What's the next freak identity? A new deep connection with Sophie Germain primes

[@Mathologer]

Let's collect some interesting comments and remarks as they come in (also check out the description of this video): VincentvanderN I don't know if this is already down here somewhere in the comments, but 2 x 2 = 2 + 2 can be used to show that there are infinitely many prime numbers. We generate a sequence a_1, a_2, a_3 etc by starting with a_1 = 3 and a_{n + 1} = a_n^2 - 2. Now we use our freak equation to show that if q is a prime divisor of some a_m, it cannot be a prime divisor of a_n for any n > m. (And then, as a result neither of an a_n with n < m, because otherwise we could repeat the proof with n in the role of m and get a contradiction.) Here is the argument. Look at the sequence a_m, a_{m+1}, ... modulo q. We have a_m = 0 mod q, a_{m+1} = - 2 mod q, a_{m + 2} = 2 mod q, and then, by the freak equation 2 x 2 = 2 + 2 (in the form 2 x 2 - 2 = 2) we get that a_{m + k} = 2 for all k >= 2. Neat, right? I believe I learned this from Proofs from the Book. charlesstpierre9502 Use 2x2=2+2 to make a formula for generating Pythagorean triples. Start with two positive integers m > n. Then (2×2)(mn)² = (2+2)(mn)² (2×2)(mn)² = 2(mn)² + 2(mn)² (2×2)(mn)² = 2(mn)² + 2(mn)² + (m⁴ - m⁴) + (n⁴ - n⁴) (2×2)(mn)² = m⁴ + n⁴ + 2(mn)² - m⁴ - n⁴ + 2(mn)² (2mn)² = (m² + n²)² - (m² - n²)² Set: a = m² - n² b = 2mn c = m² + n² Then: a² + b² = c² This will not work for: aᵏ + bᵏ = cᵏ; k > 2 franknijhoff6009 Hi Burkhard, the 3-variable equation plays a role in integrable systems. In fact, it appeared in Sklyanin's work on quadratic Poisson algebras (around 1982) providing solutions in terms of elliptic functions, and (with an extra constant) as the equation for the monodromy manifold of the Painleve II equation. Sklyanin's paper is: Some algebraic structures associated with the Yang-Baxter equation. Functional.Anal.i Prilozhen 1982 vol 16, issue 4,pp 27-34 ; look at equation (27). Furthermore, in L.O. Chekhov etc al., Painleve Monodromy Manifolds, Decorated Character Varieties and Cluster Algebras , IMRN vol 2017, pp 7639--7691 you can find in Table 1 a close variant of the 3variable equation with extra parameters. I think it would be interesting to explore this idea with exponents. For example 1+1+1+1+2+3=3^2^1^1^1^1 2¹×2¹=2¹+2¹ 3½×3½×3½=3½+3½+3½ 4⅓×4⅓×4⅓×4⅓=4⅓+4⅓+4⅓+4⅓ 5¼×5¼×5¼×5¼×5¼=5¼+5¼+5¼+5¼+5¼ ... tanA + tanB + tanC = tanA x tanB x tanC (that's the identity that features prominently in the Heron's formula video) For completeness sake and for fun let's mention 2⁴ = 4² 2 + 2 = 2 × 2 = 2² log(1+2+3)=log(1)+log(2)+log(3) log(2+2)=log(2)+log(2)=2log(2) www.mtai.org.in/wp-content/uploads/2023/09/IOQM_Sep_2023_Question-paper-with-answer-key.pdf ... 29th question in an Indian maths olympiad problem Check out this paper by Maciej Zakarczemny, On the equal sum and product problem (linked to in the description of this video) www.iam.fmph.uniba.sk/amuc/ojs/index.php/amuc/article/view/1662 Theorem 5.1. says If there are exactly two sum-equals-product identitites , then one of the following two conditions is true: 1. n - 1 is a prime and 2n - 1 \in { p, p^2, p^3, pq } , 2. 2n - 1 is a prime and n - 1 \in { p, p^2, p^3, pq } , where p, q denote prime numbers. Why don't we start out the sequence of basic sum-equals-product identities with 1=1? Well, N=N for all N :) Very tempting to add the 1=1 at the top. To not do so is very much in line with not calling 1 a prime (saves you a lot of unnecessary heart ache further down the line :) For this video I played around with generating some AI animated photos of Sophie Germain using www.myheritage.com/deep-nostalgia as well as with the ChatGPT generated rendering of Sophie Germain in the thumbnail. Here I fed the AI some of the existing pictures of Sophie Germain and asked it to play artist to come up