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"Let $a(x), b(x) \in \mathbb{R}[x]$, not both the zero polynomial and suppose that gcd[$a(x), b(x)$] = 1. Let $u(x), v(x) \in \mathbb{R}[x]$ be such that

$a(x)u(x) + b(x)v(x) = 1$

Let also $s(x)t(x) \in \mathbb{R}[x]$ be such that

$a(x)s(x) + b(x)t(x) = 1$

Prove that $a(x)$ divides $(v(x) = t(x))$"

My attempt so far:

$a(x)u(x) + b(x)v(x) = a(x)s(x) + b(x) t(x)$

$\iff v(x) - t(x) = a(x)\frac{s(x)-u(x)}{b(x)}$

Hence to show that $a(x)$ divides $(v(x) = t(x))$ we need to show that $\frac{s(x)-u(x)}{b(x)} \in \mathbb{Z}$. That is, that $b(x) $ divides $(s(x)-u(x))$

I'm not sure how to show this exactly. I'm guessing I will need to use the property that if c | ab, with a and c being relatively prime, then c |b. But I can't seem to figure out how to make it work.

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Your guess about the right strategy is correct. Dividing is not useful, it is better to leave things "flat" and write $$a(s-u)=b(v-t).$$ So $a$ divides $b(v-t)$. Since $\gcd(a,b)=1$, by the analogue of Euclid's Lemma it follows that $a$ divides $v-t$.

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