17
$\begingroup$

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 $x_1+x_2=x_3+x_4$.

$\endgroup$
14
$\begingroup$

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 $P(x)=(x-x_3)(x^2-ux+a)=x⁴+..-x_3ax$.

$\endgroup$
  • 2
    $\begingroup$ Why can we assume that $\nu=x_3+x_4$ is rational ? $\endgroup$ – Dietrich Burde Jul 21 '16 at 19:14
  • 2
    $\begingroup$ 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) $\endgroup$ – Kelenner Jul 21 '16 at 19:17
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.