am2 for s, is easily made; we have only to put m2 in the place of r, and n2 in the place of s, and when the coefficients are not affected with or s, we must put the product mn after the coefficient; this will be easily understood from the nature of algebraic frac tions. CASE III. When c is a square, or when the expressions are of the form (ax2 + bx+c2). Put (ax2 + bx+c2) = px+c, then ax2+bx+c2=p2x2+2cpx+c2, or ax2+bx-p2x2+2cpx, ... ax+b= 2cp-b px+2cp, whence x = a-p2 1. Find two numbers, whose sum shall be 16, and such, that the sum of their squares shall be a square. Let x be one number, then 16 x is the other, and we have to make (16-x), or 2x2-32x+256, a square,= (px —— 16)2 —p2x2-32px+256, and we then have 2x2-32x-p2x2 — 32px, or 2x-32-p2x-32p, whence x = 32(p-1) p2-2 If we take p=3, we shall x 94,.. the two numbers are 94, and 69. Otherwise. Put ax2+bx+c2=y. Then ax+bx-y-c2. y } 2c-br 2cmn bm2 whence rax+rb-c=sx+, and x = ar-s am2-n2 CASE IV. When a is a square, or when the expression is of the form ✔ (a2x2+bx+c). Put (a2x2+bx+c)=ax+p, or a2x2 + bx + ca2x2+2pax+p2, 1. Find a number such, that if it be increased by 2 and 5 separately, the product of the sums shall be a square. Let x be the number, then we have to make (x+2)(x +5), or +7x+10, a square, which denote by (xp), then x2+7x+10 x2-2px+p2, or 7x+10=-2px+p2, .. x= p=4, we shall have x=. Otherwise. For a2x2+bx+c=□*, put, and it becomes, by rejecting the denominator z2, a + bz+cz2=□; which is of the same form as above only a is in the place of c: hence, changing 2amn-bm2 cm2-n2 a for c, we have z= 2amn-bm2 1 .. X cm2 n2 CASE V. When neither a, nor c, are squares, but when b2— 4ac In this case it will first be necessary to show that the expression ax2+bx+c will always be resolvable into two pos is a square. or find the two values of x in it, as x-k, and x-k', then x - k, b C and x—k', will obviously be the two factors of x++; and.. a(x-k)(x-k') will be equal to the proposed expression: Now the values of x in the above equation are and a b (b2-4ac) b (b2-4ac) 2a + or putting 2a b2-4ac-d, the values of x are and 2a 2a b+d, and.. we see therefore that the proposed expression under those conditions is always resolvable into two factors. Let there be then ✔✅(ax2+bx+c)=√{(fx+g)(hx+k)}, which put equal to p(fx +g), then (fr+g)(hx+k=p2 (ƒx+g)2; or (hx+k)=p2(fx+g); whence x= p❜g-k "h-p2f" 4ac is a In addition to the above, we will here show the method of finding the factors of the formula ax2 + bx + c, when 62 square. Assume ax2+bx+c=(mx+p)(nx+q)=mnx2+(mq+np)x+pq. Hence we have by equating the coefficients, amn, b=mq+ np, .. c=pq, we have also, ac=mnpq, and by subtracting 4 times this, from the square of the equation b = mq+np, and extracting the root, we have mq-np (b2-4ac); hence b2-4ac must be a square put it =f2, then mq—np=f. Hence by addition, subtraction, &c. This symbol is used to denote the words a square Here Thus to find the factors of the formula, 6x2+13x+6. m=3, n=3, f=(169-144)=5, .. p-2, q-3. Hence the factors are 3x + 2, and 2x+3. Taking the above general expressions for the factors of the formula ax2 + bx+c, we (b+ƒ)nr2—(b −ƒ)ms2° 2mn(ms-nr2) shall find the value of x to be ; where r and s may be any numbers and m and n are factors of a, the coefficients of x. 1. Find such a value of x as will render the expression 6x+ 13x+6 a square. Here a 6, b=13, and c-6, and as this expression evidently does not belong to any of the preceding cases, it will be proper to try whether b2. 4ac is a square, which it is found to be, viz. 25; we are certain, therefore, that the expression may be represented by two factors, which are readily found to be 2x+3, and 3x+2. Put therefore 6x+13x+6, or (2x+3)(3x+2)={p(2x+3)}2, and it follows that 3x+2=p2(2x+3), whence x= 3p2 2 32p If we take p=1, then x=1, and the expression is equal to 25. Otherwise. Assume ax+b+c=(fx+g)(hx+k)=y'. Now take (f)};.. rfx+rg=she+sk; ys(hxk) S and therefore x= grks gm2 kn2 "hs fr hn" — fm2 CASE VI. When the proposed expression can be divided into two parts, one of which is a square, and the other the product of two factors. This is the last case in which any general method of proceed ing can be pointed out, and may often be serviceable when the expression does not come under either of the preceding cases: It is, however, sometimes troublesome to find whether the proposed expression can be decomposed as this case requires, or not; but if it be ascertained that it can, the expression (ax2+bx+c) may be put under the form (dx+e)2+(fx + g)(hx + k)}, and if we equate this with (dx+e)+p(fx+g), there will result (dx+e)2+(ƒx+g)(hx+k) hx+k=2p(dx+e)+p2(fx+g); x = p(2e+pg)-k 1. Find a value of x such, that 2x2+8x+7 shall be a square. This expression, after a few trials, is found to be equivalent to |