By Furui S., Sandhi M.M. (eds.)

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**Example text**

Each of the systems of two equations in two unknowns Xo and Yo admits of only trivial integral solutions, Xo = 1, Yo = 0 and Xo = - 1, Yo = 0 respectively. Hence, when A is equal to the square of an integer, equation (51) has only trivial solutions in integers Xo = ± 1, Yo = ltnen A is a negative integer, equation (51) has the same trivial integral solutions. ) Let us now consider a more general equation o. x2 - A y2 = C (73) V4 . , = is irrational. We have already seen that when C = 1 this equation always possesses an infinite number of integral solutions (see Theorem III) .

In which the positive integers Zo, z 1, Z2, ... zo> ZI > Z2 > ... > Zn > ... hold for them. But positive integers cannot form an infinite and monotonically decreasing sequence as there cannot be more than Zo terms in it. We have thus come to a contradiction by assuming 53 that equation (94) has at least one solution in integers x, y, Z, XYZ·=I= o. This serves as a proof that equation (94) does not have a solution.. Accordingly equation (93) has no solutions in positive integers [x, y, z] either, since, if otherwise, if [x, y, z] were a solution of equation (93), then [x, y, Z2] would be a solution to (94).

Anyn = c (81) where n is an integer greater than two and all the numbers ao, a2, ... , an and c are integers. At the beginning of this century, A. Thue proved that this equation possesses only a finite number of solutions in integers x and y, with the possible exception of cases when the homogeneous left-hand side is a power (1) of a homoqeneous linear binomial ab (ax + by)" = Co or (2) of a homogeneous quadratic trinomial (ax 2 + bxy + cy2)" = Co In both these instances integral solutions can exist only if Co is the nth power of some integer and, consequently, if equation (81) reduces to an equation of- the first or of the second degree respectively.