(^370) Answers to Exercises and Solutions to Problems
- 24 = (x + y + 2)” - (x3 + y3 + z3) = 3Cx2y + 6xyz j 8 = x(y + z)~ +
(x2 + YZ)(Y + z) = (y + z)(x + z)(x + y) = (3 - x)(3 - y)(x + y). Hence
(X:,YJ) = (Ll,l), (4,4,-5), (4,--5,4), (-5,4,4). - The equation can be rewritten 4(2a + 1)2 = (2b + 1)2 + 3. Since 4 and 1
are the only squares which differ by 3, the result follows. - The equation is equivalent to
0 = x2y2 + 3xsy + xy - 3xy2 + 2y3
= y[2y2 + (x2 - 3x)y + (x + 3x2)].Either y = 0 and x is arbitrary or the quadratic in y has integer zeros. For
the latter case, it is necessary that the discriminant(x2 - 3~)~ - 8(x + 3x2) = x(x - 8)(x + 1)”be a square. Hence, either x = -1, 0, 8, or else x2 - 8x = (x - 4)2 - 16
is a perfect square, i.e. x = 9. Thus, the solutions (x, y) with y # 0 are
C-1, -I), (8, --lo>, (9, -4% (9, -21).- (u + vi)3 = y + 2i leads to v(3u2 - v2) = 2, whence v divides 2. Trying
the possibilities yields (u, v) = (*l, l), (fl, -2). Since (u - vi)3 = y - 2i,
we can take x = (u + vi)(u - vi), and obtain the solutions
(2, y) = (u” + 212, u(u” - 3v2) = (2, F2), (5, Ql).- a2xy+abx+acy+ad = 0 j (ax+c)(ay+b) = bc-ad. If bc= ad, then
there are infinitely many solutions if and only if either c or b is a multiple
of a (in which case one of the left factors can be made to vanish and the
other can be arbitrary). If bc # ad, then ax + c and ay + b both divide a
nonzero integer and there are at most finitely many possibilities for both
x and y. - First solution. Suppose xy > 0. Then, since
(2 - l/)(x2 + xy + y2) = 2(xy + 4)we have that x > y. Since at least one of Ix] and ]y] differs from 1, x2 +
xy+y2>xy+4,sothatx-y<2.Sincex#y,wehavethatx=y+l,
which leads to y2 + y - 7 = 0 with no solution in integer y.
Suppose x < 0 and y > 0. Then
8 = x3 -2xy-ys+x2-2xy-y2=-(x+y)2,which is impossible. Suppose x > 0 and y < 0. Then 8 = x3-2xy-y3 > x3,
so that x must be 1. This does not work.
Hence xy = 0 and we have the solution (I, y) = (2,0), (0, -2).