with a picture of her in the same style used in this real picture of Leibniz www.alamy.com/gottfried-von-leibniz-image5071937.html). Lots of fun and quite eerie :) Not perfect, but will be interesting to find out what the machines can do in 6 months time (and at what point the machines can make better Mathologer videos than the Mathologer :( In general quite a bit of AI hate in this comment section. Strangely I've never had people complain about me using any of those generic white men with beards images of ancient mathematicians that museums are full of. Sophie Germain primes and safe primes (the 2n+1 primes) are important in cryptography, those involving prime fields. Because if the size field is a prime p, the size of its multiplicative group is p-1. For the group to be secure, the multiplicative group must have a large subgroup of prime size so p-1 must have a prime factor. So, having a prime q such that 2q+1 is also a prime is a good candidate. Even though it is not strictly required as long as p-1 has a big enough prime factor, this is what students are taught as a start. Kaprekar's constant is among the lengths corresponding to exactly two sum-equals-product identities (discussed at the end of this video :) en.wikipedia.org/wiki/6174 Cunningham chains are an interesting generalisation of Sophie Germain primes (chains of primes such that the next prime in the chain is always the previous one times 2 plus 1). en.wikipedia.org/wiki/Cunningham_chain If you are happy to also play this game with negative integers then you get more solutions. In particular, since (-1)+(-1)+1+1=0 and (-1)x(-1)x1x1=1, you can splice these two blocks into a/any sum-equals-product identity of length n to arrive at a sum-equals-product identity of length n+4. And for complex integers we've also got things like this: (1 - i) + (1 + i) = 1 - i + 1 + i = 1 + 1 -i + i = 2 + 0 = 2 = 1 + 1 = 1 - (-1) = 1 - i^2 = (1 - i)(1 + i) 2xy - (2+x+y) +1 = N-2 2xy - 2 - x - y +1 = N-2 2xy - x - y +1 = N 4xy - 2x - 2y +2 = 2N 2x(2y - 1) - 2y +2 = 2N 2x(2y - 1) - (2y - 1) = 2N-1 (2x - 1)(2y - 1) = 2N-1 Probably the easiest way to demonstrating this equivalence is to go backwards and start by expanding (2x-1)(2y-1)... Relevant Project Euler problem: 88 projecteuler.net/problem=88 Relevant Online Encyclopedia of integer sequences: A033178 @Frafour Tangentially related to 2x2 = 2+2: there is this video of Gromov (Gromov: 4 = 2 + 2 as "proof of Donaldson's theorem") kzbin.info/www/bejne/mJjIgYCYbdyrrJIsi=DmTf2wzbIr1F21f8 observing that 4 = 2+2 in 3 ways. That is, there are 3 ways to partition a 4-element set in two 2-element subsets. The fact that 3 is smaller than 4 is unique to 4, and produces a surjection from A_4 to A_3 - the alternating groups on 4 and 3 elements. This shows that A_4 is not simple, 4 is the only number where this happens, and this is responsible for many weird things in dimension 4. In another direction, I first encoutered Sophie Germain primes accidentally when learning group theory in my undergrad. There is an elementary proof of simplicity of the Mathieu groups M_11 and M_23, which actually gives a general simplicity criterion for subgroups of S_p, p prime www.jstor.org/stable/2974771?origin=crossref. When I read this I wondered if this ever works for proving simplicity of the alternating group A_p. If I remember correctly, it works precisely when p is a Sophie Germain prime! @YSCU261 The roots of a quadratic equation of the form x^2-bx+c=0 satisfy the following equations : r1+r2=b r1*r2=c So we have that the solutions of xy = x+y can be expressed as the solutions to the quadratic equation x^2-bx+b for all b this can be extended to x+y+z=xyz and so on with vieta's formulas

[@michaelheeren5845]

Adding 1 = 1 at the top would be very satisfying. Length of 1. Justified with the sum if a single number = product of a single number.

