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Let $p$ be a prime number, and let $m,k_1,\ldots,k_{p-2}$ be even numbers. Define the polynomial $h(x)=(x^2+m)(x-k_1)\cdots(x-k_{p-2})$ and $r=\min \{|h(a)|\mid a\in\mathbb{R},h'(a)=0\}$.

Under these condition I've proven that $r>0$.

Now let $n$ be an odd number large enough so that $\frac{2}{n}<r$ and let $f(x)=h(x)-\frac{2}{n}$, prove that $f$ is irreducible over $\mathbb{Q}$ and that it has exactly two complex roots.

Any hints would be appreciated, thanks!

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up vote 2 down vote accepted

It suffices to show the polyonomial $n(x+m^2)(x-k_1)\cdots(x-k_{p-2})-2$ is irreducible over $\mathbb{Q}$, notice that this polynomial is primitive. So it suffices to show that it is irreducible over $\mathbb{Z}$, this can be proved by Eisenstein criterion when modulo 2.

The above arguement only uses the assumption that $m,k_i$ are even and $n$ is odd.

If $h$ has no multiple roots, then when we draw the picture of $f$, it is easy to see that all roots of $f$ are real and distinct.

E.g. $h=(x+2^2)(x-4)$ and $2/n=2$, then $f=x^2-18$ has two different real roots.

Edit: Our assumptions are that $k_i$ are distinct and $p$ is odd and $k_i>0$, $m>0$. The arguments for irreducibility still work. Now denote $h_2=x^2+m$ and $h_1=(x-k_1)(x-k_2)(x-k_{p-2})$. Consider the equation $h_1=s/h_2$, where $s=2/n<r$, draw the picture to see that $h_1$ and $s/h_2$ intersects exactly $p-2$ different real points. We can use intermediate value theorem(continuity) to make this rigorous.

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I apologize, I've made a typing error which rendered your last example incorrect... There should be exactly 2 complex roots, that's actually the whole point of this construction... – Shai Deshe May 12 '12 at 15:38

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