Suppose p(x) is an irreducible polynomial over Q of degree n, with integer coefficients. If p(x) has two roots r1 and r2 satisfying r1r2 = 5, prove that n is even.

Attempt at solution:

  • Because the base field is Q, the field extension is separable, hence no roots of multiplicity > 1.

  • Because r1r2 = 5, I think we can show that if the degree of n is odd, then this will contradict the fact that p(x) has integer coefficients somehow, but I can't seem to do this.

Any help would be appreciated very much, thank you.


Claim. For every root $r$ of $p(x)$, $r' = 5/r$ is also a root of $p(x)$.

Proof: Let $g$ be an element of the Galois group that maps $r_1$ to $r$, i.e. $gr_1 = r$. Let $r' = gr_2$. Then $rr' = gr_1 \cdot gr_2 = g(r_1r_2) =g(5) = 5$. QED

Therefore, we can split all roots of $p(x)$ into pairs $(r, r')$ such that $rr' = 5$. Since $p(x)$ doesn't have multiple roots, $r\neq r'$ in each pair. Thus the number of roots is even. Therefore, the polynomial has even degree.

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  • $\begingroup$ Yury - thank you so much for the quick response & concise solution. Appreciate it. $\endgroup$ – Conan Wong Aug 9 '12 at 5:21
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    $\begingroup$ Presumably, 5 can be replaced with any rational number, and the same result obtains. $\endgroup$ – Gerry Myerson Aug 9 '12 at 6:08

Hint $\ $ Since $\rm\,p(x)\,$ is irreducible the Galois group G acts transitively on its roots, so G contains the involution $\rm\:x\to 5/x.\:$ If the involution has no fixed points then it $\rm\color{#C00}{pairs}$ the roots so they are $\rm\color{#C00}{even}$ in number. Else, if $\,\alpha\,$ is a fixed point then $\rm\,\alpha\, = 5/\alpha\,$ so $\rm\:\,\alpha^2 = 5\,$ so $\,\ldots$

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  • $\begingroup$ Thank you Bill - much appreciated. $\endgroup$ – Conan Wong Aug 9 '12 at 14:45

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