# What can be said about the prime decomposition of the Bezout coefficients $\beta(a,b)$?

Let $$a, b$$ be coprime rational integers. Then by Bezout's lemma we can find $$(s,t) := \beta(a,b) \in \mathbb{Z}^2$$ such that $$a*s + b*t = 1$$. My question concerns the prime factorization of the second Bezout coefficient $$t$$.

Now suppose that only $$a\in\mathbb{Z}$$ is given and our task is to choose $$b\in\mathbb{Z}$$ with $$(a,b)=1$$ such that $$t$$ is easy to factorize, eg. is n-smooth prime, for some $$n \in O(\log a)$$.

With this much freedom would it be possible to ensure that $$t$$ can be factorized easily? In general, can anything be said about the decomposition of Bezout coefficients?

edit I considered $$a s + b t = 1$$. If $$b << a$$, then we can expect $$s = 1$$, and so $$t$$ can be written as $$t = \frac{1-a}{b}$$. However, I don't know anything about the factorization of $$a$$ or $$1-a$$...

edit2: in response to FredH's comment: I'd like to find b with the additional property that $$b << a$$.

• As written, this is trivial. Choose $b = 1-a$; then $t = s = 1$. Is there some other condition you had in mind? – FredH Mar 2 at 18:46
• @FredH, yes thank you for pointing this out. I did consider that case, and in fact I'd like to have $b << a$, I just didn't want to make the question too convoluted. I see this is suboptimal, will edit it in a minute. – gen Mar 2 at 18:48