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65

Problem: Does every triangular billiard have a periodic orbit? For acute triangles, the question has been answered affirmatively by Fagnano in 1775: one can simply take the length $3$ orbit joining the basepoints of the heights of the triangle. For (generic) obtuse triangles, the answer is not known in spite of very considerable efforts of many ...


25

First of all, the Hodge conjecture is not about one particular differential form. It says that any differential form which satisfies certain conditions will be a $\mathbb{Q}$-linear combination of algebraic cycle forms. There are some particular forms which satisfy those conditions but have not been proven to be such a linear combination.1 However, it is my ...


24

This is an inadequate answer, but... These problems are difficult because (a) the space of possible solutions is vast, and (b) there seems insufficient structure to reduce the space so that searching it becomes feasible. Consider the 11-squares problem. Each square has a location and an orientation, and so can be pinned down by specifying three parameters. ...


22

This is an incredibly difficult problem. It is one of Landau's 4 problems which were presented at the 1912 international congress of mathematicians, all of which remains unsolved today nearly 100 years later.


20

In 4 dimensions, it is an open question as to whether there are any exotic smooth structures on the 4-sphere.


17

A more or less elementary example I'm quite fond of is the Erdős conjecture on arithmetic progressions, which asserts the following: If for some set $S\subseteq \mathbb{N}$ the sum $$\sum_{s\in S}\frac{1}s$$ diverges, then $S$ contains arbitrarily long arithmetic progressions. I've never seen a heuristic argument one way or the other - I believe ...


17

Try Mazur's Questions about Number.


17

Bhargava and Shankar have proved results about average $3$-Selmer ranks of elliptic curves. (See this arxiv preprint, which is the same paper cited by the Wikipedia article linked in the OP.) Their argument is via geometry of numbers (so to speak). In fact, they are able to construct families such that exactly half of them have positive sign in their ...


17

"My current understanding is that the field of one element is the most popular approach to RH." Analytic number theory, with ideas from algebraic geometry, random matrix theory, and any other areas that might be relevant, is the only approach known to have produced any concrete results toward RH. The random matrix theory in particular has produced a ...


17

Here is an argument that Tate is harder than Hodge: We know the Hodge conjecture in the codimension one case (this is the Lefschetz $(1,1)$ Theorem). On the other hand, the Tate conjecture remains open even in codimension one except in some very special cases. Also, those special cases have often been proved by reducing to the Hodge conjecture. Here ...


15

I believe whether or not the Thompson group $F$ is amenable is such question. The paper/article "WHAT is... Thompson's Group" mentions that at a conference devoted to the group there was a poll in which 12 said it was and 12 said it was not. There are in fact papers claiming (at least at the time) to have proofs for both sides. Here are some posts to get an ...


14

The case of factor $3$ is more interesting because with factor $1$ it's easy to prove that the sequence hits $1$. We can prove this by induction: For $n_0=1,2$ this is clearly true. Suppose that it is true for all starting values $1,2,\dots,n_0-1$. Then if $n_0$ is even, the next number will be $n_0 / 2 < n_0$ and thus the sequence will hit $1$. If $n_0$ ...


13

Nobody knows. I suppose if there is a polynomial-time algorithm for 3-SAT (or some other NP-complete problem) then a computer could find it and prove P = NP. And if there is a proof that P isn't NP, well, I suppose a computer could find that, too. Why - are you looking for something to work on this summer?


13

For $n=12$ and $k=12$ here is a solution: $1=\frac{12}{12+12+12+12+12+12-(12+12+12+12+12)}$ $2=\left(12 \times \frac{12}{12-12+12-12+12+12+12+12+12+12}\right)$ $3=\left(12 \times \frac{12}{\left(12-\left(12+\left(12+\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)$ ...


12

Making numbers out of 4 fours is a common problem: $$1=\frac {44}{44}$$ $$2=\frac {4\cdot 4}{4+4}$$ $$3=\frac{4+4+4}{4}$$ $$4=\frac{4-4}{4}+4$$ $$5=\sqrt{4!+\frac{\sqrt 4+\sqrt 4} 4}$$ $$6=\sqrt{\frac{4!\cdot 4-4!}{\sqrt 4}}$$ $$7=\sqrt{4!\sqrt 4+\frac 4 4}$$ $$8=\sqrt{\frac{4^4}{\sqrt4+\sqrt 4}}$$ $$9=(4-\frac 4 4)^{\sqrt 4}$$ $$10=\frac{4!} 4 - (4-\sqrt ...


12

Here's a wonderful open problem in set theory, which can be translated to a statement which you might be looking for. Suppose that $\aleph_\omega$ is a strong limit cardinal. Is $2^{\aleph_\omega}<\aleph_{\omega_1}$? We can prove that under the assumption that $\aleph_\omega$ is a strong limit cardinal, it is necessarily the case that ...


