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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 $b << a$.

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  • $\begingroup$ As written, this is trivial. Choose $b = 1-a$; then $t = s = 1$. Is there some other condition you had in mind? $\endgroup$ – FredH Mar 2 at 18:46
  • $\begingroup$ @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. $\endgroup$ – gen Mar 2 at 18:48
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Knowing a and b are coprime, two cases appear:

  • a,b both odd
  • a,b opposite parity

In the first case, if s and t are both odd, then we have odd+odd=odd which is false. So if s is even, t is odd. If s is odd, t is even.

The second case, has s and t of same parity. If s is even, then t will also be even. If s is odd, t will be odd as well.

You can continue this, mod as many primes as you want.

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