*CHAPTER XXVIII.

INDETERMINATE EQUATIONS OF THE SECOND DEGREE.

*366. The solution in positive integers of indeterminate equations of a degree higher than the first, though not of much practical importance, is interesting because of its connection with the Theory of Numbers. In the present chapter we shall confine our attention to equations of the second degree involving two variables.

*367. To shew how to obtain the positive integral values of x and y which satisfy the equation

ax2 + 2hxy + by2+ 2gx + 2fy + c = 0,

a, b, c, f, g, h being integers.

have

Solving this equation as a quadratic in x, as in Art. 127, we

ax +hy+g = ± √ (h2 — ab) y2 + 2 (hg − aƒ ) y + (g2 — ac).....(1).

Now in order that the values of x and y may be positive integers, the expression under the radical, which we may denote by py2+2qy+r, must be a perfect square; that is

py2 + 2qy + r = x2, suppose.

Solving this equation as a quadratic in y, we have

and, as before, the expression under the radical must be a perfect square; suppose that it is equal to ť2; then

t' - pz2 = q2 - pr,

where t and z are variables, and p, q, r are constants.

Unless this equation can be solved in positive integers, the original equation does not admit of a positive integral solution. We shall return to this point in Art. 374.

If a, b, h are all positive, it is clear that the number of solutions is limited, because for large values of x and y the sign of the expression on the left depends upon that of ax+2hxy + by [Art. 269], and thus cannot be zero for large positive integral values of x and y.

Again, if hab is negative, the coefficient of y' in (1) is negative, and by similar reasoning we see that the number of solutions is limited.

But 30+24y-2y2=102 - 2 (y - 6)2; hence (y-6)2 cannot be greater than 51. By trial we find that the expression under the radical becomes a perfect square when (y-6)2=1 or 49; thus the positive integral values of y are 5, 7, 13.

When y=5, x=21 or 1; when y=7, x=25 or 5; when y=13, x=29 or 25.

*368. We have seen that the solution in positive integers of the equation

ax2 + 2hxy+by+ 2gx + 2fy+c=0

can be made to depend upon the solution of an equation of the form

x2± Ny2 = ±a,

where N and a are positive integers.

=

The equation a2 + Ny2 = -a has no real roots, whilst the equation x2 + Ny3a has a limited number of solutions, which may be found by trial; we shall therefore confine our attention to equations of the form x2 - Ny2 =± a.

*369. To shew that the equation x3- Ny3-1 can always be solved in positive integers.

Let N be converted into a continued fraction, and let

p'

If the number of quotients in the period is even, is an even convergent, and is therefore greater than N, and therefore greater than 2; thus p'q-pq'=1. In this case p” – Nq2 = 1, and therefore x = p', y=q' is a solution of the equation x2 - Ny2 = 1. p'

Since is the penultimate convergent of any recurring period, the number of solutions is unlimited.

If the number of quotients in the period is odd, the penultimate convergent in the first period is an odd convergent, but the penultimate convergent in the second period is an even convergent. Thus integral solutions will be obtained by putting x=p', y = q', p'

where is the penultimate convergent in the second, fourth, q' sixth,......recurring periods. Hence also in this case the number of solutions is unlimited.

*370. To obtain a solution in positive integers of the equation x2 - Ny2 = - 1.

As in the preceding article, we have

p'2 - Nq'2 = p'q- pq'.

If the number of quotients in the period is odd, and if

p'

is an odd penultimate convergent in any recurring period,

p'

and therefore p'q − pq' ·

P < q q

- 1.

In this case p22 – Nq2 = − 1, and integral solutions of the equation - Ny3=-1 will be obtained by putting x =p', y = q', p'

where is the penultimate convergent in the first, third, fifth...

recurring periods.

Н. Н. А.

20

Example. Solve in positive integers x2 - 13y2 = ±1.

Here the number of quotients in the period is odd; the penultimate con18 vergent in the first period is ; hence x=18, y=5 is a solution of

x2 - 13y2= -1.

By Art. 364, the penultimate convergent in the second recurring period is

hence x=649, y = 180 is a solution of x2 - 13y2=1.

By forming the successive penultimate convergents of the recurring periods we can obtain any number of solutions of the equations

*371. When one solution in positive integers of x- Ny2 = 1 has been found, we may obtain as many as we please by the following method.

-

Suppose that x=h, y=k is a solution, h and k being positive integers; then (h2 - Nk2)" = 1, where n is any positive integer. x2 - Ny2 = (h2 - Nk2)”.

Thus

.: (x+y√N) (x − y √N) = (h + k√/N)" (h − k√N)".

Put x+y√N=(h+k√N)", x-y√N = (h-k√N)"; ... 2x = (h + k√N)" + (h − k√N)" ;

2y√N = (h+k√N)" — (h− k √N)". ·

The values of x and y so found are positive integers, and by ascribing to n the values 1, 2, 3,..., as many solutions as we please can be obtained.

Similarly if x=h, y=k is a solution of the equation x2 - Ny2 = − 1, and if n is any odd positive integer,

x2 - Ny2 = (h2 – Nk2)”.

Thus the values of x and y are the same as already found, but n is restricted to the values 1, 3, 5,.......

*372. By putting x = ax', y=ay the equations x2- Ny2 = ±a2 become - Ny' = 1, which we have already shewn how to

solve.

*373. We have seen in Art. 369 that

p'2 — Nq2 = − r„ (Pq′ − p'q) = ±r„.

Hence if a is a denominator of any complete quotient which occurs in converting N into a continued fraction, and if

=

is

the convergent obtained by stopping short of this complete quotient, one of the equations x2-Nya is satisfied by the values x =p', y=q.

Again, the odd convergents are all less than even convergents are all greater than

N, and the

N; hence if?

is an even

p'

convergent, x = p', y=q' is a solution of a* - Ny2 = a

is an odd convergent, x = p', y = q' is a solution of x2

; and if

Ny2

= - a.

*374. The method explained in the preceding article enables us to find a solution of one of the equations x2 - Ny2 = a only when a is one of the denominators which occurs in the process of converting N into a continued fraction. For example, if we convert 7 into a continued fraction, we shall find that

and that the denominators of the complete quotients are 3, 2, 3, 1.

and if we take the cycle of equations

x2-7y=-3, x2-7y2 = 2, x2-7y=-3, x3-7y = 1,

we shall find that they are satisfied by taking

for x the values 2, 3, 5, 8, 37, 45, 82, 127,..............

and for y the values 1, 1, 2, 3, 14, 17, 31, 48,......

*375. It thus appears that the number of cases in which solutions in integers of the equations x2 - Nya can be obtained with certainty is very limited. In a numerical example it may, however, sometimes happen that we can discover by trial a