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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

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closed as too broad by BlueRaja - Danny Pflughoeft, Arnaud D., Cesareo, Mees de Vries, Carl Mummert Jan 11 at 12:33

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not. $\endgroup$ – Blue Jan 11 at 5:21
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    $\begingroup$ That's a good metaphor! $\endgroup$ – YuiTo Cheng Jan 11 at 5:23
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    $\begingroup$ The reason nobody has created such a list may be the fact that most (nontrivial) "iff" theorems belong on your list. It would be more interesting to see a list of "iff" theorems where both directions are nontrivial. $\endgroup$ – bof Jan 11 at 5:39
  • $\begingroup$ @bof done, see math.stackexchange.com/questions/3069590/… $\endgroup$ – YuiTo Cheng Jan 11 at 7:57
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    $\begingroup$ You cannot make such a question more specific. You are asking for a list of things of a certain nature, where that set is known to be enormous. Any answer is either just one among many with no way to determine a clear "best" or requires thousands of pages to sufficiently answer in one go. This is textbook classic VTC:TB. $\endgroup$ – Nij Jan 11 at 9:57
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"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.

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Let $n$ be a positive integer. The equation $x^n+y^n=z^n$ is solvable in positive integers $x,y,z$ iff $n\le2$.

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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.

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All planar graphs are $n$-colorable iff $n\ge4$.

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Integers $a * b = 944871836856449473$ and $b > a > 1$

iff

$a = 961748941$ and $b = 982451653$

If is trivial multiplication. Only if requires large prime factorization.

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    $\begingroup$ The hard direction only requires primality testing, not factorization. $\endgroup$ – bof Jan 11 at 9:10
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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 $K_5$.

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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.

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    $\begingroup$ Assuming they're all non-empty. $\endgroup$ – Arnaud D. Jan 11 at 9:14
  • $\begingroup$ Actually, the projections always being surjective is also equivalent to the axiom of choice. $\endgroup$ – Marc Paul Jan 11 at 12:30
  • $\begingroup$ @MarcPaul Would you mind explain a bit on why that is the case? I thought the projection of a product set onto its component looks pretty constructive to me. $\endgroup$ – BigbearZzz Jan 14 at 22:00
  • $\begingroup$ @BgbearZzz Axiom of choice is equivalent to the statement that a product of non-empty sets is non-empty. So if AoC fails, the image of the projection map can be empty. And indeed, equipping each set with the discrete topology, and making sure that at least one of the sets is infinite, you obtain a contradiction to the 'easy' direction of the statement. $\endgroup$ – Marc Paul Jan 14 at 22:59
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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.)

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    $\begingroup$ There may be chances that Goldbach's conjecture is wrong... $\endgroup$ – YuiTo Cheng Jan 11 at 9:20
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$\mathbb R^n$ has the structure of a real division algebra iff $n=1,2,4$ or $8$.

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