substituting in the original equation 1+x for x and (q-kx) for y, where k is some integer.

3. Form Ax2 + Bx + C = y2.

This can be reduced to the form in which the second term is

Diophantus, however, treats this case separately and less fully. According to him, a rational solution of the equation Ax2 + Bx + C = y2 is only possible

(a) when A is positive and a square, say a2;
(B) when C is positive and a square, say c2;
(y) when B2- AC is positive and a square.

In case (a) y is put equal to (ax— m), and in case (B) y is put equal to (mx-c).

Case (y) is not expressly enunciated, but occurs, as it were, accidentally (IV. 31). The equation to be solved is 3x+18—x2 = y2. Diophantus first assumes 3x + 18 − x2 = 4 x2, which gives the quadratic 3x+18= 5 x2; but this is not rational'. Therefore the assumption of 4x2 for y2 will not do, ' and we must find a square [to replace 4] such that 18 times (this square+1)+(2)2 may be a square'. The auxiliary equation is therefore 18 (m2 + 1) + y2, or 72m2 +81 = a square, and Diophantus assumes 72 m2+81-(8m+9)2, whence m=18. Then, assuming 3x+18-x2 = (18)2x2, he obtains the equation 325x2-3x-18= 0, whence, that is,.

(2) Double equation.

=

The Greek term is διπλοϊσότης, διπλῆ ἰσότης οι διπλῆ ἴσωσις. Two different functions of the unknown have to be made simultaneously squares. The general case is to solve in rational numbers the equations

mx2+xx + α = u2

nx2 + ẞx + b = w2)

The necessary preliminary condition is that each of the two expressions can be made a square. This is always possible when the first term (in x2) is wanting. We take this simplest case first.

Diophantus has one general method taking slightly different forms according to the nature of the coefficients.

(a) First method of solution.

This depends upon the identity

2

{{ (p + q) }2 — { } (p −q)}2 = pq.

If the difference between the two expressions in x can be separated into two factors p, q, the expressions themselves are equated to {(p+q)}2 and {(p−q)}2 respectively. As Diophantus himself says in II. 11, we'equate either the square of half the difference of the two factors to the lesser of the expressions, or the square of half the sum to the greater'.

We will consider the general case and investigate to what particular classes of cases the method is applicable from Diophantus's point of view, remembering that the final quadratic in x must always reduce to a single equation.

Subtracting, we have (a-B)x+(a−b) = u2 — w2.

Separate (x-ẞ)x + (a−b) into the factors

P, {(α −ẞ)x + (a —b)} / p.

(α-B)2x2 + 2x { (x − ẞ) (a − b) — p2 (x+B) }

+(ab)2-2 p2 (a+b) + p1 = 0.

In order that this equation may reduce to a simple equation, either

(1) the coefficient of x2 must vanish, or x-ẞ = 0,

or (2) the absolute term must vanish, that is,

or

p1 − 2p2 (a + b) + (a−b)2 = 0,

{p2 - (a+b)} 2

24ab,

so that ab must be a square number.

As regards condition (1) we observe that it is really sufficient if an2 = ẞm2, since, if xx+a is a square, (xx+a) n2 is equally a square, and, if Ba+b is a square, so is (Bx+b) m2, and vice versa.

That is, (1) we can solve any pair of equations of the form

Multiply by n2, m2 respectively, and we have to solve the equations

x m2 n2x + an2 = u'

x m2 n2x+bm2 = w'2)

Separate the difference, an2-bm2, into two factors p, q and

Any factors p, q can be chosen provided that the resulting value of x is positive.

therefore (15-13)265-24x; that is, 24x= 64, and x = §.

Taking now the condition (2) that ab is a square, we see that the equations can be solved in the cases where either a and b are both squares, or the ratio of a to b is the ratio of a square to a square. If the equations are

and factors are taken of the difference between the expressions as they stand, then, since one factor p, as we saw, satisfies the

The difference is 5x+5=5(x+1); the solution is given by (x+3)2 = 10x+9, and x = 28.

Another method is to multiply the equations by squares such that, when the expressions are subtracted, the absolute term vanishes. The case can be worked out generally, thus. Multiply by d2 and c2 respectively, and we have to solve

Then x is found from the equation

xd2x+c2 d2 = (px + q)2,

which gives p2x2+2x(pq-2x d2)+q2-4c2d2 = 0,

or, since

Pq= xd2-Bc2,

p2x2 - 2x(αd2 +ẞc2)+q2 − 4 c2d2 = 0.

In order that this may reduce to a simple equation, as Diophantus requires, the absolute term must vanish, so that =2cd. The method therefore only gives one solution, since is restricted to the value 2 cd.

Ex. from Diophantus:

8x+4= u2)

6x+4= w2)

(IV.39)

Difference 2x; q necessarily taken to be 24 or 4; factors therefore, 4. Therefore 8x + 4 = (x+4)2, and x = 112.

(B) Second method of solution of a double equation of the first degree.

There is only one case of this in Diophantus, the equations being of the form

hx + n2 = u2).

(h+f)x+n2 = w2)

Suppose ha+n2 = (y+n)2; therefore ha= y2+2 ny,

and

hx

It only remains to make the latter expression a square, which is done by equating it to (py—n)2.

The case in Diophantus is the same as that last mentioned (IV. 39). Where I have used y, Diophantus as usual contrives to use his one unknown a second time.

(1)

p2x2 + ax + α = u2 }, where, except in one case, a = b. p2x2+ẞx+b= w2]'