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Let $f(x)\in \Bbb Q[x]$ be an irreducible polynomial over $\Bbb Q$.

Show that there exists no complex number $\alpha$ such that $f(\alpha)=f(\alpha+1)=0$.

Following @quasi;

Let $f(x)=a_0+a_1x+a_2x^2+\dots +a_nx^n$.

Define $g(x)=f(x+1)-f(x)$

Then $g(\alpha)=0$.Also $g(\bar \alpha)=0\implies g(x)=(x-\alpha)^a(x-\bar \alpha)^b h(x);h(x)\in \Bbb C[x]$ .

What to do now?Please help.

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    No, it's not correct. How do you know $g(x) \in \Bbb Q[x]$? – 2017-01-18
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    Note -- you never used the given info that $f(\alpha)=f(\alpha+1)=0$. – 2017-01-18
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    Let $g(x) = f(x+1) - f(x)$. Note that g cannot be the zero polynomial (why?). The polynomial $g$ has what obvious root? If $\text{deg}(f) = n$, what can you say about $\text{deg}(g)$? – 2017-01-18
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    Note that $\alpha$ is specified as complex, but that doesn't mean it's non-real, so you can't claim that $\alpha$ and $\bar \alpha$ are distinct roots of $f$. – 2017-01-18

3 Answers 3

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This is a conceptual proof: if $f(a)=f(a+1)=0$, then $a$ is a root of both $f(x)$ and $f(x+1)$. Both polynomials are irreducible, hence minimal polynomials of $a$. By uniqueness, we obtain $f(x)=f(x+1)$ as polynomials, hence by plugging in $x=a+1$ we get $0=f(a+1)=f(a+2)$. Continue with $a+2,a+3, \dotsc$ and you will obtain infinitely many roots. Contradiction!

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    Excellent ,just love it – 2017-01-19
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    Maybe one should note that this proof immediately generalizes to the fact that the difference $d$ of two roots of an irreducible rational polynomial is never a rational number.. Just replace any $1$ by $d$ and you obtain infinitely many roots $a,a+d,a+2d,...$. Of course this statement fails in positive characteristic as the Artin-Schreier polynomial shows. – 2017-01-19
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Suppose $f$ is irreducible in $\mathbb{Q}[x]$ and $\alpha$ is a complex number such that $f(\alpha)=f(\alpha+1)=0$.

The goal is to derive a contradiction.

Let $n = \text{deg}(f).\;\,$Since f is irreducible in $\mathbb{Q}[x]$ and $\alpha$ is a root of $f$, there is no nonzero polynomial in $\mathbb{Q}[x]$ of degree less than $n$ for which $\alpha$ is a root.

Let $g(x) = f(x + 1) - f(x)$.

If $g$ is the zero polynomial, then for every root $r$ of $f$, $r+1$ would also be a root, hence $f$ would have infinitely many roots, which is silly -- $f$ is a nonzero polynomial so can't have infinitely many roots.

Note that $f(x+1)$ has the same leading coefficient as $f(x)$, hence $\text{deg}(g) < n$.

Since $f(\alpha)=f(\alpha+1)=0$, $\alpha$ is a root of $g$.

But then $g$ is a nonzero polynomial in $\mathbb{Q}[x]$ of degree less than $n$, having $\alpha$ as a root, contradiction.

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The two polynomials $f(x)$ and $f(x+1)$ have rational coefficients, so their GCD is a polynomial $h(x)$ that also has rational coefficients. Since $f$ is irreducible and $h$ divides $f$, $h$ must either be a nonzero constant or $h=f$. From the hypotheses we get $h(\alpha)=0$, so $h$ cannot be a nonzero constant. So $h=f$. Then $f(x)$ divides $f(x+1)$ ; since those polynomials have the same degree, there is a constant $c$ with $f(x+1)=cf(x)$. Looking at the highest degree monomials, we get $c=1$. So $f(x+1)=f(x)$ which is impossible.