[@jaromir.adamec]

I'd guess it's because also eg. 13=13 or 423=423 ... so, it would defy any further pattern... 😅

[@Mathologer]

Correct. Very tempting to add the 1=1 at the top. To not do so is very much in line with not calling 1 a prime (saves you a lot of unnecessary heart ache further down the line :)

[@Xanthe_Cat]

You can call 1 a prime of inconvenience - most of the time you would be justified in regarding 1 as not prime except in rare instances, such as I was reminded of recently, exempli gratia, back in 1732 Euler gave a list of Mersenne exponents starting with 1…

[@mscha]

Why stop there? How about length 0? Unfortunately, that won't work. With 0 terms, you'd get 0 = 1 (using the additive and multiplicative identities.)

[@aaaaaa8410]

@@Xanthe_Cat 1 is a REALLY inconvenient prime, when writing down prime factors. Also it would make itself not a prime because of 1 = 1*1*1*1*1....

[@phyarth8082]

log(1+2+3)=log(1)+log(2)+log(3)=log(1*2*3)

[@jiaan100]

Having the third part is silly. The second and third is just a log rule, but even worse the third gives up the joke

[@sheikhalsumaiya7230]

What a weekend! 3b1b + mythologer in the same weekend!

[@SobTim-eu3xu]

Geniuses think the same)

[@ಭಾರತೀಯ_ನಾಗರಿಕ]

Mythologer?? 🤔 Is that supposed to be a pun? Or just autocorrect?

[@howchess7631]

Mathologer​@@ಭಾರತೀಯ_ನಾಗರಿಕ

[@chriswebster24]

Best weekend ever! Ha ha

[@blakelapierre]

ok, but where is the explain of almost prime in the video description

[@ianstopher9111]

Speaking of Sophie Germain, my son and wife were looking up facts about the number 47, when I mentioned that 47 is the end of the Cunningham chain of the first kind with the smallest Sophie Germain prime as the initial number.

[@authenticallysuperficial9874]

As one does

[@ianstopher9111]

@@authenticallysuperficial9874 I just so happened to reinvent Cunningham chains while awake in bed. I did not know about them 2 weeks ago, but read up on them when I discovered I was pipped by a few centuries.

[@evanev7]

always something uncanny about that ai generated yassified sophie germain

[@MetalMint]

Her hair isn't even curly 😭

[@shoam2103]

This is almost click bait. Actually, I was curious if it was a female or just a male mathematician with long curly hair and feminine features. So I actually clicked to check the name. Does that constitute click bait?

[@shoam2103]

0:45 "start of infinite identities" whatever the case, this is going to be an interesting video!

[@nightytime]

​@@shoam2103it might be, but not to the point of necessarily being malicious. Sophie Germain is a woman.

[@Mathologer]

For this video I played around with generating some AI animated photos of Sophie Germain using www.myheritage.com/deep-nostalgia as well as with the ChatGPT generated rendering of Sophie Germain in the thumbnail. Here I fed the AI some of the existing pictures of Sophie Germain and asked it to play artist to come up with a picture of her in the same style used in this real picture of Leibniz www.alamy.com/gottfried-von-leibniz-image5071937.html). Lots of fun and quite eerie :)

[@heathrobertson2405]

love to see a video about Sophie Germain's work, she doesn't get nearly enough attention

[@Wecoc1]

That's true, the primes page on Wikipedia doesn't exist in any other language yet.

[@mekaindo]

​@@Wecoc1wait really?

