1. When the number of unknown quantities exceeds the number of
equations, the equations will admit of innumerable solutions, and are
therefore said to be _indeterminate_. Thus if it be required to find
two numbers such that their sum be 10, we have two unknown quantities
x and y, and only one equation, viz. x + y = 10, which may evidently
be satisfied by innumerable different values of x and y, if fractional
solutions be admitted. It is, however, usual, in such questions as
this, to restrict values of the numbers sought to positive integers,
and therefore, in this case, we can have only these nine solutions,
x = 1, 2, 3, 4, 5, 6, 7, 8, 9;
y = 9, 8, 7, 6, 5, 4, 3, 2, 1;
which indeed may be reduced to five; for the first four become the
same as the last four, by simply changing x into y, and the contrary.
This branch of analysis was extensively studied by Diophantus, and is
sometimes termed the Diophantine Analysis.
2. Indeterminate problems are of different orders, according to the
dimensions of the equation which is obtained after all the unknown
quantities but two have been eliminated by means of the given
equations. Those of the first order lead always to equations of the
form
ax [+-] by = [+-]c,
where a, b, c denote given whole numbers, and x, y two numbers to be
found, so that both may be integers. That this condition may be
fulfilled, it is necessary that the coefficients a, b have no common
divisor which is not also a divisor of c; for if a = md and b = me,
then ax + by = mdx + mey = c, and dx + ey = c/m; but d, e, x, y are
supposed to be whole numbers, therefore c/m is a whole number; hence m
must be a divisor of c.
Of the four forms expressed by the equation ax [+-] by = [+-]c, it is
obvious that ax + by = -c can have no positive integral solutions.
Also ax - by = -c is equivalent to by - ax = c, and so we have only to
consider the forms ax [+-] by = c. Before proceeding to the general
solution of these equations we will give a numerical example.
To solve 2x + 3y = 25 in positive integers. From the given equation
we have x = (25 - 3y)/2 = 12 - y - (y - 1)/2. Now, since x must be a
whole number, it follows that (y - 1)/2 must be a whole number. Let us
assume (y - 1)/2 = z, then y = 1 + 2z; and x = 11 - 3z, where z might
be any whole number whatever, if there were no limitation as to the
signs of x and y. But since these quantities are required to be
positive, it is evident, from the value of y, that z must be either 0
or positive, and from the value of x, that it must be less than 4;
hence z may have these four values, 0, 1, 2, 3.
If z = 0, z = 1, z = 2, z = 3;
Then x = 11, x = 8, x = 5, x = 2,
y = 1, y = 3, y = 5, y = 7.
3. We shall now give the solution of the equation ax - by = c in
positive integers.
Convert a/b into a continued fraction, and let p/q be the convergent
immediately preceding a/b, then aq - bp = [+-]1 (see CONTINUED
FRACTION).
([alpha]) If aq - bp = 1, the given equation may be written
ax - by = c(aq - bp);
:. a(x - cq) = b(y - cp).
Since a and b are prime to one another, then x - cq must be divisible
by b and y - cp by a; hence
(x - cq) / b = (y - cq)/a = t.
That is, x = bt + cq and y = at + cp.
Positive integral solutions, unlimited in number, are obtained by
giving t any positive integral value, and any negative integral value,
so long as it is numerically less than the smaller of the quantities
cq/b, cp/a; t may also be zero.
([beta]) If aq - bp = -1, we obtain x = bt - cq, y = at - cp, from
which positive integral solutions, again unlimited in number, are
obtained by giving t any positive integral value which exceeds the
greater of the two quantities cq/b, cp/a.
If a or b is unity, a/b cannot be converted into a continued fraction
with unit numerators, and the above method fails. In this case the
solutions can be derived directly, for if b is unity, the equation may
be written y = ax - c, and solutions are obtained by giving x positive
integral values greater than c/a.
4. To solve ax + by = c in positive integers. Converting a/b into a
continued fraction and proceeding as before, we obtain, in the case of
aq - bp = 1,
x = cq - bt, y = at - cp.
Positive integral solutions are obtained by giving t positive integral
values not less than cp/a and not greater than cq/b.
In this case the number of solutions is limited. If aq - bp = -1 we
obtain the general solution x = bt - cq, y = cp - at, which is of the
same form as in the preceding case. For the determination of the
number of solutions the reader is referred to H.S. Hall and S.R.
Knight's _Higher Algebra_, G. Chrystal's _Algebra_, and other
text-books.
