Wikipedia:Reference desk/Mathematics - Wikipedia

Prove or disprove: These numbers are composite for all n>=2 such that these numbers are positive

118.170.47.16 (talk) 08:18, 13 September 2024 (UTC)Reply

Re   #3:  

Re   #4:  

Re   #6:  

Re   #7:  

Re   #8:  

Re #23:  

Re #24:  

Re #30:  

Re #36:  

These are all special cases of the difference of two nth powers.  --Lambiam 10:23, 13 September 2024 (UTC)Reply

Do your own homework. SamuelRiv (talk) 15:34, 13 September 2024 (UTC)Reply

Based on other posts with this type of number theoretic-focused content coming from IPs in the same geographic area (north Taiwan), I'm pretty sure this isn't meant to be homework. WP:CRUFT, perhaps, but this is the Reference Desk, and it seems to me that it's a lot less of an issue to have it here than elsewhere. GalacticShoe (talk) 16:52, 13 September 2024 (UTC)Reply

For all of these which are not listed as always composite or having a prime, I tested   and didn't find any primes.

1. Prime:   is probably prime. Note that   is divisible by   when   is even,   when  , and   when  , so when there is such a prime then  . Thanks to RDBury for helping establish compositeness originally for  .
2. Unknown:   is divisible by   when   is odd,   when  , and   when  , so if there is such a prime then  .
3. Always composite:   can be factorized.
4. Always composite:   can be factorized.
5. Always composite:   is divisible by   when   is even,   when  , and   when  .
6. Always composite:   can be factorized.
7. Always composite:   can be factorized.
8. Always composite:   can be factorized.
9. Unknown:   is divisible by   when   is odd,   when  ,   when  , and   when  , so if there is such a prime then  .
10. Unknown:   is divisible by   when   is even,   when  ,   when  , and   when  , so if there is such a prime then  .
11. Unknown:   is divisible by   when   and   when  , and it becomes a difference of squares if   is even, so if there is such a prime then  .
12. Always composite:   is divisible by   when   is odd, and it becomes a difference of squares if   is even.
13. Unknown:   is divisible by   when   is even, so if there is such a prime then   is odd.
14. Prime:   is probably prime. Note that   is divisible by   when   is odd and   when  , and it becomes a difference of cubes if  , so when there is such a prime then  .
15. Always composite:   is divisible by   when   is odd and   when   is even.
16. Always composite:   is divisible by   when   is odd, and it becomes a difference of squares if   is even.
17. Always composite:   is divisible by   when   is odd and   when   is even.
18. Always composite:   can be factorized.
19. Prime:   is probably prime. Note that   is divisible by   when   is even,   when  ,   when  ,   when  ,   when  , and   when  , so when there is such a prime then  .
20. Prime:   is probably prime. Note that   is divisible by   when   is odd,   when  , and   when  , so when there is such a prime then  .
21. Prime:   is probably prime. Note that   is divisible by   when   is even,   when  , and   when   so when there is such a prime then  .
22. Always composite:   is divisible by   when   is odd, and it becomes a difference of squares if   is even.
23. Always composite:   can be factorized.
24. Always composite:   can be factorized.
25. Always composite:   is divisible by   when   is odd and   when   is even.
26. Prime:   is probably prime. Note that   is divisible by   when   is odd,   when  , and   when  , so when there is such a prime then  .
27. Always composite:   is divisible by   when   is odd and   when   is even.
28. Always composite:   is divisible by   when   is odd, and it becomes a difference of squares if   is even.
29. Always composite:   is divisible by   when   is odd, and it becomes a difference of squares if   is even.
30. Always composite:   can be factorized.
31. Always composite:   is divisible by   when   is odd and   when   is even.
32. Always composite:   is divisible by   when   is odd and   when   is even.
33. Unknown:   is divisible by   when   is odd,   when  ,   when  , and   when  , so if there is such a prime then  .
34. Always composite:   is divisible by   when   is odd, and it becomes a difference of squares if   is even.
35. Always composite:   is divisible by   when   is odd and   when   is even.
36. Always composite:   can be factorized.
37. Unknown:   is divisible by   when   is odd,   when  , and   when  , so if there is such a prime then  .
38. Always composite:   is divisible by   when   is odd and   when   is even.
39. Unknown:   is divisible by   when   is odd,   when  , and   when  , so if there is such a prime then  .
40. Always composite:   is divisible by   when   is even,   when  , and   when  .
41. Always composite:   is divisible by   when   is odd, and it becomes a difference of squares if   is even.
42. Always composite:   is divisible by   when   is odd, and it becomes a difference of squares if   is even.

