I accidentally asked this on science area first, sorry. How do the Zariski and metric topologies on the complex numbers interact or complement each other when mathematicians are studying algebraic geometry or several complex variables or in other areas of mathematics? Thanks. Rich (talk) 20:43, 1 March 2025 (UTC)[reply]

I guess the basic answer is that the metric topology (or analytic topology) is much stronger than the Zariski topology, and therefore more "intuitive". However, in many cases there are deep connections between the two topologies (e.g., Serre's GAGA and Chow's theorem). Tito Omburo (talk) 21:22, 1 March 2025 (UTC)[reply]

Ok thank you. it's heavy reading but i'll tackle it. One question left is : is metric topology ALWAYS strictly finer than Zariski that every open set in Zariski is also always open in metric topology? Because in several complex variables zero sets(the closed sets) don't have to be isolated if I remember rightly.Rich (talk) 06:59, 2 March 2025 (UTC)[reply]

It's always strictly finer because the Zariski topology is defined by polynomial functions, which are continuous in the metric topology. Tito Omburo (talk) 10:50, 2 March 2025 (UTC)[reply]

On the x-y plane a circle of radius R1 is centered on (x,y) = (0,0). Also centred on 0,C is another circle of radius R2 where y < R1 and R2 > R1 + C. A line at angle a (and therefore would cross 0,0 if projected back) goes from the first circle to the second circle. How do I calculate the length of this line, given angle a and radii R1 and R2? [Edited to overcome objection by RDBury] Dionne Court (talk) 13:13, 2 March 2025 (UTC)[reply]

I'm not entirely sure I understand the question. First, it can be confusing to label anything 'y' if you're working in the x-y plane; it's better to use 'c'. And it's not clear how the line is defined, is it from any point on the first circle to any point on the second circle? If so then the line would not cross the origin. So can I take it that you're defining the line to be the line though (0, 0) at angle a to the x-axis? If that's the case the I'm pretty sure the problem will be much easier if you convert polar coordinates. The equation of an arbitrary circle can be found at Polar coordinate system#Circle. Note also that the line will intersect both circles up to twice, so you have to specify which points you're talking about, otherwise you get up to four possible lengths depending on how you interpret the problem. --RDBury (talk) 19:25, 2 March 2025 (UTC)[reply]

If I understand the question correctly, the equations of these circles are

and

while the ray is given by the parametric equation

The ray emerges from the first circle at

The point of intersection with the second circle is a bit trickier. Substitution of a generic point of the ray in the circle's equation gives

a quadratic equation in Solving it gives two solutions of which only the larger, say should be positive, and in fact larger than The length of the segment between the circles is then equal to  ​‑‑Lambiam 23:27, 2 March 2025 (UTC)[reply]

Thanks Lambian. If my 77 year old brain is working today, this leads to:-

x2 = c.sin(a) + (c^2 - sin^(a) - c^2 + R2^2)^0.5 and the ray intercepts at u = x2.cos(a), v = x2sin(a).

Then the length of the ray at angle a lying between the two circles is ( (R1.cos (a) - u.cos(a))^2 + (R1.sin(a) - v.sin(a))^2 )^0.5. Dionne Court (talk) 04:05, 3 March 2025 (UTC)[reply]

In the quadratic formula above, "c^2 − sin^(a)" should be "c^2 × sin^2(a)". (The variable c has the dimension "length" while sin(a) is dimensionless. Only for quantities of equal dimensions is adding or subtracting a meaningful operation.) Also, the length of the ray segment is ((R1.cos(a) − u)^2 + (R1.sin(a) − v)^2)^0.5 = ((R1.cos(a) − x2.cos(a))^2 + (R1.sin(a) − x2.sin(a))^2)^0.5 = ((R1 − x2)^2.cos^2(a) + (R1 − x2)^2.sin^2(a))^0.5 = ((R1 − x2)^2.(cos^2(a) + sin^2(a))^0.5 = ((R1 − x2)^2)^0.5 = ❘R1 − x2❘ = x2 − R1.

