After Fermat's last theorem, what is the next most basic sounding proof that has never been found?

But you have no proof, how do you know its basic? You could just as legitimately call it the most complicated sounding proof that has never been found

The way OP phrased his post sucks. The guy who posted Goldbach's conjecture read that as "the most basic theorem/conjecture which has never been proved" and I have to agree with his choice.

"Basic sounding" you fucking retard, nowhere is it necessitated that it actually be basic in practice.

Now hop back in your fucking clown car and roll back home where you can get back to sucking on your mother's titties.

You've got a long way to go in life if you think a 'basic sounding proof' is not something done in practice. How are you misinterpreting this? I'd recommend going through some proof-heavy book, I've never used it but sci likes to recommend 'How to Prove It' by Velleman

That's already a bit advanced for Veeky Forums, but the Jacobian conjecture only requires some calculus to be formulated but is notoriously difficult to prove.

>but the Jacobian conjecture only requires some calculus to be formulated but is notoriously difficult to prove.
What I took from this is that mathematicians are too lazy to solve it.

Collatz conjecture

Yeah probably this.
You score incredibly high on the spectrum dude if you really do interpret things this literally.

Yes, you could infer this by his mentioning of Fermat's Last Theorem which is a very simply posed but incredibly difficult to answer question.

cause you didn't {18}

The statement is basic, not the proof. The fact that the thread is about this criterion should've been clear from the OP, since Wiles' proof is not basic by any means.

>sustained world peace only requires basic English to describe

> this means it hasn't been achieved because people are too lazy

Wew

Basic theorem of the basics

Is it possible to make cuts from one end of a pizza to another such that every resulting piece has the same amount of pizza but no two pieces are the same shape?

Four Colour Theorem.

*ducks*

I know, I know, it's been proved. I've read enough to convince myself it's true. But I want another, better, proof, one that helps me understand more about why and more about other mathematics that informs it.

For one cut, obviously not.
My first thought was [pic related] for two cuts, but does the second cut count as not being "from one end to the other"?
Hmm.

>My first thought was [pic related] for two cuts, but does the second cut count as not being "from one end to the other"?
No, of course not. That would be too easy.

"Every Jordan curve (that is, a curve that is closed without self-intersecting points) divides the plane in two: the interior and the exterior of the curve."

What it says is so damn trivial that you wouldn't expect such machinery to prove it.

This is going to bother me for a long time. My desk is now covered in pizza scribbles. Good problem, user.

Here's the simplest possible candidate that breaks symmetry. It almost worked except for the little intersection in the top left.

collatz conjecture
pretty much any unsolved diophantine equation

Woodin's Ω-conjecture.

Collatz conjecture.

Sofa problem.

Poor choice of words but interesting question. For "next most basic sounding proof that has never been found", let's instead read, "open problem or conjecture which is simplest for anyone to understand, yet has never been proven."

I guess Goldbach is simpler, but I personally (also) nominate the question (or equivalent conjecture: "a perfect cuboid exists") "Does a perfect cuboid exist?" This goes to another user's point that some open problems with the simplest statements, yet remain stubbornly open, involve diophantine equations - just as FLT did.

A perfect cuboid is an alleged mathematical object of which no example has been found, nor has it been shown to be an impossible construction. Simply put, a perfect cuboid is an ordinary right rectangular prism, right rectangular parallelpiped, cuboid, or (getting technical now) a "box" with the following properties:

All three edge lengths,
All three of the face diagonals, and
THE Spatial Diagonal --- that is, all SEVEN of these lengths is a nonzero natural number.

The perfect cuboid is a stronger case of the Euler brick, which only entails the first two of those three lines, or six equations not involving the spatial diagonal. Infinitely many Euler bricks exist.

The above three-line statement can be rephrased as a system of four diophantine equations with seven unknowns a-g as I like to call them, in familiar terms of the pythagorean theorem. No solution element (a,b,c,d,e,f,g) has been found, nor has such an element been shown not to exist.

