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.