64th Putnam 2003

Problem B4

az4 + bz3 + cz2 + dz + e has integer coefficients (with a ≠ 0) and roots r1, r2, r3, r4 with r1+r2 rational and r3+r4 ≠ r1+r2. Show that r1r2 is rational.



Put h = r1+r2, k = r3+r4. Comparing coefficients of z3 in polynomial and (z2 - hz + r1r2)(z2 - kz + r3r4) we see that h + k is rational. But h is rational, so k is rational. Comparing coefficients of z2 we have that hk + r1r2 + r3r4 is rational. Hence also h(r1r2 + r3r4). But comparing z, hr3r4 + kr1r2 is rational. Subtracting, (h-k)r1r2 is rational. But h-k is non-zero rational, so r1r2 is rational.



64th Putnam 2003

© John Scholes
8 Dec 2003
Last corrected/updated 8 Dec 03