12

Taken from http://arxiv.org/abs/0902.3961 through this question http://mathoverflow.net/questions/21003/polynomial-bijection-from-mathbb-q-times-mathbb-q-to-mathbb-q . Is $f(x,y)=x^7+3y^7$ injective on $\Bbb Q \times \Bbb Q \ $ ?


12

This problem is hard in the sense that it is still unproven. I will provide a set of references, but little conclusive work (as far as I know) has been done on any of them. This is a conjecture of Hardy; he later generalized it to say: if a, b, c are relatively prime, a is positive, and $(a+b)$ and c are not both even, and $b^2 - 4ac$ is not a perfect ...


12

I improved on Laczkovich's solution by using a different orientation of the 4 small central triangles, by choosing better parameters (x, y) and using fewer triangles for a total of 64 triangles. The original Laczkovich solution uses about 7 trillion triangles. Here's one with 50 triangles:


11

The $n$-body problem is an ancient problem that was originally a problem of Euclidean geometry (which was originally identified with the study of "space" --- physical space and abstract geometry were not conceptually separated until the modern era). The problem was to determine the motion of $n$ celestial interacting through gravity. It traces its origins to ...


11

One of my favourites because it's just so simple to state. Discussed in this MO question and related to an also very interesting, though solved question/puzzle which I posted here a while ago. I'm not sure when the problem was first stated but it could certainly have been understood by even the earliest mathematicians. Can a disk be tiled by a finite ...


11

Richard Guy's Unsolved Problems in Number Theory, Problem F24, mentions only that Dan Hoey has verified this conjecture for $2^n$ up to $n = 2,500,000,000$. Guy tends to be fairly complete in his references. Since he doesn't give any others I doubt there's much more out there. Several other related questions on decimal representations of powers of ...


11

Diophantine equations are not special; any branch of mathematics in which it is possible to ask sufficiently strong questions will suffer a similar fate due to the existence of problems like the halting problem and the possibility of asking a question equivalent to the halting problem (or something similar). For example, group theory has such problems: it ...


11

The first four perfect numbers have been known for at least 2300 years. Here are their prime factorizations: $$\begin{align} 6 & = 2^{\hphantom{2}}\cdot 3 \\ 28 & = 2^2\cdot 7 \\ 496 & = 2^4\cdot 31 \\ 8128 & = 2^6\cdot 127 \end{align} $$ Any knucklehead, looking at these, or perhaps even just looking at the first two, could make the ...


11

Seems like $2$ would do it: $$ 1: 2^2 - 2 - 2/2 $$ $$ 2:2^2 - 2^2 + 2 $$ $$ 3: 2 + 22/22 $$ $$ 4: 2^{2^2}/2^2 $$ $$ 5: 2^2 - 2/2 + 2 $$ $$ 6: 2^2 + 2 - 2 + 2 $$ $$ 7:2^2 + 2 + 2/2 $$ $$ 8:2^{2}(2) + 2 - 2 $$ $$ 9:2^2(2) + 2/2 $$ $$ 10:22/2 - 2/2 $$ $$ 11 : (2^2)!/2 - 2/2 $$ $$ 12: 2^{2^2} - 2^2 $$ That should do it. Thanks to Phira for $10$ and ...


11

solution for n = 1, k = 12: $$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 = 1 $$ $$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1 = 2 $$ $$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1 = 3 $$ $$ 1 \times 1 ...


11

It is currently unknown whether $P = NP$. All purported proofs have either been shown to be faulty, are under review, or are dismissed by the community at large without being considered. For more information on the last one, you might want to read Scott Aaronson's blog post detailing indicators why a purported proof is likely to be wrong and won't ...


11

I'm pretty sure the in-text question is "no," which means the title question is pretty tough to answer cleanly. But just for the sake of mentioning it, it's easy to check that $n$ would have to be a power of 2 for this quantity to prime: Indeed, if we re-write it as $36^n+1$, then the argument is verbatim the same as for Fermat primes: If $n$ were not a ...


11

One problem with classification of objects of unrestricted size/complexity is set theoretic, and will be related to such strong set theoretic axioms related to accessible cardinals, measurable cardinals, and Vopenka's principle, and may depend on the continuum hypothesis and (certainly) by the axiom of choice. While I'm not an expert on these issues, I'll ...


10

Note that we can write your iteration scheme as, for $i \geq 0$ $$\begin{align} x(0,i) & = \text{given sequence of nonnegative integers}\\ x(n+1,i) & = \begin{cases} 0 & i = 0 \\ \left|x(n, i) - x(n,i-1)\right| & i > 0\end{cases}\end{align}$$ The first implication of this definition is that Property 1 the value of $x(n,i)$ is ...



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