Using vector notation, putting z = (cos(a), sin(a)) and using ‖z‖ = 1, we get a simpler calculation:

‖(u,v) − (x,y)‖ = ‖x2.z − R1.z‖ = (x2 − R1).‖z‖ = x2 − R1.  ​‑‑Lambiam 07:43, 3 March 2025 (UTC)[reply]

Simple question, I’ve the following expression : (y² + x×2032123)÷(17010411399424)

for example, x=2151695167965 and y=9 leads to 257049 which is the perfect square of 507

I want to find 1 or more set of integer positive x and y such as the end result is a perfect square (I mean where the square root is an integer). But how to do it if the divisor 17010411399424 is a different integer which thar time is non square and/or 2032123 is replaced by a semiprime impossible to factor ? 2A0D:E487:133F:E9BF:C9D5:9381:E57D:FCE8 (talk) 21:35, 3 March 2025 (UTC)[reply]

We can generalize to finding solutions to for fixed (in your case, 2032123 and 17010411399424 respectively.) Rearranging yields . As long as there is some such that , you can generate infinitely many solutions by taking and and working backwards to get . Of course, some solutions correspond to negative values, but you can always just increase and/or decrease as needed. To find if there is such satisfying in the first place, you could just check values between 1 and inclusive until you find one, without needing to factorize. GalacticShoe (talk) 04:00, 4 March 2025 (UTC)[reply]

I need only positive solutions and where y<A

May you give a step by step example please?

Also, what do you mean by checkin values between 1 and inclusive until you find one ? How to do it ? Becuase I suppose that if A is 2000 bits long that this can t be done at random

2A0D:E487:35F:E1E1:51B:885:226F:140F (talk) 05:07, 4 March 2025 (UTC)[reply]

Has there been a proof that every next successive prime is always less than twice its previous prime (i.e.: )? I am not sure if this statement is implemented in the Prime gap article.Almuhammedi (talk) 08:14, 4 March 2025 (UTC)[reply]

This is Bertrand's postulate, which was proved by Chebyshev in 1852. It is mentioned in the section Prime gap § Upper bounds.  ​‑‑Lambiam 08:25, 4 March 2025 (UTC)[reply]

Thank for reply and I also found it in the article.Almuhammedi (talk) 08:30, 4 March 2025 (UTC)[reply]

Give two integers m>=1, n>=2, is there always a number of the form (m*generalized pentagonal number+1) which is Fermat pseudoprime base n? 220.132.216.52 (talk) 15:39, 4 March 2025 (UTC)[reply]

This is a list for 1<=m<=64 and 1<=n<=64, 0 if the corresponding generalized pentagonal number is larger than the 16777216th generalized pentagonal number (0 is the 0th generalized pentagonal number, 1 is the 1st generalized pentagonal number).

--220.132.216.52 (talk) 21:54, 5 March 2025 (UTC)[reply]

Note that (assuming Bunyakovsky conjecture is true), (m*generalized pentagonal number+1) is always composite if and only if m is in {24, 25, 27, 32, 49}, and if m is not in {24, 25, 27, 32, 49}, then there are infinitely many primes of the form (m*generalized pentagonal number+1), see Wikipedia:Reference desk/Archives/Mathematics/2023 May 20#For which natural numbers n.2C this sequence contains infinitely many primes.3F 220.132.216.52 (talk) 12:42, 10 March 2025 (UTC)[reply]

For m in {24, 25, 27, 32, 49}, there are no primes of the form (m*generalized pentagonal number+1) because:

• 24 * generalized pentagonal numbers + 1 are always squares
• 25 * generalized pentagonal numbers + 1 are always generalized pentagonal numbers
• 27 * generalized pentagonal numbers + 1 are always triangular numbers
• 32 * generalized pentagonal numbers + 1 are always generalized octagonal numbers
• 49 * generalized pentagonal numbers + 1 are always in OEIS: A144065 (generalized pentagonal numbers-1, n*(n+5)/6 with (positive or negative or 0) integer n)

And all generalized polygonal numbers (other than the trivial cases, i.e. the generalized pentagonal numbers 1, 2, 5, 7, the triangular numbers 1, 3, the generalized octagonal numbers 1, 5) are composite. (and the only prime in OEIS: A144065 is 11) 220.132.216.52 (talk) 12:53, 10 March 2025 (UTC)[reply]

I understand pentagonal numbers and the image at the top of Pentagonal number is an easily understood graphical representation of pentagonal numbers. Given that, I can believe the given formula (though I haven't attempted to verify it to myself yet). However Generalized pentagonal numbers have no "meaning" to me; we take the pentagonal number formula and use it on a strange sequence of "n"s. Why should we consider that is a useful thing to do? Is there some way of visualising the sequence akin to the graphic mentioned earlier? The sequence starts 0, 1, 2, 5, 7, 12, 15 and I can see 0, 1, 5, 12 being pentagonal, but there is nothing apparently pentagonal about 2, 7, 15. So can anyone explain and/or provide some graphics for the sequence? -- SGBailey (talk) 11:58, 6 March 2025 (UTC)[reply]