GalacticShoe (talk) 08:10, 15 September 2024 (UTC)Reply

I'm guessing this is about the best one can do without actually discovering that some of the values are prime. I did some number crunching on the first sequence 5ⁿ+788 with n<1000. All have factors less than 10000 except for n = 87, 111, 147, 207, 231, 319, 351, 387, 471, 487, 499, 531, 547, 567, 591, 639, 687, 831, 919, 979. You can add more test primes to the list, for example if n%30 = 1 then 5ⁿ+788 is a multiple of 61, but nothing seems to eliminate all possible n. Wolfram Alpha says the smallest factor of 5⁸⁷+788 is 1231241858423, so it's probably not feasible to carry on without something more sophisticated than trial division. --RDBury (talk) 19:12, 15 September 2024 (UTC)Reply

Thanks, RDBury. I've been using Alpertron, it's good at rapidly factorizing or finding low prime factors. GalacticShoe (talk) 22:20, 15 September 2024 (UTC)Reply

Yes, Alpertron is very good and you can give it a file to factor, or specify a formula to factor. Bubba73 ^{You talkin' to me?} 05:44, 22 September 2024 (UTC)Reply

Not quite sure where to ask this but I decided to put it here. I apologize if this doesn't belong here.

I was recently reading about hyperbolic geometry and when reading the article Limiting parallel, I came across the statement "Distinct lines carrying limiting parallel rays do not meet." What exactly does it mean for a line to carry a ray? Is this standard mathematical terminology? TypoEater (talk) 18:14, 13 September 2024 (UTC)Reply

Yes, this is the place for such a question, though you might also complain at Talk:Limiting parallel that the phrase is unclear. —Tamfang (talk) 22:51, 13 September 2024 (UTC)Reply

I find the whole article unclear and confusing. Is it me?  --Lambiam 23:54, 13 September 2024 (UTC)Reply

Given the country's unusual shape, Croatia's geographical centre is rather awkwardly placed; if you Google <croatia geographical centre> you get lots of forums and similar content discussing the idea that its centre is actually outside the country, in western Bosnia and Herzegovina. From this I'm left wondering: (1) Are there any reliable sources for this claim? I couldn't find any. If so, it would be a good addition to geography of Croatia. (2) The geographical centre article discusses different methods of calculating the geographical centre of a region and the potential problems resulting therefrom. If Croatia's centre really can be defined to be in Bosnia, do all definitions put it there, or do some definitions put it in Croatia? (3) Is there any concept of most-centred-within-boundary? [This is the biggest reason I came to the Maths desk; I'm wondering if topologists would care about it?] Let us assume that Croatia's centre is outside Croatia: is there a term that refers to the Croatian location that is least-off-centre? It wouldn't necessarily be the Croatian location closest to the centre (imagine a narrow salient that would be closest but severely off-centre), but I suppose it could. I'm thinking of the point where, if you balanced a flat map of Croatia on it, the map would topple most slowly. Obviously some points are better than others — a point at the country's southern tip would be worse than places farther northwest — so I wonder if it's reasonable to define the best place when no place is ideal. Nyttend (talk) 07:51, 16 September 2024 (UTC)Reply

PS, imagine Croatia like a balancing bird toy. If you broke off the bird's head, you probably couldn't balance it at all, but you'd do a lot better balancing on the body (or even the tail) than at the wingtips. Croatia lacks the "bird head", but you're probably better-off balancing in the northwest than anywhere else. Nyttend (talk) 07:55, 16 September 2024 (UTC)Reply

There was a discussion not that long ago here: Wikipedia:Reference desk/Archives/Miscellaneous/2024 July 18 § Lake Lats and Longs. It contains, at least implicitly, answers to some of the questions. Since the notion of centre is not well-defined, neither is that of "least off-centre". The location nearest to a given point outside the area is on its boundary. The interior point furthest from the boundary works for most actual country shapes, including Croatia, for which this is a point roughly 20 km east of zagreb. — Preceding unsigned comment added by Lambiam (talk • contribs) 16:06, 16 September 2024 (UTC)‎Reply