[@VincentvanderN]

I don't know if this is already down here somewhere in the comments, but 2 x 2 = 2 + 2 can be used to show that there are infinitely many prime numbers. We generate a sequence a_1, a_2, a_3 etc by starting with a_1 = 3 and a_{n + 1} = a_n^2 - 2. Now we use our freak equation to show that if q is a prime divisor of some a_m, it cannot be a prime divisor of a_n for any n > m. (And then, as a result neither of an a_n with n < m, because otherwise we could repeat the proof with n in the role of m and get a contradiction.) Here is the argument. Look at the sequence a_m, a_{m+1}, ... modulo q. We have a_m = 0 mod q, a_{m+1} = - 2 mod q, a_{m + 2} = 2 mod q, and then, by the freak equation 2 x 2 = 2 + 2 (in the form 2 x 2 - 2 = 2) we get that a_{m + k} = 2 for all k >= 2. Neat, right? I believe I learned this from Proofs from the Book

[@Mathologer]

This is super neat. Thanks for sharing! That's exactly the kind of insight I was fishing for.

[@VincentvanderN]

​@@Mathologer I've been in love with the 2 x 2 = 2 + 2 equation since childhood. As a more adult mathematicians I have been looking for 'categorizations' of this equation, i.e 2-dimensional vectorspaces V where V \tensor V = V \oplus V in some natural, or meaningful way. So far I came up emptyhanded. The two natural candidates: the general equation V \tensor V = S^2 V + \bigwedge^2 V applied to the dim V = 2 case and the decomposition into irreducible representations of V \tensor V in case V is the defining rep of the group SL(2) both reduce to 2 + 2 = 3 + 1. (One can argue that they are in fact the same example, just viewed as special cases of different general phenomena). I find it quite annoying how 2 + 2 = 3 + 1 seems to beat 2 x 2 = 2 + 2 to the punch every time, but maybe you or one of the other readers knows an example. I have one example that sort of works, but in an artificial way. View C as a two-dimensional R-algebra. Then, if I am not mistaken, C \tensor_R C is isomorphic (as an R-algebra) to Mat(2, R), so simple and hence we get 2 x 2 = 4 rather than 2 + 2. However. If we interpret C \tensor_R C as a C-algebra in the same way that we can interpret C \tensor_R A as a C-algebra for any R-algebra A, then we get, as a C-algebra that C \tensor_R C = C \oplus C. So in a way this is an example of 2 x 2 = 2 + 2, although here it is perhaps fairer to say that we have an example of 2 x 2 = 1 + 1 for some very fat notion of 1

[@Mathologer]

Interesting and fun :) "I've been in love with the 2 x 2 = 2 + 2 equation since childhood." Same here :)

[@julianpilbrow4963]

I think your math tutorials are excellent. My anecdotal story is that while copying a maths table across while sitting in Epsom Library wearing a helmet and a red cape, a Korean mother of a little girl who was sitting nearby, looked me in the face and called me "Asshole!"

[@whenthingsfly4283]

A new mathologer video? Awesome :D

[@Naomi_Boyd]

I have a nice little puzzle for the mathologers out there. Inscribe a right angle triangle within a circle and use rotational symmetry, along with circle theorems, to derive the quadratic formula. I did this completely by accident and was very pleased with the result. It is not a very difficult puzzle, but I think it is quite fun and a good lesson about how a conversion of geometry can be used to restate a difficult problem.

[@tomasebenlendr6440]

9:21 Let's find length with more than 10 identities of type 1+...+1+A+B=A*B*1*....*1 (that turns out to be relatively easy): We need just A*B-A-B = const for sufficiently many pairs of A and B. So we rewrite A*B-A-B = (A-1)*(B-1)-1. Thus we take some number with many divisors (M) and take all different pairs of numbers C and D, such that C*D = M. Then we take A=(C+1) and B=(D+1) and we get indentities of length A*B-A-B + 2 = C*D + 1 = M + 1. EDIT: should keep watching before posting, this is exactly what is explained later (around 14:45)

[@Mathologer]

Very good anyway :)

[@User_2005st]

Mr. Burkard Polster, Mathematician, I really say from the bottom of my heart, you are a very valuable analyst in explaining the theorems of the mathematical world. Every time I watch your videos, a window opens to me deep into the infinite world of mathematics. Your 19 yrs fan :')

[@Mathologer]

Glad you think so and thank you very much for saying so :)

[@SaturnCanuck]

Always love your videos, a great sunday afternoon

[@MrCheeze]

Combo Class has a video as well on the {2,2}, {1,2,3}, {1,1,2,4}, ... infinite family. But they're very different and complement each other nicely.

