# Tag Info

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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.

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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. ...

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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 ...

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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 ...

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"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 ...

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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 ...

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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

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

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?

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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 ...

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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 ...

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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:

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Making numbers out of 4 fours is a common problem: $$1=\frac {44}{44}$$ $$2=\frac {4\cdot 4}{4+4}$$ $$3=\frac{4+4+4}{4}$$ $$4=\frac{4-4}{4}+4$$ $$5=\sqrt{4!+\frac{\sqrt 4+\sqrt 4} 4}$$ $$6=\sqrt{\frac{4!\cdot 4-4!}{\sqrt 4}}$$ $$7=\sqrt{4!\sqrt 4+\frac 4 4}$$ $$8=\sqrt{\frac{4^4}{\sqrt4+\sqrt 4}}$$ $$9=(4-\frac 4 4)^{\sqrt 4}$$ $$10=\frac{4!} 4 - (4-\sqrt ... 12 Here's a wonderful open problem in set theory, which can be translated to a statement which you might be looking for. Suppose that \aleph_\omega is a strong limit cardinal. Is 2^{\aleph_\omega}<\aleph_{\omega_1}? We can prove that under the assumption that \aleph_\omega is a strong limit cardinal, it is necessarily the case that ... 11 Richard Guy's Unsolved Problems in Number Theory, Problem F24, mentions only that Dan Hoey has verified this conjecture for 2^n up to n = 2,500,000,000. Guy tends to be fairly complete in his references. Since he doesn't give any others I doubt there's much more out there. Several other related questions on decimal representations of powers of ... 11 Diophantine equations are not special; any branch of mathematics in which it is possible to ask sufficiently strong questions will suffer a similar fate due to the existence of problems like the halting problem and the possibility of asking a question equivalent to the halting problem (or something similar). For example, group theory has such problems: it ... 11 I'm pretty sure the in-text question is "no," which means the title question is pretty tough to answer cleanly. But just for the sake of mentioning it, it's easy to check that n would have to be a power of 2 for this quantity to prime: Indeed, if we re-write it as 36^n+1, then the argument is verbatim the same as for Fermat primes: If n were not a ... 11 Seems like 2 would do it:$$ 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 $$That should do it. Thanks to Phira for 10 and ... 11 One problem with classification of objects of unrestricted size/complexity is set theoretic, and will be related to such strong set theoretic axioms related to accessible cardinals, measurable cardinals, and Vopenka's principle, and may depend on the continuum hypothesis and (certainly) by the axiom of choice. While I'm not an expert on these issues, I'll ... 10 The problem with the Collatz conjecture has to do with the possibility of having long "alternating" chains. Consider a sequence of numbers a0, a1, a2, ... obtained by applying the Collatz function, starting with some integer a0. Consider just some element of this sequence, an: as you note, if an is odd, then an+1 = 3an+1 is even, so that ... 10 It isn't --- the Tate conjecture is harder than the Hodge conjecture. For example, the Tate conjecture implies that all absolutely Hodge classes are algebraic. Thus, in the presence of Deligne's conjecture that all Hodge classes are absolutely Hodge (which is known for abelian varieties), the Tate conjecture implies the Hodge conjecture. There is no similar ... 10 solution for n = 1, k = 12:$$ 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 ...

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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 $$C<\mathrm{rad}\,(ABC)^{\kappa ... 10 Note that we can write your iteration scheme as, for i \geq 0$$\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}$$The first implication of this definition is that Property 1 the value of x(n,i) is ... 10 The first four perfect numbers have been known for at least 2300 years. Here are their prime factorizations:$$\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 ...

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Is $\pi \cdot e$ rational? What is the minimal number of people in a party, such that there are necessarily either at least 5 mutual strangers or at least 5 mutual acquaintances? Is there a positive non-integer $x$ such that both $2^x$ and $3^x$ are integers? Does every closed curve in the plain contain 4 vertices of a square? Can you factor an integer in ...

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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 ...

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There is already a similar scheme the Polymath projects which you can find at the blog and wiki

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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 ...

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