The visual center and algorithms to approximate the pole of inaccessibility are discussed in this 2016 Mapbox post. I thought that if you'd include just a spherical geometry of the Earth you'd get even more interesting questions, but it appears at a glance (? not sure?) the algorithm in the link already generalizes nicely to noneuclidean geometries and higher dimensions. SamuelRiv (talk) 17:46, 16 September 2024 (UTC)Reply

Through any point on Earth there is (by the mean value theorem) a great circle that bisects the population of Croatia. Among such circles, consider the segments that cut the territory of Croatia into exactly two pieces. I propose the midpoint of the shortest such segment as a centre. (shamelessly OR) —Tamfang (talk) 01:29, 18 September 2024 (UTC)Reply

If the border of Croatia is sufficiently fractal-like, there may be no great circle that cuts the area into just two connected pieces.  --Lambiam 09:19, 18 September 2024 (UTC)Reply

Doesn't need to be fractal; a C-shape with overlapping ends (ie a spiral with just over a full turn) can't be halved (by area) into only two pieces by an infinite straight line. There may be a coral atoll somewhere that approaches this? 213.143.143.69 (talk) 12:37, 18 September 2024 (UTC)Reply

There's quite a few simply connected shapes that create uncomfortable solutions for this algorithm (so far splitting 3+ pieces, you can also just have a wide lobe with two long thin tails side-by-side; the shortest area or even-population bisector straight line would have to cut through both tails; the longer unbroken line that cuts the lobe can be stopped by making the lobe an S-turn; which can only be resolved (to retain contiguity) by removing the requirement of a great circle.

(The spiral atoll example need not be a problem, since the great circle need not be required to cut through the entire atoll, but just one segment.)

It's a good idea, but it needs a bit more refinement to get a sensible-and-unique definition in all cases. SamuelRiv (talk) 16:32, 18 September 2024 (UTC)Reply

Seems to me your solution to the spiral problem (for which thanks) applies to any simply connected shape. —Tamfang (talk) 05:46, 23 September 2024 (UTC)Reply

Could it be meaningful to measure the density of such circles passing through a given neighborhood? I think I see an approach or two, but it needs more thought. The maximum by this measure is my new favorite center. —Tamfang (talk) 05:43, 23 September 2024 (UTC)Reply

Consider any point within the boundary. Calculate the maximum distance from that point to any other point within the boundary. Then choose the point where that maximum distance is least. This will give a 'centre' that is in some way the closest possible to the rest of the area. For convex shapes it is equivalent to the circumcentre (centre of the bounding circumcircle). Would this be unique for concave shapes? -- Verbarson  ^talk_edits 11:38, 19 September 2024 (UTC)Reply

Not necessarily; a symmetric C-shape where the ends curl back nearly to the centre would have two possible solutions, one on each tip. -- Verbarson  ^talk_edits 13:27, 19 September 2024 (UTC)Reply

1) For a practical problem like this, a finite set of 'diverse' solutions (or a continuous and symmetric set) is not really an issue -- just choose any one point; 2) you can also include the shortest path between two points within the concave polygon. A central consideration above is the calculation time.

Another possibility is to get the approximate convex skull, the largest enclosed convex polygon, which reduces you to an easy incenter calculation and you're done. SamuelRiv (talk) 18:24, 19 September 2024 (UTC)Reply

Also, for cases like long bulbous C-shapes, we can adapt if we prefer the narrow midway 'neck' as opposed to the widest of the 'bulbs'. (To illustrate the scenario: consider North and South American -- what the cartographer calls the natural center of the supercontinent is either around the Isthmus of Panama, or it's at the geographic center of their choice of the gigantic North or South continent (or they can put the same label on both). One could find all the local extrema of the largest enclosed convex polygon at any point (assuming they are discrete sets, so the shape is not a perfect circular atoll for example, and if the approximate algorithm for the convex skull can be so adapted) and then choose the median polygon incenter (however you want to resolve it for even numbers). SamuelRiv (talk) 18:40, 19 September 2024 (UTC)Reply

I assume the distance is measured on an interior path. —Tamfang (talk) 05:28, 23 September 2024 (UTC)Reply

