# Tag Info

## Hot answers tagged open-problem

21

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.

18

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 ...

16

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 ...

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 $n_0$ ...

12

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 ...

12

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?

12

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

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 ...

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

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

Making numbers out of 4 fours is a common problem: $$1=\frac {44}{44}$$ $$2=\frac {4\cdot 4}{4+4}$$ $$3=\frac{4+4+4}{4}$$ $$4=\frac{4-4}{4}+4$$ $$5=\sqrt{4!+\frac{\sqrt 4+\sqrt 4} 4}$$ $$6=\sqrt{\frac{4!\cdot 4-4!}{\sqrt 4}}$$ $$7=\sqrt{4!\sqrt 4+\frac 4 4}$$ $$8=\sqrt{\frac{4^4}{\sqrt4+\sqrt 4}}$$ $$9=(4-\frac 4 4)^{\sqrt 4}$$ $$10=\frac{4!} 4 - (4-\sqrt ... 11 Richard Guy's Unsolved Problems in Number Theory, Problem F24, mentions only that Dan Hoey has verified this conjecture for 2^n up to n = 2,500,000,000. Guy tends to be fairly complete in his references. Since he doesn't give any others I doubt there's much more out there. Several other related questions on decimal representations of powers of ... 10 One problem with classification of objects of unrestricted size/complexity is set theoretic, and will be related to such strong set theoretic axioms related to accessible cardinals, measurable cardinals, and Vopenka's principle, and may depend on the continuum hypothesis and (certainly) by the axiom of choice. While I'm not an expert on these issues, I'll ... 10 solution for n = 1, k = 12:$$ 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 ...

10

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:

9

If you will permit me to respond in meme: Firstly, one cannot go around asserting a proof of the 3x+1 conjecture without sounding like a crackpot. The first thing I would do is stop claiming this, and instead say things like I'd like to discuss some ideas I've had which relate to the 3x+1 problem. One of the greatest mathematicians of all time, Paul ...

9

The problem with the Collatz conjecture has to do with the possibility of having long "alternating" chains. Consider a sequence of numbers a0, a1, a2, ... obtained by applying the Collatz function, starting with some integer a0. Consider just some element of this sequence, an: as you note, if an is odd, then an+1 = 3an+1 is even, so that ...

9

If one wants to avoid epsilons and constants in the formulation of the conjecture one can use this one instead. If i) $\mathrm{rad}\,(n)$ is the product of the distinct primes in $n$, ii) $A,B,C$ are three positive coprime integers, iii) $A+B=C\$, iv) $\kappa >1$, then, with finitely many exceptions we have $$C<\mathrm{rad}\,(ABC)^{\kappa ... 7 In Serge Lang's Algebra, he says: "One of the most fruitful analogies in mathematics is that between the integers and the ring of polynomials over a field". He then proves the abc conjecture for polynomials, and for good measure he proves Fermat's Last Theorem for polynomials. In other words, Lang is saying that if something is true for the ring of ... 7 It isn't --- the Tate conjecture is harder than the Hodge conjecture. For example, the Tate conjecture implies that all absolutely Hodge classes are algebraic. Thus, in the presence of Deligne's conjecture that all Hodge classes are absolutely Hodge (which is known for abelian varieties), the Tate conjecture implies the Hodge conjecture. There is no similar ... 7 Note that we can write your iteration scheme as, for i \geq 0$$\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 ...

7

I won't try to address the 'Is there validity to this argument...?' except to say that it is number 65 out of 67 at the moment on The P-versus-NP page My guess is that, if the proof doesn't work, it will take a couple of weeks (if people who are able to judge care enough), but much longer the more substance it has. The last really serious proof attempt ...

7

A cursory look at the paper reveals that the author is not a professional mathematician. The algorithm looks very simple, and the bibliography is extremely short. So according to Scott Aaronson's criteria (mentioned by svenkatr), the probability that this particular approach works is small. This is aggravated by the statement in the website that the ...

7

It is not known whether there are infinitely many primes of the form $a^2 + 1.$ Your numbers are the subset with $a = 6^n.$ So it is not known that your set is infinite. The strongest results are in two variables, the well-known result that infinitely many primes are of the form $x^2 + y^2$ (indeed all primes $p \equiv 1 \pmod 4$) and the result of Iwaniec ...

7

There is a compiled list titled "SOME OPEN PROBLEMS ABOUT DIOPHANTINE EQUATIONS" and this problem happens to be listed as Problem#16 on page 3. Check it here (Originally thought it was Problem #15, but I stand corrected it is indeed Problem#16).

6

For $n=9$ and $k=9$ here is a solution: $1=\left(9+\frac{9}{\left(9 \times \frac{9}{\left(9+\left(9+\left(9-99\right)\right)\right)}\right)}\right)$ $2=\frac{9}{\left(9+\frac{9}{\left(9-\frac{9}{\left(9 \times \frac{9}{99}\right)}\right)}\right)}$ $3=\left(9-\left(9+\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+99\right)}\right)}\right)\right)\right)$ ...

6

"Birch and Swinnerton Dyer conjecture" usually refers to an amazing formula that predicts exactly the leading term of the L-function at $s=1$ (a real number $c$ and an integer $k$ such that the leading term is $c(s-1)^k$). The prediction of $k$ only, the conjecture that it is an "analytic rank" equal to the rank of the group of rational points on the curve, ...

5

For $n=4$ and $k=5$ here is a solution: $\frac{4}{\left(4+\left(4 \times \left(4-4\right)\right)\right)}=1$ $\left(4-\left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=2$ $\left(4+\frac{4}{\left(4-\left(4+4\right)\right)}\right)=3$ $\left(4+\left(4+\left(4-\left(4+4\right)\right)\right)\right)=4$ ...

5

The problem with such a claim is that there are so many such claims made every year, not just for the P vs NP problem but other open problems as well. I am not an expert in this topic, so I won't comment on the content of the paper you have linked to. In order to check claims of this sort, an expert will probably have to spend at least a few days (perhaps ...

5

Consider a Kähler manifold $X=X_n$ of dimension $n$, for example a projective algebraic manifold . It is in particular a topological space and as such has cohomology groups $H^r(X,\mathbb C)$. A very down-to-earth way of studying it is through De Rham cohomology. Moreover the Kähler structure (= a special type of Riemannian structure) on $X$ permits ...

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