Given that $p(x)$ is a polynomial of the form $x^n + a_{n-1}x^{n-1}+ ... + a_1 x + a_0$, where $a_1, a_2, ... , a_n$ are integers

Question

I am given that $p(x)$ is a polynomial of the form $x^n + a_{n-1}x^{n-1}+ ... + a_1 x + a_0$, where $a_1, a_2, ... , a_n$ are integers.

I am also given that $x = \frac{a}{b}$, $\gcd(a,b) = 1$ (thus $\frac{a}{b}$ are in their lowest terms), $p(\frac{a}{b}) = 0$, and $|b| > 1$

Use the factorization of $b$ to contradict the assumption that $a$ and $b$ are coprime.

We have been using the Bezout theorem quite a bit in class so I first started off with that then took it to the power of $n$, so $(ax + by)^n$, then attempted to use the binomial theorem in an attempt to group the $b^i$ terms together, but after that I am unsure how to proceed in proving that $a$ and $b$ are coprime.

Any hint or help is appreciated. Thank you.

Answer 1

The Bezout theorem is not needed (directly) here.

Hint: If $p\left(\frac ab\right)=0$ then $b^np\left(\frac ab\right)=0.$ But $$b^np\left(\frac ab\right)=a^n + a_{n-1}a^{n-1}b+a_{n-2}a^{n-2}b^2+\cdots+a_0b^{n}.$$

so $a^n$ is divisible by $b.$ Show this contradicts $\gcd(a,b)=1$ and $|b|>1.$

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I suppose you can use Bezout to prove that $\gcd(a^n,b)=1$ if $\gcd(a,b)=1.$ Specifically, $$1=(ax+by)^n = a^nX+bY$$ where $X=x^n$ and $Y=\sum_{k=1}^{n}\binom{n}{k}b^{k-1}x^k(ay)^{n-k}$