Either interior or straight-line. It may depend on whether you want to site a power station or an airport? -- Verbarson  ^talk_edits 07:47, 23 September 2024 (UTC)Reply

If I raise a permutation matrix to a high power it is still a permutation matrix (all values 0 or 1). If I raise a stochastic matrix to a high power it is still a stochastic matrix (all values <=1). But there are a lot of matrices raised to a high power have elements with very large values. Is there a name for the property of a matrix that when raised to a high power remain reasonably valued? RJFJR (talk) 22:12, 19 September 2024 (UTC)Reply

The spectrum of such a matrix is inside the closed unit disk (a necessary but not sufficient condition). A matrix whose spectrum is in the open unit disk always satisfies this property. I'm not sure there is a general name for this property though. Tito Omburo (talk) 23:07, 19 September 2024 (UTC)Reply

That's it. Thank you! RJFJR (talk) 02:08, 20 September 2024 (UTC)Reply

Note that when the spectral radius is strictly less than   the powers of the matrix tend to the null matrix, which is not very exciting. I also vaguely remember seeing a theorem that states or implies that when the spectral radius is equal to   the growth of the entries is polynomial in the value of the exponent.  --Lambiam 21:55, 20 September 2024 (UTC)Reply

1. Since the expression in the title is a definition, we can conclude, that A exists if and only if both B and C exist as well.

2. However, we can't conclude that if B or C exist then A exists. Check: A denotes the total price, B denotes the price of the first product, and C denotes the price of the second product: if only the price   of the first product exists, we still can't conclude - that the price   of the second product exists - nor that the total price   exists. The defintion   only tells us (besides the relation  ), that the total price   exists if and only if both - the price   of the first product exists - and the price   of the second product exists.

3. For concluding, that if any value of the above three values A,B,C exists then all of them exist, it's sufficient to write down three definitions:

4. Question: as I've indicated, it's sufficient (to write down all three defintions), but is there any shorter notation, to make sure that if any value of the above three values A,B,C exists then - all of them exist - and satisfy  

HOTmag (talk) 13:26, 20 September 2024 (UTC)Reply

Are you asking about a shorter notation? I may be looking mighty foolish, but isn't it circular and therefore meaningless to write the system of three definitions like you have above? Or rather, wouldn't writing down the first be exactly as meaningful as writing down all three, since both situations only relate A, B, C to one another? I get they're definitions and not merely statements, but I'm not really seeing the difference here. You'd have to define B, C in terms other than A to reach a more meaningful domain. Again, I could look totally foolish right now, since I've never answered a question here before. Remsense ‥ 论 13:34, 20 September 2024 (UTC)Reply

It seems you haven't read 2#. HOTmag (talk) 13:40, 20 September 2024 (UTC)Reply

I don't really understand it, no. You're defining A, B, C as the prices of products, but then you're writing abstract definitions of them in terms of each other. Am I missing something? Oh, did you mean to say A is the total? It makes more sense to me that way. Remsense ‥ 论 13:43, 20 September 2024 (UTC)Reply

In any case, I still don't understand what the extra two definitions achieve: either B and C are defined outside of A or they're not, right? If we wanted to know them in terms of A, we already got that in the first definition. Remsense ‥ 论 13:45, 20 September 2024 (UTC)Reply

I've just added an addition to 2#, to make it clear. HOTmag (talk) 13:49, 20 September 2024 (UTC)Reply

I return to my initial question then, are you just looking for a shorter notation for this? Remsense ‥ 论 13:52, 20 September 2024 (UTC)Reply

Yep. I've just made it clear in 4# (thanks to your question). HOTmag (talk) 13:55, 20 September 2024 (UTC)Reply

Sorry for being slow on the uptake. I'm not sure if this needs to fit into any particular system or paradigm: anything wrong with  . Sorry if that hurts anyone to see, haven't flexed these muscles in a while Remsense ‥ 论 14:31, 20 September 2024 (UTC)Reply

Not to bug you, but only since I'm relatively unsure of myself in this area—was this answer something like what you were looking for? Remsense ‥ 论 20:10, 20 September 2024 (UTC)Reply

You limit the set of the Bs and the Cs to be the positive integers, but my question is general, without limiting anything. HOTmag (talk) 02:05, 22 September 2024 (UTC)Reply

