p(x) = ax3 + bx2 + cx + d is a polynomial with integer coefficients and a non-zero. We have x p(x) = y p(y) for infinitely many pairs (x, y) of unequal integers. Show that p(x) has an integer root.
Solution
Suppose that x p(x) = y p(y) (and x and y are unequal). Put u = x + y, v = x2 + y2 and we get (after dividing by x - y), 2avu + b(u2 + v) + 2uc + 2d = 0 (*), so v = -(bu2 + 2uc + 2d)/(2au + b). But v ≥ u2/2, so -(bu2 + 2uc + 2d)/(2au + b) ≥ u2/2. But that can only be true for finitely many values of u. Hence for some value k of u, there are infinitely many solutions x, y.
But we can also write (*) as p(u) = (b+2au)xy. For u = k, that holds for more than one value of xy (because for given k = x+y and xy, there is only one value of x, y). But that implies that p(k) = 0 and (b+2ak) = 0, so p has the integer root k.
The official solution looks wrong to me. Thanks to El Idrissi el Mrhari Moulay Abdellah for this variant
© John Scholes
jscholes@kalva.demon.co.uk
8 Aug 2003
Last corrected/updated 20 Oct 2003