Putnam 2003, Problem B-4

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:
### First solution:  
Put \( g = r_1 + r_2 \), \( h = r_3 + r_4 \), \( u = r_1 r_2 \), \( v = r_3 r_4 \). We are given that \( g \) is rational. The following are also rational:  
\[
-\frac{b}{a} = g + h, \quad \frac{c}{a} = gh + u + v, \quad -\frac{d}{a} = gv + hu.
\]  
From the first line, \( h \) is rational. From the second line, \( u + v \) is rational. From the third line, \( g(u + v) - (gv + hu) = (g - h)u \) is rational. Since \( g \neq h \), \( u \) is rational, as desired.  


### Second solution:  
This solution uses some basic Galois theory. We may assume \( r_1 \neq r_2 \), since otherwise they are both rational and so then is \( r_1 r_2 \).  

Let \( \tau \) be an automorphism of the field of algebraic numbers; then \( \tau \) maps each \( r_i \) to another one, and fixes the rational number \( r_1 + r_2 \). If \( \tau(r_1) \) equals one of \( r_1 \) or \( r_2 \), then \( \tau(r_2) \) must equal the other one, and vice versa. Thus \( \tau \) either fixes the set \( \{r_1,r_2\} \) or moves it to \( \{r_3,r_4\} \). But if the latter happened, we would have \( r_1 + r_2 = r_3 + r_4 \), contrary to hypothesis. Thus \( \tau \) fixes the set \( \{r_1,r_2\} \) and in particular the number \( r_1 r_2 \). Since this is true for any \( \tau \), \( r_1 r_2 \) must be rational.  

**Note:** The conclusion fails if we allow \( r_1 + r_2 = r_3 + r_4 \). For instance, take the polynomial \( x^4 - 2 \) and label its roots so that \( (x - r_1)(x - r_2) = x^2 - \sqrt{2} \) and \( (x - r_3)(x - r_4) = x^2 + \sqrt{2} \).