You gave the example of prices, so that's what I picked the positive reals* based on. Clearly, you can replace the set with whatever you want. Remsense ‥ 论 02:08, 22 September 2024 (UTC)Reply

If A denotes the total price, B denotes the price of the first product, and C denotes the price of the second product, does your definition let us deduce the other definition:   which I would like to deduce, bearing in mind that not all products have a price? Note: Since the latter is a defintion of C, then the existence of A and of B must be derived from the very existence of C. HOTmag (talk) 02:21, 22 September 2024 (UTC)Reply

It seems the bit about "deducing a definition" articulates a fundamental confusion you have about what you're trying to accomplish. Definitions are stated, not deduced. Remsense ‥ 论 05:34, 22 September 2024 (UTC)Reply

By asking whether a new definition can be deduced from an old definition, I mean whether a new claim, that was presented before as an additional definition, can be derived as consequence, from a given assumption that was presened before as an old definition.

In our case, the given assumption, was presented before as an old definition: A^def
₌B+C. The new claim, was presened before as an additional definition:   The question is, whether we can deduce the latter from the former, i.e. whether we can deduce the new claim from the given assumption. HOTmag (talk) 13:08, 22 September 2024 (UTC)Reply

If the values we assign to B and C are logically independent from one another, then no. That's what logical independence means.Remsense ‥ 论 13:17, 22 September 2024 (UTC)Reply

This is totally confused. The addition operation   is special in that it is operand-wise strictly monotonic and therefore has, at least in the integers and reals (but not in the natural numbers) an operand-wise inverse, which we can denote using the subtraction operation  . In general, this is not possible.

Defining some quantity   by an equation of the form   only makes sense if all terms in the right-hand side are defined. It is not just that this fails to define   if   or   is not defined. It just does not make sense. And if   is not defined, writing something like   only increases the confusion.

There is another issue in which the definedness of a defined term depends on the definedness of another term, namely when defining a function. Suppose we define a new function   using existing known functions with a limited domain. For example, we may define real-valued function   on the real numbers by the equation

 

For   to be defined by this equation for some given value of   it is necessary that both   and   are defined. This is more a matter of common sense than anything else, and I see no need for some notational device to express this dependency.  --Lambiam 22:29, 20 September 2024 (UTC)Reply

If A denotes the total price, B denotes the price of the first product, and C denotes the price of the second product, does the definition A^def
₌B+C make senee? If it does, can you deduce the definition:   HOTmag (talk) 02:13, 22 September 2024 (UTC)Reply

You can't define something that already has a meaning. So if A, B and C all have independent meanings, as shown by the use of "denote", then none of them can be defined in terms of the other two. It may be true that A = B + C based on these meanings, but that's not a definition. In any case, you can't deduce a definition. If A has no independent meaning then you can define it to be anything, B + C, B - C, or B * C. --RDBury (talk) 05:29, 22 September 2024 (UTC)Reply

Well, I'm presenting an analogous question (taken from a discipline very close to arithmetic), in my following thread. I hope my new question explains also my old question, but if my old question is still not clear, you can ignore it, and focus on my new question. HOTmag (talk) 13:38, 22 September 2024 (UTC)Reply

For example: S is a set of the following vectors:

A=(1,1,0),

B=(1,0,0),

C=(0,1,0).

Note: A=B+C, and B=A-C, and C=A-B, so the set S is linearly dependent.

Using A,B,C only, i.e without using S, what's the shortest description, claiming that the set S is linearly dependent but every proper sub set of S is linearly independent? HOTmag (talk) 13:30, 22 September 2024 (UTC)Reply

@HOTmag: I'd just say 'S has k linearly independent elements' (in the give example k=2). --CiaPan (talk) 14:46, 22 September 2024 (UTC)Reply

Is your response a suggestion of rephrasing my question?