[@ComboClass]

Thanks for the mention! I love when I stumble across a comment like this, especially on a channel as great as Mathologer :)

[@Mathologer]

Combo class, that's a channel I had not encountered before. Do they say anything of substance that I don't cover?

[@MrCheeze]

@@Mathologer The video's called "These Simple Equations Are Levels of an Infinite Pattern" - somewhat different target audience, so it's mostly a slower exploration of finding the identities themselves. He does point out that it's the "integers only" restriction that makes there only be finitely many solutions for each size.

[@Mathologer]

Just watched your video. Great fun :)

[@Mathologer]

Thanks for that. Just watched the video. Great fun :)

[@franknijhoff6009]

Hi Burkhard, the 3-variable equation plays a role in integrable systems. In fact, it appeared in Sklyanin's work on quadratic Poisson algebras (around 1982) providing solutions in terms of elliptic functions, and (with an extra constant) as the equation for the monodromy manifold of the Painleve II equation.

[@Mathologer]

That's great and really the first really interesting response to my request. Would you have a reference?

[@franknijhoff6009]

Sklyanin's paper is: Some algebraic structures associated with the Yang-Baxter equation. Functional.Anal.i Prilozhen 1982 vol 16, issue 4,pp 27-34 ; look at equation (27). Furthermore, in L.O. Chekhov etc al., Painleve Monodromy Manifolds, Decorated Character Varieties and Cluster Algebras , IMRN vol 2017, pp 7639--7691 you can find in Table 1 a close variant of the 3variable equation with extra parameters. BTW it seems we met in 2000 in Adelaide.

[@Mathologer]

That's great, thank you very much !!!

[@Mathologer]

I moved to Melbourne (from Adelaide) in 2000, but, yes, very possible that we've met there. Maybe at an event organised by Nalini in honour of Martin Kruskal? Did you stay in Kathleen Lumley College?

[@TheMichaelmorad]

to check how many sum=prod identities are there for a given length n, you first check what is the maximal number of non-one numbers in your identity. you do this by taking a list of n ones and seeing how many ones can you replace by twos s.t. the product is less than the sum. now you manually check cases for each number of non-1 numbers below the suprimum. so for 10 you've got 2 identities: 4,4,1,1,1,... and the trivial case 2,10,1,1,1,...

[@Mathologer]

Very good :)

[@manishadhikari4132]

Sophie Germain primes and safe prime are important in cryptography, those involving prime fields. Because if the size field is a prime p, the size of its multiplicative group is p-1. For the group to be secure, the multiplicative group must have a large subgroup of prime size so p-1 must have a large prime factor. So, having a prime q such that 2q+1 is also a prime is a good candidate. Even though it is not strictly required as long as p-1 has a big enough prime factor, this is what students are taught as a start.

[@Mathologer]

Nice insight, thanks for sharing :)

[@TinySpongey]

Copying this from a comment I added to the "last call" message it's probably more appropriate here. Don't know if this one had been mentioned before but... the expression X+Y=XY reminded me of something. Rearranging it you get XY/(X+Y)=1 the expression on the left is a formula for the value of two resistors in parallel (or capacitors in series) in electronics. This results in an equivalent problem (for two numbers) of finding two natural numbers such that 1/X + 1/Y = 1. This generalises into a different problem: find a series of natural numbers such that the sum of their reciprocals is 1. Though I haven't looked into it it doesn't seem to be that difficult to characterise all solutions to this. There is always the trivial example of N values of 1/N and cases where N is composite where some can be grouped together for example (4*1/4 and 1/2+1/4+1/4). Are those the only solutions? If we allow infinite series then other possibilities appear: for example (1/2^n) for n=1,2,... When I came to write I got a nagging familiarity somewhere. Let me show you what I mean. Take the reciprocal case with three integers X, Y, Z. I/X+I/Y+I/Z. Give them a common denominator and see what happens... (YX+XZ+ZY)/XYZ. This generalises to more than three integers and each term in the numerator is the product of all but one of the integers. That's when I saw a parallel with Vieta's formulas for polynomials. I think that the orignial video and the case I mentioned above can be characterised as finding equal values for the coefficients of a certain degree of polynomial with integer roots. For example (X+2)*(X+2) = X^2+4X+4. The reciprocal version for 3*1/3 can be exemplified by (X+3)^3 = X^3+3X^2+3X+9. The symmetry of binomial coefficiencts makes it possible to find many answers where the roots are all equal. What is not at all clear is what happens if the roots are not all equal.

