math

For those who don't know this book of Topology is rated very positively and Free ! Some professors teaching this branch of Mathematic may be able to confirm this.

Topology is an important and interesting area of mathematics, the study of which will not only introduce you to new concepts and theorems but also put into context old ones like continuous functions. However, to say just this is to understate the significance of topology. It is so fundamental that its influence is evident in almost every other branch of mathematics. This makes the study of topology relevant to all who aspire to be mathematicians whether their first love is (or will be) algebra, analysis, category theory, chaos, continuum mechanics, dynamics, geometry, industrial mathematics, mathematical biology, mathematical economics, mathematical finance, mathematical modelling, mathematical physics, mathematics of communication, number theory, numerical mathematics, operations research or statistics. Topological notions like compactness, connectedness and denseness are as basic to mathematicians of today as sets and functions were to those of last century.

For the reader who has not previously studied an axiomatic branch of mathematics such as abstract algebra, learning to write proofs will be a hurdle. To assist you to learn how to write proofs, in the early chapters I often include an aside which does not form part of the proof but outlines the thought process which led to the proof.

Since 1985 with the last update dating from this year... Amazing commitment !

The music's really good also.

I really hoped Matt Parker would have shown how to construct these divisibility rules, so I came up with my own method. Find prime P and natural number N such that P*N = a *10^b + c
The smaller a and c are, the better. b determines where you split the number.

Example: P=313 N=16 a=5 b=3 c=8

313*16=5008  
313*16=5 *10^3 + 8  

Now we can test for 223795=313*13*11*5
Split 223795 after the b:th number, 3rd in this case.
223795/10^3=223.795 decimal point separates the components A, B

Multiply the A=223 by c and subtract from the rest B=795 multiplied by a

B*a-A*c=  
795*5-223*8=2191  

Repeat if needed till you get to small enough number.

2191/10^3=2.191  
191*5-2*8=939  

which is easy to see that's 3*313
Some bad combinations don't reduce the starting number but they are at least always divisible by P. Those cases could be called Parker divisibility rules.

You can also see that when c is negative, A*c is added rather than subtracted, which explains why some method add or subtract like Vsauce vs James.
This is just a funny trick to simulate division and modulus by 10^b to get smaller number, while preserving the congruence.

It's possible I made mistake somewhere, but was able to get correct answers with other few examples.

So what's the problem.

eBay, purveyor of buying and selling the stuff from your Grandma's house, or from China. Has a IMHO terrible fee structure which works like this.

You sell the item you want (S), then it gets taxed (T). That taxed sale then has two additional fees applied to it, one for transaction (E), another for international (I). Finally a static fee is needed for the listing itself (F). Oh and all of this is taxed.

This results in the following Formula to calculate how much you will actually make from a sale (P).

S - ST - ([ES*(1+T)])T - ([IS*(1+T)])T - F(1+T) = P

This is stupid, and I shouldn't be taxed on my tax, but whatever. I just need to math my way out of this. It took all evening but I was able to take the formula above and isolate S

P+F+F(1*T)

------------------------------------------- = S

1-T-E-E*(2T+T^2 )-I-I(2*T+T^2 )

The result of this is a sale prince which accounts for eBay's fees

I've attached a screenshot to my excel sheet to prove that it works. I'm going to bed.

I have been exploring this particular prime 13238717 which is sum of two squares and has Pythagorean triple.
I found this interesting property and so far I haven't found any texts about what I wrote below.

This is just my conjecture, I have no formal proof and I have only tested few small primes.
I haven't found any counter examples yet, but I have checked only few dozen primes and couple composites by hand.

I modified ChatGPT script which lists numbers that have both forms P^2=a^2+b^2 and P=c^2+d^2 and it appears to generate the exact same sequence as: A004431
5 10 13 17 20 25 26 29 34 37 40 41 45 50...

Numbers P seem to always have both following forms Hypotenuse numbers (Pythagorean triples) A009003: P^2=a^2+b^2
Numbers that are the sum of 2 squares A001481: P=c^2+d^2

Wolfram notes, "-- one side of every Pythagorean triple is divisible by 3, another by 4, and another by 5."
I noticed if P is prime and have both forms: one of the Pythagorean sides (a or b) whichever is divisible by 4 has the exact factors that construct both of the square sum components 'c' and 'd', with the exception of extra factor '2'.

Here's the conjecture put out more formally based solely on my observations:
Pythagorean sides: a,b and square sum componentsc,d are natural numbers and n#, m# are prime factors.

a^2 + b^2 = P^2 (pyth. triple)  
c^2 + d^2 = P (sum of two squares) 

(a or b) mod 4 = 0
(a or b) factors are = 2* (n1*n2*n3...) * (m1*m2*m3...)
c = n1*n2*n3...
d = m1*m2*m3...
(a or b) = 2*(c*d)

Here's couple examples:
primes 2 and 5 are trivial exceptions as 1 isn't a prime factor.