For square and triangular numbers, n² and n(n+1)/2, you get the same set of numbers if you plug in negative n. This is not true for pentagonal numbers though. In the positive direction in goes 0, 1, 5, 12, 22, ..., and in the negative direction it goes ... 40, 26, 15, 7, 2, 0. (See (sequence A005449 in the OEIS).) I don't know if there's a natural geometric definition of the second set of numbers; the article has a section "Generalized pentagonal numbers and centered hexagonal numbers" which tells us that each centered hexagonal number is the sum of a (regular) pentagonal number and the corresponding (offset by one) negative pentagonal number. This section is unsourced though and I'm not convinced it's anything more than a mathematical coincidence. To me, the real use of generalized pentagonal numbers is Euler's Pentagonal number theorem which gives a relatively simple recurrence relation for the Partition numbers, see that article for details. Note that the definition of the partition numbers apparently has nothing to do with polygonal or geometric numbers of any kind, so it's really kind of an accident that numbers involved in the theorem were related to a sequence that was already well known. The pentagonal number theorem is important because it's a much easier (if more complex) way to compute these numbers than directly from the definition. (Finding an even easier asymptotic formula was of great interest in the early part of the 20th century, see the section "Approximation formulas" in the article.) The partition numbers have connections to other areas of mathematics such as representation theory. One could, I suppose, define generalized n-gonal numbers for any n in the same way, but afiak there isn't much in the way of applications for them. --RDBury (talk) 16:36, 6 March 2025 (UTC)[reply]

Now you have pointed out that it is p(n) and p(-n), the input sequence has become obvious - well it was before, I just didn't see it (!!!). I now observe that p(-n) = p(n) + n . This can be illustrated by drawing the p(n) pentagons and adding a duplicate row below the bottom edge. Thus

 *       *           *
 o     *   *       *   *
        * *      *  * *  *
        o o       *     *
                   * * *
                   o o o
 1,2    5,7        12,15

Thanks. -- SGBailey (talk) 11:06, 7 March 2025 (UTC)[reply]

Now that I've thought about it some, there are two "natural" geometric arrangements in which both p(n) and p(-n) show up in Franklin's bijective proof of the pentagonal number theorem. These are exactly the arrangements (i.e. Ferrers diagrams) that don't cancel themselves out, so they're the ones that turn up in the generating function. And if you ever want to waste some time with a bit of mathematical doodling I sure you can find many other pleasing geometrical ways to compose and decompose both pentagonal and negative pentagonal numbers. If t(n) = n(n+1)/2 is the nth triangular number, and s(n) = n² then p(-n) = s(n)+t(n) and p(n) = s(n)+t(n-1). These correspond to Ferrers diagrams in Franklin's proof.) Or p(n)+p(-n) = 3s(n), where the right hand side gives you a variation on the Centered hexagonal number where there's a small triangle in the center instead of a single dot:

         * * * *
        * * * * *
       * * * * * *
        o o o o o
         o o o o
          o o o

I don't know if such results are particularly significant, but if you're bored on a rainy spring afternoon... --RDBury (talk) 17:31, 7 March 2025 (UTC)[reply]

Hello good people,

I was reading an article in the New Scientist: 'Amazing' spinning needle proof unlocks a whole new world of maths. It is a proof of the Kakeya conjecture. I hardly understand what the conjecture is, but that is not my question. Something grabbed my attention because I understand all words but do not understand what it means. It was this: "The proof could even help reveal the origin of prime numbers". What is meant by "the origin of prime numbers"? Wimdw (talk) 16:34, 12 March 2025 (UTC)[reply]

I would attach no significance to this choice of phrasing. Rather the Kakeya problem is related to Szemerédi's theorem in additive number theory. Tito Omburo (talk) 16:40, 12 March 2025 (UTC)[reply]

Thank you, but I merely was (and remain) puzzled and don't expect that a mere different choice of phrasing will help to reduce my puzzlement. Wimdw (talk) 11:13, 13 March 2025 (UTC)[reply]

More details can be found in Erdős–Szemerédi theorem. The core point is that there is superficially some relationship between the Kakeya problem and arithmetic progressions, but these superficial connections turn out to run very deep indeed. Tito Omburo (talk) 11:27, 13 March 2025 (UTC)[reply]