Some months ago, I did a little basic research on this problem to amuse myself, and I came up with and easily proved a few useful lemmas (or being pretentious, "lemmata") concerning the problem of the perfect cuboid. I even bought a very simple, short book about Pythogorean triples by Sierpinski because it relates directly to my inquiry.

cont.

I mis-wrote earlier when I said that the euler brick entails six EQUATIONS. Instead, it entails six UNKNOWNS, in THREE equations. The situation of the perfect cuboid entails one MORE UNKOWN, and a FOURTH equation.

As for the toolkit of lemmata I worked up, I came up with the following. Preliminary triviality: every alleged perfect cuboid is an Euler brick by definition, and thus every property that an Euler brick has is shared by any possible candidate of a perfect cuboid. Thus, statements about Euler bricks themselves tell us about these possible (or impossible?) perfect cuboids. Thus:

1) In an Euler brick, no two edge lengths are equal. In other words, there is always a "short" one (a), a "middle" one (b), and a "long" one (c).
2) There is no such thing as an Euler brick which has an edge length which is equal to one (this is a useful observation to establish certain strict inequalities, as opposed to "or equal to" inequalities, concerning the 6 or 7 unknowns a-g).
3) In an Euler brick, no two of the face diagonals are equal. Moreover, they "correspond" to the un-equal edge lengths in the expected ways. (pf: use 1 and 2 and manipulate both equations and inequalities). They are thus (d,e,f), from shortest to longest.
4) In spite of the above, it can sometimes happen that an Euler brick's longest edge length c is longer its shortest face diagonal d, /and vice verse/. (pf. inspect the long-known Euler bricks with edge lengths (240,252,275) vs. (44,117,240) as diverse examples of each).
5) "a-g^2 satisfy a chart of inequalities". pf: a long, ordered logical train of thought in a 14x14 table. I have about 80% filled in but I got stuck at certain regions which are directly related to the c>d or d>c possibility-knowledge. So I've left this incomplete for some months, haven't drilled back in.
6) I later began to think of an Euler Brick as having finite (48) representations in R^3. This led me via geometry to conjecture that c-b =/= b-a (unproven).

post more yuka chan

I love you, Euler brick user. Keep on trucking.

Pic related is where I left off with my train of thought on one item, some months ago. Notice especially the red cells which say that such-and-such are inconsistent counter-examples (the c-d thing), while the blank bits are the bits I got stuck on and haven't taken back up.

Here, a,b and c are the shortest, middle and longest of the edge lengths, d, e and f are the shortest, middle and longest face diagonals, and g is the spatial diagonal, while their squares are obviously their squares. The table proposes to examine necessary inequalities as they related to euler bricks or perfect cuboids. The numbered items entail a linear, logical train of thought which is not terribly different from sudoku (although this context is clearly different): the idea was to logically block-out such-and-such truths by exactly the lemmas I've just stated, and other logical operations.

The c-d thing is terribly interesting to the problem from a surface level, and I wonder if it has been discussed at length in any location with reference to the question of a perfect cuboid.

Furthermore, my investigation led me, for some reason whose exact nature currently escapes me as I am quite tired, to consider pythagorean triples (triangles) whose legs have a difference of one, and so are "almost-isosceles". This is taken up usefully by Nyblom in a great little note on number theory which is fun to work through:

citeseerx.ist.psu.edu/viewdoc/download;jsessionid=0128D38E20C21AD95E6D91239BAE40E8?doi=10.1.1.376.2966&rep=rep1&type=pdf

As a matter of fact, Nyblom ends up saying much the same thing that is said in Sierpinski's tiny Pythagorean Triples (a cheap little book!): I cannot state it just-so for the above reason, but the point is that both authors identify a relationship (1-1 correspondence?) between these so-called "AIRA" triangles (3,4,5), (20,21,29), etc, and the so-called /square triangular numbers/, that is numbers that are simultaneously both square and triangular.

whoops, here's the pic.