# Sum of roots rational but product irrational

Suppose that $x_1,x_2,x_3,x_4$ are the real roots of a polynomial with integer coefficients of degree $4$, and $x_1+x_2$ is rational while $x_1x_2$ is irrational. Is it necessary that $x_1+x_2=x_3+x_4$?

For example, the polynomial $x^4-8x^3+18x^2-8x-7$ has roots $$x_1=1-\sqrt{2},x_2=3+\sqrt{2},x_3=1+\sqrt{2},x_4=3-\sqrt{2}.$$ It holds that $x_1+x_2$ is rational while $x_1x_2$ is irrational, and we have $x_1+x_2=x_3+x_4$.

Suppose that all the roots are non zero. Put $x_1+x_2=u$, $a=x_1x_2$, $x_3+x_4=v$, $b=x_3x_4$, we suppose that $u\in\mathbb{Q}$, then it is also the case for $x_3+x_4$, as the sum of all the roots is rational, and that $a,b$ are irrationals (as the product $ab$ is rational, if $a$ is irrational, then it is also the case for $b$). The polynomial with roots $x_1,x_2,x_3,x_4$ is $$P(x)=(x^2-ux+a)(x^2-vx+b)=x^4-(u+v)x^3+(a+uv+b)x^2-(av+bu)x+ab$$

By hypothesis, we get that $a+b+uv$, $av+bu$, and $ab$ are in $\mathbb{Q}$. Writing

$av+bu=(a+b)v-bv+bu$, we see that $b(u-v)$ is rational. As $b$ is not, this imply $u=v$.

If now $x_3x_4=0$, then it is not possible that $x_3=x_4=0$,(in this case $P(x)=x^4-ux^3+ax^2$) as $a\not \in \mathbb{Q}$, and if $x_4=0$ and $x_3\not = 0$, we get that $x_3\in \mathbb{Q}$, and $x_3\in \mathbb{Q}$ is not possible again, as $P(x)=(x-x_3)(x^2-ux+a)=x⁴+..-x_3ax$.

• Why can we assume that $\nu=x_3+x_4$ is rational ? – Dietrich Burde Jul 21 '16 at 19:14
• The sum $x_1+x_2+x_3+x_4$ is rational, as the $x_k$ are the roots of a polynomial with rational coefficients (and $x_1+x_2$ is rational) – Kelenner Jul 21 '16 at 19:17

Since $x_1x_2$ is irrational, there is an automorphism $\sigma$ of $\overline {\Bbb Q}$ that changes $x_1x_2$ into something else. Since $\sigma$ acts on a permutation on the roots, we must have $\sigma(x_1)=x_i$ and $\sigma(x_2) = x_j$ where $i,j \in \{1;2;3;4\}$ and $i \neq j$, and importantly, $x_1x_2 \neq x_ix_j$

Since $x_1+x_2$ is rational it is fixed by $\sigma$ and so $x_1+x_2 = x_i+x_j$. If say $x_i = x_1$, then from this we get $x_j=x_2$, and then $x_ix_j = x_1x_2$ which is impossible. And so we must have $\{i;j\} = \{3;4\}$, and so, $x_1+x_2 = x_3+x_4$.

Another way to tell this story is that we have shown that if $x_1+x_2$ is rational and $x_1+x_2-x_3-x_4 \neq 0$ then $x_1x_2$ is rational.

In fact, let $x_1,x_2,x_3,x_4$ be indeterminates and consider the Galois extension $K = \Bbb Q(x_1,x_2,x_3,x_4)^{S_4} \subset M = \Bbb Q(x_1,x_2,x_3,x_4)$.

Let $L = K(x_1+x_2)$. By the fundamental theorem of Galois theory, $L$ is the subfield of $M$ that is fixed by a certain subgroup $H$ of $S_4$. This subgroup $H$ is the set of permutations of $\{1;2;3;4\}$ that fixes the unordered pair $\{1;2\}$ (because it only has to fix $x_1+x_2$ in addition to the elementary symmetric polynomials), so $H$ is the subgroup $\{id ; (12) ; (34) ; (12)(34) \}$

Since $x_1x_2$ is also fixed by $H$ this means that $x_1x_2 \in K(x_1+x_2)$ : you can express $x_1x_2$ as a rational fraction in terms of $x_1+x_2$ and the elementary symmetric polynomials (i.e. the rational coefficients of your polynomial).

Then, what we have proved says that the denominator of that fraction has to be a power of $(x_1+x_2-x_3-x_4)$.

Indeed, someone can just waltz in and trivialize this problem by saying that $(x_1+x_2-x_3-x_4)x_1x_2 = (x_1+x_2)(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4) - (x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4) - (x_1+x_2)(x_1+x_2)(x_3+x_4)$.

So if $x_1+x_2 \neq x_3+x_4$ and $x_1+x_2$ is rational this gives you a formula proving that $x_1x_2$ is rational too.