existence of negative root

Question

I look if there exists a non-zero polynomial $p(x)= a_0 + a_1x + a_2x^2 + \cdots a_nx^n$ with positive integers coefficients : $\forall i, a_i\in \mathbb N $ such that $p(-\sqrt 2)=0$

Answer 1

The minimal polynomial for $-\sqrt{2}$ over $\mathbb{Z}$ is $x^2-2$. Thus any such polynomial having $-\sqrt{2}$ as a root must be divisible by $x^2-2$. Using the notation from your question, write $p(x)=a_0+a_1x+\ldots+a_nx^n$ where each $a_i\in\mathbb{N}$. Supposing $p(-\sqrt{2})=0$, we have $$ p(x)=q(x)(x^2-2) $$ for some $q(x)\in\mathbb{Z}[x]$. Write $q(x)=b_0+b_1x+\ldots+b_{n-2}x^{n-2}$ where each $b_i\in\mathbb{Z}$. Expanding the product, we have $$ p(x)=(b_0x^2+b_1x^3+\ldots+b_{n-2}x^n)-2(b_0+b_1x+\ldots+b_{n-2}x^{n-2}). $$ By our assumption, all coefficients of $p(x)$ are positive, so the coefficient on any $x^j$ for $2\leq j\leq n-2$ is $b_{j-2}-2b_j$ and satisfies $b_{j-2}-2b_j>0$. This means $b_{j-2}>2b_j$ for each $2\leq j\leq n-2$. The coefficient on $x^n$ is $b_{n-2}$, so $b_{n-2}>0$. This forces every coefficient of $q(x)$ to be positive, and in particular, $b_0>0$. But the constant term of $p(x)$ is $-2b_0$, contradicting our assumption that $p(x)$ had all positive coefficients.