**Problem**: (2003 Putnam - B4) Let

$ f(z) = az^4+bz^3+cz^2+dz+e = a(z-r_1)(z-r_2)(z-r_3)(z-r_4) $

where $ a, b, c, d, e $ are integers, $ a \neq 0 $. Show that if $ r_1+r_2 $ is a rational number and $ r_1+r_2 \neq r_3+r_4 $, then $ r_1r_2 $ is a rational number.

**Solution**: Some important results that you don't need to prove - if $ x $ and $ y $ are rational, then

$ x+y $, $ x-y $, $ xy $, and $ \frac{x}{y} $

are rational (note the last one is only true for $ y \neq 0 $).

We start of by using Vieta's Formulas, which tells us that

$ S_1 = r_1+r_2+r_3+r_4 $

$ S_2 = r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4 = (r_1+r_2)(r_3+r_4)+r_1r_2 $

$ S_3 = r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4 = r_1r_2(r_3+r_4)+r_3r_4(r_1+r_2) $

are all rational. But since we know $ r_1+r_2 $ is rational, we must have $ r_3+r_4 = S_1-(r_1+r_2) $ be rational as well. Then

$ (r_1+r_2)(r_3+r_4) $

and

$ r_1r_2+r_3r_4 = S_2-(r_1+r_2)(r_3+r_4) $

are rational. Let $ k = r_3+r_4-r_1+r_2 \neq 0 $, which we know is rational. Then

$ S_3 = r_1r_2(r_1+r_2+k)+r_3r_4(r_1+r_2) = (r_1r_2+r_3r_4)(r_1+r_2)+r_1r_2k $,

but $ (r_1r_2+r_3r_4)(r_1+r_2) $ is rational, so $ r_1r_2k $ must be rational as well. Finally, since $ k $ is rational and nonzero, we have $ r_1r_2 $ rational. QED.

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Comment: This problem required slugging through some rational/irrational arguments, but turned out to be not so bad. As long as you recognized the use of Vieta's Formulas, you should have been able to work out the details to get to the solution.

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Practice Problem: Find a polynomial $ p(z) $ with $ r_1+r_2 = r_3+r_4 $ such that the above argument does not hold.

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