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372 Answers to Exercises and Solutions to Problems

We can now solve for x and y.

8(x + ,/jj) = 8u3 = (z” + 3%~) + (3z2 + w)&

8(x - fi) = 8v3 = (z” + 3%~) - (3%2 + w)t/ii?

so that

8x = z3 + 3%~ = z(z2 + 3w)

8,/j = (3%2 + w)fi 3 64y = (3z2 + w)~w.

Hence, if there is a solution, we must have

(x, y, z) = (z(z2 + 3w)/8, ~(3%~ + ~)~/64, z)

for suitable % and w. On the other hand, (*) yields solutions since

(*I

= (1/2)[d(%3 + 320%) + (3%2 + w)&

23 + SW%) - (322 + w)&J

= (1/2)f/z+J;;)3+ t-

= (VW + 6) + (% - 441

= t.

We need conditions to ensure that x, y, % are integers. Suppose that z = 2s
is even. Then 0 E (3%2 + w)“w = (12~~ + W)~W (mod 64) 3 w = 2r is even
=+ (6s2 + r)2r z 0 (mod 8) =+ r is even j 4120. Suppose % is odd. Then

z(z2 + 3w) E 0 (mod 8) 3 w E 5 (mod 8).

Hence, the integer solutions of the given equation are given by (*) where
either (z, w) = (2s, 4t) or (2, w) = (2k + 1,8m + 5).

Examples:

(i) % = 1, w =^5 yields (z,y,%) = (2,5,1). In this case, u = (l/2)(1+&)

and u3 = 2 + 4.

(ii) w = 0 yields the obvious solution x = s3, y = 0, % = 2s.

(iii) % = 6, w = -4 yields u = 3 + i, v = 3 - i, (x, y, z) = (18, -676,6).

11. Let u = v/w be a given rational value for the polynomial. The equation
    3x2 - 5x + (4 - u”) = 0 has discriminant ( 12v2 - 23w2)/w2 and so will have
    rational solutions when 12v2 - 23w2 is a perfect square.