Theorems in the form of "if and only if" such that the proof of one direction is extremely EASY to prove and the other one is extremely HARD I believe this is a common phenomenon in mathematics, but surprisingly, no such list has been created on this site. I don't know if it's of value, just out of curiosity, I want to see more examples. So my request is: 
Theorems in the form of "if and only if" that the proof of one direction is extremely EASY, or intuitive, or make use of some standard techniques (e.g. diagram chasing), while the other one is extremely HARD, or counterintuitive, or require a certain amount of creativity. The 'if and only if' formulation should be as natural as possible.
Thanks in advance.
Edit: I know this question is somewhat ill-posed. A better one: Theorems in the form of “if and only if” such that the proofs of BOTH directions are nontrivial
 A: All planar graphs are $n$-colorable iff $n\ge4$.
A: Integers $a * b = 944871836856449473$ and $b > a > 1$
iff
$a = 961748941$ and $b = 982451653$
If is trivial multiplication. Only if requires large prime factorization.
A: The difficult part is not as difficult as some of the other answers, but perhaps the "if and only if" formulation is more natural: Kuratowski's theorem that a graph is planar if and only if it does not have a subgraph which is a subdivision of either $K_{3,3}$ or $K_5$.
A: The product of topological spaces is compact if and only if all of them are compact. One direction is trivial because the projections are continuous and surjective functions. The another direction, well, is the axiom of choice.
A: An even integer $n$ is the sum of two primes iff $n>2$.
("If" is Goldbach's conjecture, which is still open. "Only if" is trivial.)
A: $\mathbb R^n$ has the structure of a real division algebra iff $n=1,2,4$ or $8$.
A: Let $n$ be a positive integer. The equation $x^n+y^n=z^n$ is solvable in positive integers $x,y,z$ iff $n\le2$.
A: "A compact 3-manifold is simply-connected if and only if it is homeomorphic to the 3-sphere."
One direction [only if] is the Poincaré conjecture.  The other direction shouldn't be too bad hopefully.
A: The bisectors of two angles of a triangle are of equal length if and only if the two bisected angles are equal.  If the two angles are equal, the triangle is isosceles and the proof is very easy.  However proving that a triangle must be isosceles if the bisectors of two of its angles are of equal length seems to be quite difficult.