1^2+2^2=5  
1^2+1^2=2

Prime: 13 (first non-trivial prime case)

5^2 + 12^2 = 13^2 (pyth. triple) 
2^2 + 3^2 = 13 (sum of two squares)  

12 factors are 2 2 3
12 mod 4 = 0
c=2
d=3
12 = 2* (2*3)

Prime: 821

429^2 + 700^2 = 821^2 (pyth. triple)  
14^2 + 25^2 = 821 (sum of two squares)  
  
700 mod 4 = 0
700 factors are 2 (5 5) (2 7)
c = 5*5 = 25
d = 2*7 = 14
700 = 2* (25*14)

prime: 13238717

1315508^2 + 13173195^2 = 13238717^2 (pyth. triple)
181^2 + 3634^2 = 13238717 (sum of two squares)  
  
1315508 mod 4 = 0
1315508 factors are 2 (2 23 79) (181)
c=181
d=2*23*79=3634
1315508 = 2* (181*3634)

Some composites have multiple ways to write sum of two squares, which each have different (a or b) counterpart, but not necessarily divisible by 4. composite: 260

(four valid pythagorean side pairs)  
132^2+224^2 = 64^2+252^2 = 100^2+240^2 = 156^2+208^2 = 260^2 
(two valid square sums)
8^2+14^2 = 2^2+16^2 = 260

8^2+14^2:
224 mod 4 = 0
224 factors 2 (2 2 2) (2 7)
c= 2*2*2 = 8
d= 2*7 = 14
224/(8*14) = 2
  
2^2+16^2:
64 mod 4 = 0
64 factors 2 (2) (2 2 2 2)
c = 2
d = 2^4 = 16
64 = 2* (2*16)

composite: 58

40^2+42^2=58^2 (pyth. triple)  
3^2+7^2=58  (sum of two squares) 

42 mod 4 = 2 
Not 0 mod 4 congruent, unlike primes. 
Might be result of both c, d being odd.
42 factors 2 3 7
c=3
d=7
58 = 2* (3*7)

Generally it seems that there's always at least one Pythagorean component where (a or b) = c*d*2, but I haven't quite figured why this is the case.

I reckon it has something to do with the fact that mod 4 congruence of 4k+1 doesn't change when you square it: (4k+1)^2 = 8k*(2k+1)+1
Additionally the fact that when sum of two squares is prime or odd, exactly one of the components is always odd, which may explain why (a or b) isn't always divisible by 4 with composites.

ChatGPT wrote a counterexample finder I checked up to 100 000.

To my knowledge there isn't straight up equation that would spit out a Pythagorean triple or sum of square solutions for any integer.
There might be some other way to prove or disprove that a or b = c*d*2 when P is prime, but so far I couldn't think of any.
It might be something obvious that I am missing, or it's simply all about congruence rules of additions and multiplications.

Again, this is just what I've found from my few tests and I don't have any formal proof. I couldn't find any papers or posts specifically about this. This is nothing too important, but I found it interesting enough to share.
I just write bad python code out of interest in number theory without an university degree.

Thanks for the solution goes to: @0v0
Here's summary how I understood this:
Brahmagupta–Fibonacci identity

P=c²+d²
Squaring 'P' results in:
P²=(c²-d²)² + (2cd)²

These 'P²' sum components are equivalent to the Pythagorean legs 'a' and 'b':
a=c²-d²
b=2*c*d

Hence, '(a or b)' always contains the factors '2' and all factors of 'c' and 'd'. 

Additionally, the 2*c*d divisibility by 4 is result of P being odd.
2*c*d being divisible by 4 is true for any odd number as one of the sum of square components 'c' or 'd' must be even.

c=2*k
b = 2*c*d = 4*k*d

In which Vihart constructs a 6-pentagon polylink out of (U.S.) Smarties. This candy as called "Rockets" in the U.K., Canada, and most other places.

One of the most interesting uses of diffusion models I've seen thus far.

Main video: https://www.youtube.com/watch?v=Gojd8mTl3Do

Extra video: https://www.youtube.com/watch?v=d5IMSxRgeZk

The second instalment of Vihart's recent scutoid series. I hope there are more to come.

One of the OG mathematics communicators on YouTube is back.

This site contains the mathematical archives of Alexander Grothendieck from 1949 to 1991. There you will find manuscripts, "typescripts", sometimes printed documents, in a classification essentially faithful (when it existed) to the author's classification . Of the 28,000 (approximately) pages in this collection, only 18,000 (approximately) are currently accessible

I hope some of us will benefit from his knowledge.

Related publication:

• Global positioning: The uniqueness question and a new solution method
• https://doi.org/10.1016/j.aam.2024.102741

Open source repo in link, and most recent patch was just a few months ago! Single player had several nice puzzles, took a nice 30 minutes. Includes 2 player mode and a level creator.

Note it's 'hold leftclick and swipe' to cut an edge in the browser/pc.

You may have seen that cool new arXiv feature 'experimental HTML' - this is about stuff like that! Latex (and hence a lot of math research) is not well suited to screen readers, but HTML is. If you'd like to learn more about how your paper can be in the format, or just about how to make research more accessible, this could be useful!