Existence of $\alpha\in \Bbb C$ such that $f(\alpha)=f(\alpha+1)=0$. 
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 $g(\bar \alpha)=0\implies g(x)=(x-\alpha)^a(x-\bar \alpha)^b h(x);h(x)\in \Bbb C[x]$
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What to do now?Please help.

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!
