# 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 [closed]

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.

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

• One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not.
– Blue
Jan 11 '19 at 5:21
• That's a good metaphor! Jan 11 '19 at 5:23
• 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.
– bof
Jan 11 '19 at 5:39
• @bof done, see math.stackexchange.com/questions/3069590/… Jan 11 '19 at 7:57
• 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.
– Nij
Jan 11 '19 at 9:57

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 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.

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.

• Jan 11 '19 at 9:08

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

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

iff

$$a = 961748941$$ and $$b = 982451653$$

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

• The hard direction only requires primality testing, not factorization.
– bof
Jan 11 '19 at 9:10

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 $$K_5$$.

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.

• Assuming they're all non-empty. Jan 11 '19 at 9:14
• Actually, the projections always being surjective is also equivalent to the axiom of choice. Jan 11 '19 at 12:30
• @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. Jan 14 '19 at 22:00
• @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. Jan 14 '19 at 22:59

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.)

• There may be chances that Goldbach's conjecture is wrong... Jan 11 '19 at 9:20

$$\mathbb R^n$$ has the structure of a real division algebra iff $$n=1,2,4$$ or $$8$$.