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$$c_1,c_2 \text{ are polynomial's }g(x)=x^2+ax+b \text{ roots } \Leftrightarrow \begin{cases} g(c_1)=c_1^2+ac_1+b=0 \\ g(c_2)=c_2^2+ac_2+b=0 \end{cases}$$ Prove that for every polynomial with integer coefficients $f(c_1)+f(c_2)=k$, so that $k\in \mathbb{Z}$.

Should I approach this problem with Bezout's theorem or maybe simply by factoring $g(x)$?

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You need to specify $a,b \in \mathbb Z$ and your final $a$ is different from the first one - I think you mean $f(c_1)+f(c_2)\in \mathbb Z$ –  Mark Bennet May 5 at 17:35
    
You are correct. I edited the first post, that notation issue is fixed now. –  vilbur May 5 at 19:54
    
I've updated my answer to be a little more specific. –  Mark Bennet May 5 at 20:10

2 Answers 2

up vote 3 down vote accepted

I suppose $a$ and $b$ are integers.

By euclidian division, every polynom $f$ can be written $f=pg+q$, where $p$ and $q$ are polynomials with integer coefficients (integers and not rationals because $g$ is unitary), and $\deg q \leq 1$. That is, $q(x)=ux+v$.

Hence

$$f(c_1)+f(c_2)=p(c_1)g(c_1)+q(c_1)+p(c_2)g(c_2)+q(c_2)=q(c_1)+q(c_2)$$

Or

$$f(c_1)+f(c_2)=u(c_1+c_2)+2v$$

But $c_1+c_2=-a$ is an integer, and so are $u$ and $v$.

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If deg$(f)\geqslant 3$, doesn't it mean that deg$(q)$ can be larger than $1$? –  vilbur May 6 at 16:36
    
No. But $\deg p$ may be large. It's like euclidian division on integers, when $a=bq+r$, and $0 \leq r < b$. See polynomial long division. –  Jean-Claude Arbaut May 6 at 17:35

I would show that $c_1^r+c_2^r$ is an integer for all $r$ using a recurrence based on $g(x)$

Let $k_r=c_1^r+c_2^r$ then we have from $g(x)=x^2+ax+b=0$, the roots of $g$ also satisfy the equation $$h_r(x)=x^{r-2}g(x)= x^r+ax^{r-1}+bx^{r-2}=0$$

Now $$h_r(c_1)+h_r(c_2)=k_r+ak_{r-1}+bk_{r-2}=0$$ and you can work out the rest. The notation is somewhat clumsy, but you should be able to follow what is going on.

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