[@fredg.sanford634]

Thanks!

[@Mathologer]

Thank you very much !

[@Igor_Zdrowowicz]

"lots of 4's" - I'm a genius! "but that's not it" never mind...

[@Bigandrewm]

It's not quite the same as the addition/multiplication talked about here, but in music theory there is a special class of tone rows that spit out the same result with a specific transposition vs. an inversion.

[@Mathologer]

I was not aware of this. Just read up on it. Very interesting, also mathematically :)

[@seedmole]

It makes sense that 2+2=2x2 is such an important identity, because it's the result of the two main defining traits of regular numbers: the additive identity and the multiplicative identity. And when restricted to integers, it really has a lot of weight.

[@Alan_Clark]

In any triangle, tanA + tanB + tanC = tanA x tanB x tanC.

[@Mathologer]

Yes, check out the video on Heron's formula that I mention in the intro. It's based in part on this formula :)

[@sasha-2574]

simple enough to prove this identity in an elementary trigonometry course

[@Alan_Clark]

@@sasha-2574 True. More difficult would be to prove it using Euclidean geometry.

[@Chrisuan]

you had me at welcome to a new mathologer video

[@seiedmohammadrezafatemi3878]

Watching this video I am convinced that it doesn’t worth for everything to have a pattern. Sometimes happy coincidences scattered around is more elegant

[@Mathologer]

You are not wrong there :)

[@EzraTeter]

I personally think that you should go more into details about Sophie Germain's life. For example, she was inspired by the story of Archimedes death where a Roman soldier speared him in rage when the geometry-obsessed man insisted, "Do not disturb the circles!" She also might have saved Gauss' life when she intervened with a French general in charge of the siege of the city where he lived during the Napoleonic wars. The story of her unveiling with Laplace is also quite interesting.

[@Mathologer]

Well, not much is known about her life and what little is known is just one click away on the internet. My priority/mission in my videos is to talk about things that go way beyond what wikipedia knows or to do a much better job at explaining something well-known than anybody else :)

[@Number_Cruncher]

It's not entirely related. But this identity came into my mind, when I watched your video. For 2x2 matrices there is a cool trace identity tr(MN)+tr(M^{-1}N)=tr(M)tr(N). I think M needs to be an element of SL(2). But otherwise it's generic and it also has this sum product relation.

[@Mathologer]

Maybe also check out the recent Mathologer video on the Power of A+B=AB. At the end I've got a few things to say about matrices satisfying this equation :)

[@QuestforaMeaningfulLife]

"1 is an honorary prime." Love it.

[@Merilix2]

When I was 15 old, I used 2*2=2+2 to teach myself about induction by trying to proof a*a = a+a. This method was usually teached by using successful proof examples but I wanted to get a contradiction.

[@RibbleMaths_YifanDu]

Cool video! Really had to watch it immediately when I saw it!❤

[@Mathologer]

Glad you liked it!!

[@RibbleMaths_YifanDu]

To the great Mathologer: 1. Theorem: Σa_n1 Proof: Πa_n-Σa_n+1=Π(a_n-1)>0 Q.E.D. μ 2. l really like the QEDcat and designed a 2D-QEDcat origami model from duo-color paper!

[@Mathologer]

Very good :) 2D-QEDcat origami model sounds like fun. Can you show me?

