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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.

Crystal McMillian
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1 Answers1

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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 $|b|>1.$

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 $Y=\sum_{k=1}^{n}\binom{n}{k}b^{k-1}x^k(ay)^{n-k}$

Thomas Andrews
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  • Ah, then I can say $b|a^n$, this implies $b|a$, meaning $gcd(a,b) = b$ but since $abs(b) > 1$, it contradicts $gcd(a,b) = 1$. – Crystal McMillian Sep 26 '19 at 17:58
  • No, $b\mid a^n$ does not imply that $b\mid a.$ @CrystalMcMillian That would be true if we knew $b$ was prime. – Thomas Andrews Sep 26 '19 at 18:10
  • Then $b|a^n$ implies $a^n = qb$ for some integer $q$, then $gcd(a, \frac{a^n}{q}) = 1$, umm.. – Crystal McMillian Sep 26 '19 at 18:26
  • $b \mid a^n$ does tell you something useful though, because $b$ has a prime factorization. – John Hughes Sep 26 '19 at 18:50
  • @JohnHughes Since $b|a^n$, and since both $b$ and $a^n$ are integers that can be written via prime factorization, say $a^n = c_1^{nk1} c_2^{nk2} ... * c_r^{nkr}$, meaning $a = c_1^{k1} c_2^{k2} * ... * c_r^{kr}$, and $b = d_1^{l1} * d_2^{l2} * ... * d_s^{ls}$, $k,r,l,s$ are integers with $s \leq r$, some of the prime factors in $b$, aka $d_i$, must also exist in $a$ and overlap with $c_i$. Thus $gcd(a, b) \ge d_1 * d_2 * ... * d_s$. But since $abs(b) > 1$, this implies $gcd(a, b) > 1$, which is a contradiction. I believe this solves it? – Crystal McMillian Sep 26 '19 at 22:18