INDETERMINATE EQUATIONS OF FIRST DEGREE 165

412. The values of x; + nb for various values of # form an
arithmetic sequence whose central term, corresponding to #n = 0,
is 2, and whose common difference is .

The values of y; — na form an arithmetic sequence whose
central term is y; and whose common difference is — a.

413. It is sometimes interesting to consider the solutions for
which x and y are both positiv. We will assume that ¢ and b
are both positiv.

Since x; + nb is positiv, x; > — nband — x:1/b < m

Since y; — na is positiv, y; > nae and 7 < y,/a

Hence — x1/b < n < yifa

That is, for x and vy both to be positiv n must lie between
— x1/b and y1/a.

It is easy to show also, conversely, that if n does lie within
these limits, both x and vy will be positiv.

Example. We saw that (11 4 97, 19 — 49z) is a complete
solution of the equation 49x 4+ 19y = 900.

Herea = 49,0 =19, %, = 11,3 =19

Since a and b are positiv, by the rule workt out above, for
x and y both to be positiv # must lie between — 11/19 and
19/49.

The only integer between these limits is 0.

Therfor the only solution of the given equation in positiv
integers is that for which » = 0, namely (11,19).

414. The reader may possibly be interested to discuss some
of the other 15 of the 16 possible cases that arise according
as x, v, @, b are each positiv or negativ.

415. The case when ¢ = 0 is included in the preceding dis-
cussion, but it may be interesting to consider it separately.
Then our equation is ax + by = 0.

Th. If (x, y) is a solution of the equation ax + by = 0,

where a || b, then (x, y) is of the form (nb," — na), where n is
some integer, and conversely, for every value of n (nb, — na) is a
solution.

If (x, y) is a solution, ax + by = 0
Hence ax = — by and therfor ax > b.