IMO 1992 SL 19

Origin: IRE

Problem

Let f(x) = x8 + 4x6 + 2x4 + 28x2 + 1. Let p > 3 be a prime and suppose there exists an integer z such that p divides f(z). Prove that there exist integers z1, z2, . . . , z8 such that if g(x) = (x −z1)(x −z2) \cdot \cdot \cdot (x −z8), then all coefficients of f(x) −g(x) are divisible by p.

Solution

Observe that f(x) = (x4 + 2x2 + 3)2 −8(x2 −1)2 = [x4 + 2(1 − \sqrt 2)x2 + 3+2 \sqrt 2][x4 +2(1+ \sqrt 2)x2 +3−2 \sqrt 2]. Now it is easy to find that the roots of f are x1,2,3,4 = \pmi  i 4\sqrt 2 \pm 1

and x5,6,7,8 = \pmi  4\sqrt 2 \pm 1

. In other words, xk = \alphai + \betaj, where \alpha2 i = −1 and \beta4 j = 2. We claim that any root of f can be obtained from any other using rational functions. In fact, we have x3 = −\alphai −3\betaj + 3\alphai\beta2 j + \beta3 j , x5 = 11\alphai + 7\betaj −10\alphai\beta2 j −10\beta3 j x7 = −71\alphai −49\betaj + 35\alphai\beta2 j + 37\beta3 j , from which we easily obtain that

\alphai = 24−1(127x+5x3+19x5+5x7), \betaj = 24−1(151x+5x3+19x5+5x7). Since all other values of \alpha and \beta can be obtained as rational functions of \alphai and \betaj, it follows that all the roots xl are rational functions of a particular root xk. We now note that if x1 is an integer such that f(x1) is divisible by p, then p > 3 and x1 \inZp is a root of the polynomial f. By the previous consideration, all remaining roots x2, . . . , x8 of f over the field Zp are rational functions of x1, since 24 is invertible in Zp. Then f(x) factors as f(x) = (x −x1)(x −x2) \cdot \cdot \cdot (x −x8), and the result follows.