az^{4} + bz^{3} + cz^{2} + dz + e has integer coefficients (with a ≠ 0) and roots r_{1}, r_{2}, r_{3}, r_{4} with r_{1}+r_{2} rational and r_{3}+r_{4} ≠ r_{1}+r_{2}. Show that r_{1}r_{2} is rational.

**Solution**

Put h = r_{1}+r_{2}, k = r_{3}+r_{4}. Comparing coefficients of z^{3} in polynomial and (z^{2} - hz + r_{1}r_{2})(z^{2} - kz + r_{3}r_{4}) we see that h + k is rational. But h is rational, so k is rational. Comparing coefficients of z^{2} we have that hk + r_{1}r_{2} + r_{3}r_{4} is rational. Hence also h(r_{1}r_{2} + r_{3}r_{4}). But comparing z, hr_{3}r_{4} + kr_{1}r_{2} is rational. Subtracting, (h-k)r_{1}r_{2} is rational. But h-k is non-zero rational, so r_{1}r_{2} is rational.

© John Scholes

jscholes@kalva.demon.co.uk

8 Dec 2003

Last corrected/updated 8 Dec 03