This may have been asked already but I am lacking keywords probably.
I am looking for examples of proven equivalences $A \Leftrightarrow B$ where one direction, say $A \Rightarrow B$, is very difficult compared to the other direction.
To avoid trivial examples, I am requiring that neither $A$ and $B$ are definitions (or easy restatements of definitions).
The example can be research level, in particular in number theory, algebra and functional analysis.
I think I have an example. Suppose $a,b,c \in \mathbb N, gcd(a,b)=1, a^2+b^2=c^2\iff \exists m,n \in \mathbb N, a=2mn, b=m^2-n^2, c=m^2+n^2$
One direction is easy.
$a^2+b^2=c^2 \implies (2mn)^2 +(m^2-n^2)^2=m^4-2m^2n^2+n^4+4m^2n^2=m^4+2m^2n^2+n^4=(m^2+n^2)^2$. So given $m$ and $n$ ,we necessarily have a primitive Pythagorean tripe.
How do you get your $m$ and $n$ given an arbitrary primitive triple?
$gcd(a,b)=1 \implies$ they aren't both even. They can't both be odd since the sum of the square of two odd numbers is never a perfect square.
WLOG $a$ is even, $b$ is odd $\implies c$ is odd.
$a^2=c^2-b^2=(c-b)(c+b)$
Now $g(b,c)=1$ for otherwise $gcd(a,b)>1$ which is not allowed. $(c-b)$ and $(c+b)$ are both even, but their halfs are relatively prime, i.e. gcd ([c-b]/2, [c+b]/2)=1. else we would have $gdcd(b,c)>1$
Sincd LHS is perfect square, so is RHS. This suggests $c-b=2n^2$ and $c+b=2m^2$ for some positive integers $m$ and $n$.
So $a^2=4m^2n^2\implies a=2mn, b=(2m^2-2n^2)/2=m^2-n^2$. By similar arguments $c=m^2+n^2$.