2010/052) For which real values of p and q are the roots of x^3-px^2+11x-q=0 three consecutive integers

the three consecutive integers can be taken to be

s = a-1, r = a, t = a +1

coefficient of x = 11 = rs + rt + sr = a(a-1) + a(a+1) + (a-1)(a+1) = 3a^2 - 1 = 11

so a = 2 or -2
a=2 gives
roots are 1,2,3

so (x-1)(x-2)(x-3) = x^3- 6x^2 + 11x -6 or p = q = 6

a = -2 gives
roots as -1 ,-2,-3

(x+1)(x+2)(x+3) = x^3+6x + 11+ 6 so p=q = - 6

so 2 solutions are

1) p= q = 6
2) p =q = - 6