5. If an equation were proposed involving three unknown quantities, as
ax + by + cz = d, by transposition we have ax + by = d - cz, and,
putting d - cz = c', ax + by = c'. From this last equation we may find
values of x and y of this form,
x = mr + nc', y = mr + n'c',
or x = mr + n(d - cz), y = m'r + n'(d - cz);
where z and r may be taken at pleasure, except in so far as the values
of x, y, z may be required to be all positive; for from such
restriction the values of z and r may be confined within certain
limits to be determined from the given equation. For more advanced
treatment of linear indeterminate equations see COMBINATORIAL
ANALYSIS.
6. We proceed to indeterminate problems of the second degree: limiting
ourselves to the consideration of the formula y^2 = a + bx + cx^2,
where x is to be found, so that y may be a rational quantity. The
possibility of rendering the proposed formula a square depends
altogether upon the coefficients a, b, c; and there are four cases of
the problem, the solution of each of which is connected with some
peculiarity in its nature.
_Case_ 1. Let a be a square number; then, putting g^2 for a, we have
y^2 = g^2 + bx + cx^2. Suppose [root](g^2 + bx + cx^2) = g + mx; then
g^2 + bx + cx^2 = g^2 + 2gmx + m^2 x^2, or bx + cx^2 = 2gmx + m^2 x^2,
that is, b + cx = 2gm + m^2x; hence
2gm - b cg - bm + gm^2
x = --------, y = [root](g^2 + bx + cx^2) = --------------,
c - m^2 c - m^2
_Case_ 2. Let c be a square number = g^2; then, putting [root](a + bx
+ g^2 x^2) = m + gx, we find a + bx + g^2x^2 = m^2 + 2mgx + g^2 x^2,
or a + bx = m^2 + 2mgx; hence we find
m^2 - a bm - gm^2 - ag
x = -------, y = [root](a + bx + g^2 x^2) = --------------.
b - 2mg b - 2mg
_Case_ 3. When neither a nor c is a square number, yet if the
expression a + bx + cx^2 can be resolved into two simple factors, as f
+ gx and h + kx, the irrationality may be taken away as follows:--
Assume [root](a + bx + cx^2)=[root]{(f + gx)(h + kx)} = m(f + gx),
then (f + gx)(h + kx) = m^2(f + gx)^2, or h + kx = m^2(f + gx); hence
we find
fm^2 - h (fk - gh)m
x = --------, y = [root]{(f + gx)(h + kx)} = ----------;
k - gm^2 k - gm^2
and in all these formulae m may be taken at pleasure.
_Case_ 4. The expression a + bx + cx^2 may be transformed into a
square as often as it can be resolved into two parts, one of which is
a complete square, and the other a product of two simple factors; for
then it has this form, p^2 + qr, where p, q and r are quantities which
contain no power of x higher than the first. Let us assume [root](p^2
+ qr) = p + mq; thus we have p^2 + qr = p^2 + 2mpq + m^2q^2 and r =
2mp + m^2q, and as this equation involves only the first power of x,
we may by proper reduction obtain from it rational values of x and y,
as in the three foregoing cases.
The application of the preceding general methods of resolution to any
particular case is very easy; we shall therefore conclude with a
single example.
_Ex._ It is required to find two square numbers whose sum is a given
square number.
Let a^2 be the given square number, and x^2, y^2 the numbers required;
then, by the question, x^2 + y^2 = a^2, and y = [root](a^2 - x^2).
This equation is evidently of such a form as to be resolvable by the
method employed in case 1. Accordingly, by comparing [root](a^2 - x^2)
with the general expression [root](g^2 + bx + cx^2), we have g = a, b
= 0, c = -1, and substituting these values in the formulae, and also
-n for +m, we find
2an a(n^2 - 1)
x = -------, y = ----------.
n^2 + 1 n^2 + 1
If a = n^2 + 1, there results x = 2n, y = n^2 - 1, a = n^2 + 1. Hence
if r be an even number, the three sides of a rational right-angled
triangle are r, (1/2r)^2 - 1, (1/2r)^2 + 1. If r be an odd number,
they become (dividing by 2) r, 1/2(r^2 - 1), 1/2(r^2 + 1).
For example, if r = 4, 4, 4 - 1, 4 + 1, or 4, 3, 5, are the sides of a
right-angled triangle; if r = 7, 7, 24, 25 are the sides of a
right-angled triangle.