Intuition Of Bézout's Theorem

Question

I've been wondering for a while now for the intuition of Bézout's theorem.

$ax+by=\gcd(a,b)=d$

$dpx+dqy=d(px+qy)$

Now why there exists pairs of $x$ and $y$ such that $px+qy$ is $1$ ($p$ and $q$ are co-primes)

Answer 1

Depending on what you’re looking for as “intuition”:

To use notions reminiscent of linear algebra, one can think of coprime integers $p$ and $q$ as “independent”, or “orthogonal”, and Bézout’s identity tells us that “independent” integers span the space of integers: $p(nx)+q(ny)=n$ for any integer $n$.

If $p$ and $q$ have a common factor, then they can be considered as “dependent”, or “colinear”, and can thus only span a “subspace” of the integers, i.e., integers that are multiples of $\text{gcd}(p,q)$.

Answer 2

Your question is a little unclear. To me, it certainly isn't obvious at first glance why $px+qy=1$ must have integer solutions $x,y$ when $p,q$ are coprime. However, I can explain why this is a natural conjecture to make.

If $p$ and $q$ are not coprime, then $p,q$ have some common divisor greater than $1$, meaning that $px+qy$ must also be divisible by that divisor whenever $x,y$ are integers. Hence, $px+qy$ cannot be equal to $1$ in this case. Since we've just found a necessary condition for $px+qy=1$ to have integer solutions - namely, that $p,q$ must be coprime - it's natural to ask whether this is also a sufficient condition, that is, whether $p,q$ being coprime guarantees the existence of a solution. And the answer, conveniently enough, turns out to be "yes"!