If it's intended to be an answer, then please note: My condition requires to be "using A,B,C only, i.e without using S". Additionally, where does your description claim, that S is linearly dependent? HOTmag (talk) 14:56, 22 September 2024 (UTC)Reply

One way to characterise the set is "a set of vectors, any one of which can be written in terms of the others in a unique way". The set is just A, B and C, i.e. any property of them is a property of the set of them. --2A04:4A43:900F:F4C3:49F4:4EFB:C442:608F (talk) 15:15, 22 September 2024 (UTC)Reply

Since my condition requires to be "using A,B,C only, i.e without using S", so I guess you mean the following: "Each vector, can be written as a unique linear combination of the other vectors". Thanks. HOTmag (talk) 15:52, 22 September 2024 (UTC)Reply

Assuming the vectors are  , form the matrix   whose columns are  . The stated condition is then:

1. the matrix of   minors of   is zero, and
2. each of the columns of the matrix of   minors of V has a non-zero entry.

- Tito Omburo (talk) 17:36, 22 September 2024 (UTC)Reply

Your description, both uses sets, and also becomes longer than the original one indicated in the title. HOTmag (talk) 08:41, 23 September 2024 (UTC)Reply

You can say, "each of the sets {A,B}, {A,C} and {B,C} is linearly independent".  --Lambiam 21:05, 22 September 2024 (UTC)Reply

You also have to add that the set {A,B,C} is linearly dependent, but then the description - both uses sets, and also becomes longer than the original one indicated in the title. HOTmag (talk) 08:40, 23 September 2024 (UTC)Reply

There is a unique linear combination generating 0, and in this linear combination all coefficients are nonzero.2404:2000:2000:8:FDE8:8311:95E3:654D (talk) 00:00, 23 September 2024 (UTC)Reply

Yes, and by symbolic notation the description even becomes shorter. Thanks. HOTmag (talk) 08:42, 23 September 2024 (UTC)Reply

How can I calculate the height and width in meters of a future geographic map of the world based on its desired scale in cm and km? 212.180.235.46 (talk) 15:35, 23 September 2024 (UTC)Reply

For a map of the world, you will have to specify which map projection it will use, and then to which part(s) of the projection the scale is to apply (and for some projections, how much of the globe to include). Mapping a spherical surface onto a plane surface cannot be done with a constant scale factor. Eg: you might specify a Mercator projection, between latitudes +/-85deg, with a scale of 1:10,000,000 at the equator, which would give you a map about 4m wide along the equator, and (judging by the illustration for Mercator) slightly more than that high. Other projections will vary considerably. -- Verbarson  ^talk_edits 18:07, 23 September 2024 (UTC)Reply

My own map with van der Grinten projection and equatorial scale of 1:20 000 000 for example is 1.18 m high and 1.94 m wide. Out of curiosity, I started to think about dimensions of imaginative humongous world maps. With that as benchmark, if my calculations are correct, a 1:160 000 scale world map, for example, would be 147 m long and 242 m wide. 212.180.235.46 (talk) 20:35, 23 September 2024 (UTC)Reply

The scale of a map is the ratio between the distance between two points on the map and that between the corresponding points on the ground. Any map projection necessarily distorts the distances, so the scale you get if you take New York and San Francisco as the two points is generally substantially different from what you get if you take Berlin and Moscow. For maps of the whole Earth it is non-trivial to decide on a "nominal scale". In spite of the name, the van der Grinten projection is not a geometric projection in which points on the map are found by following rays projected along a straight line out of a shrunk globe. The equatorial and central-meridional scales are well defined, though. The length ℓ_eq of the equator is approximately 40,000 km. If the equatorial scale is chosen to be 1 : S_eq, the width of the map is ℓ_eq / S_eq. So for S_eq = 160,000 this comes out as about 250 m. The length ℓ_mer of a meridian, measured from one pole to the other, is about 40,000 km. Letting S_cm stand for the central-meridional scale factor, the height of the map will be ℓ_mer / S_cm. If these scale factors are chosen to be equal, the map will have a height of half its width. For a circular map as shown in the article, S_cm = 2 S_eq.  --Lambiam 10:14, 24 September 2024 (UTC)Reply

Simple question : let’s say I have a pairing friendly curve having a very large trace, and that I have a pairing with points A∈  and B∈  such as the optimal ate pairing e(A,B)=y, then is it possible to fully determine 2 point I and J such as e(I,J) equals a target multiple of the finite’s field element y ?

Or does it requires full pairing or full generalized Miller’s inversion and thus would be impossible in practice on a curve like bn254 ? 2A01:E0A:401:A7C0:9CB:33F3:E8EB:8A5D (talk) 15:51, 26 September 2024 (UTC)Reply