# If $\alpha$ is algebraic number then so is $\alpha+1$

I have to prove that if $$a$$ is algebraic number then so is $$\alpha+1$$. I've tried to construct the polynomial for $$\alpha+1$$ using the polynomial for $$\alpha$$ but that didn't lead me to anything. Let the $$W(x)=a_n x^n+a_{n-1} x^{n-1} +\dots + a_1x+a_0$$ be a polynomial and $$W(\alpha)=0$$. Consider $$W(\alpha+1)=a_n (\alpha+1)^n+a_{n-1} (\alpha+1)^{n-1} +\dots + a_1(\alpha+1)+a_0$$ $$W(\alpha+1)=a_n(\alpha^n+{n\choose 1}a^{n-1}+\dots+{n\choose n-1}\alpha +1)+\\a_{n-1}(\alpha^{n-1}+{n-1\choose 1}a^{n-2}+\dots+{n-1\choose n-2}\alpha +1)+\dots +\\a_2\alpha+a_1+a_0$$ Then taking first element of every bracket I can obtain $$W(\alpha)$$ which is $$0$$. Now I'm left with the rest of the stuff and don't know where to go from there. Is this strategy any good? If not - what would be a proper way to prove this. If I'm doing it correctly -what's next?

• If $W(x)$ is a polynomial with root $\alpha$ then $W(x-1)$ is a polynomial with root $\alpha+1$. Jun 17, 2020 at 21:27

If you are to substitute $$x \mapsto x - 1$$ in your characteristic polynomial $$W(x)$$ then you obtain a polynomial $$W'(x) = a_n (x-1)^n + \dots a_1 (x -1) + a_0.$$ It should be clear that $$W'(\alpha + 1) = W(\alpha) = 0$$. If you want to find the coefficients $$a'_k$$ of $$W'(x)$$ you will have to use binomial expansion similar to the way you have already done to calculate them.