@CalvinKhor I'm asking this out of confusion - why are the usually seen proofs of Bezout's identity aka the theorem regarding a linear combination of two numbers returning their GCD try and show that the least positive value of that the linear combination can attain is the GCD of the two numbers?? Is it out of random testing that such an observation was derived and later put inot use for proving the theorem?

I am asking so since I was a bit confused while trying to prove the theorem myself. Had it been something like the division algorithm, I could understood the logic very much; this one in the first step is giving me trouble.

@Spectre i found that surprising at first, and I don't have any good explanation other than random testing to build intuition could have lead to it, yes

mgiht want to check the main chat, more people who work nearer to number theory are there :)

@CalvinKhor Yeah thanks... but I doubt if they'll understand my level and my flow of intelligence... only last week had I nearly dug up the grave for a lil bit of my reputation to fall into by attempting to answer a question... I'll just share the link of it (my answer had been deleted)

I think that the question was kicked out... so perhaps I must tell it to you myself

So it was a question that gathered a lot of downvotes

Something like a long procedure was written up by the OP as an attempt to infiltrate into RSA codes

I hadn't been able to reap rep for a long while so I decided I dig right into it and hit the problem down, be it an illegal reason or not...

@CalvinKhor I think the intuition there is to use well-ordering principle for integers.

@Koro I understood using that should help one round in upon an element in the set $S = \lbrace ax + by : ax + by > 0, x, y \in \mathbb{Z} \rbrace$

But that isn't what's giving me the trouble

To connect the chain of logic is what's giving me a lil bit of confusion

(Consider a number $N=pq,p,q∈\mathbb{P}$ ($\mathbb{P}$ being the set of primes) - now define $H$ as the closest prime to $N$ that is greater than $N$, $L$ defined the same way but less than $N$.

Now define $Z = LH$

The define another number $x = Z - n$

Now we see that $x$ will have powers of $2$ in the prime factorisation (I had forgotten to mention that $p$ and $q$ are both odd)

So let $k = \frac{x}{2^{v_2(x)}}$

It is seen that $k$ will have only two prime factors

What the OP needed to know was the reason for this to happen (and something else as well, I had forgotten it)

I put up an answer that at first talked about correcting the question to make the procedure work properly, since when $2$ is taken for $p$ or $q$, contradictions can arise

The second part was where I was a but flawed; I tried to write a proof as if to explain the phenomenon but in vain, so I left it as a partial answer

And I saw that the backlashes might begin once I saw the single downvote, so I deleted my answer and left the problem for good 😂

@Koro do you have any idea about this question?

At first I felt something was off with the question since there are a lot of unknowns, the solution (if at all there was one) would be a lot exhaustive, or if it never existed, still lives in a mathematician's library as an unsolved problem

Byee then

@Spectre I'm afraid, I don't know which question you're referring to. I thought the question was about about proving Bezout's identity while responding to Mr. Calvin Khor.