[@RibbleMaths_YifanDu]

Hello Mathologer! Here is the link to the 2D QEDcat origami model designed by me: m.youtube.com/@RibbleMaths_YifanDu/community It is posted in my math channel. I learned how to design duo-color origami models in the fantastic books 'Duo-color Origami' and 'Multi-color Origami' by the great Chinese origami artist Mi wu (Chinese name:郭嵩). I also designed a few other color models.

[@Mathologer]

Very good! Thanks for sharing :)

[@3Max]

One question that came to my mind at 9:30 (it definitely wasn't "who cares?" !) -- consider the number of non-1 terms in the identities. For each identity-length N, consider all identities, and find the largest in terms of "how many non-1 entries does it have". Wonder if there's anything there, but couldn't find anything in OEIS... maybe i'll look into this later.

[@MrShogunfish]

I'm not a fan of the transition slides with the slightly twitching eyes, they're unsettling to look at and in general the use of AI makes me uncomfortable.

[@rudyj8948]

Personally I didn't mind the use of Ai art, but I'm this case the slide hurt my eyes to look at

[@Mathologer]

For this video I played around with generating some AI animated photos of Sophie Germain using www.myheritage.com/deep-nostalgia as well as with the ChatGPT generated rendering of Sophie Germain in the thumbnail. Here I fed the AI some of the existing pictures of Sophie Germain and asked it to play artist to come up with a picture of her in the same style used in this real picture of Leibniz www.alamy.com/gottfried-von-leibniz-image5071937.html). Lots of fun and quite eerie :)

[@nealmcb]

I actually really enjoyed the blinking eyes. Well done!

[@onejumpman9153]

This is not quite the same as sum = product, but it might be related: Consider the pair of pairs of numbers (1, 5), (2, 3). The sum of the first pair equals the product of the second, and the sum of the second equals the product of the first. Aside from the obvious (2, 2), (2, 2) and the trivial (0, 0), (0, 0), I can't find any other pairs of pairs of positive integers for which this relationship (ab = c+d and a+b = cd) holds. If we allow negative numbers, then (-1, n), (-1, -n+1) is a general solution. I have no idea what non-integer solutions exist. I have no idea if this is an already-explored topic, or whether it's of any significance. But I find it quite interesting.

[@Mathologer]

Have a look at section 8 in this paper www.iam.fmph.uniba.sk/amuc/ojs/index.php/amuc/article/view/1662/906

[@BritishBeachcomber]

OMG this is math for math sake. I just love pure math. Thank you Mathologer

[@BritishBeachcomber]

OMG I love this identity. It is so obvious, when explained, but not intuitive until then.

[@Mathologer]

Glad that you like this so much :)

[@Fluff-gl6yr]

Yesterday I was scribbling some stuff on paper and I remember thinking “hmm a+b = ab is an interesting equation, I wonder what’s up with it” and then I briefly checked to see if mathologer had any videos on the topic and went on with my life. I check back today, and I’m treated to this uncannily timely video 😂

[@Mathologer]

Lucky you :)

[@sortsome]

11:30, "leave in the comments if you like being kidnapped". OMG, it scalated quickly!!

[@duckimonke]

awesome video! it's nice to see these types of problems be recognised

[@Mathologer]

Glad you can appreciate this sort of off the beaten track video :)

[@wesleydeng71]

As in the Goldbach's conjecture, the larger the even number is there tends to be more ways to write it as sum of two primes. So, it is quite reasonable to assume there will be no more such cases. But just as the Goldbach's conjecture, it will hard to prove vigorously.

[@Idonai]

Thank you for another great video! I was wondering if you would ever do a video about the Collatz conjecture and how such a problem could ever be proven in theory.

[@Mathologer]

Sure, on my list of things to do :) Having said that there are already quite a few reasonable Collatz conjecture videos out there ...

[@knooters]

I remember sniffing at this, inspired by the 4 digit case - which appeared in a Norwegian math olympiad some years ago. I didn't find as much as you did. One little detail I thought was a bit funny, was that you can replace any 4, 6 or 8 with 2*2 = 2+2, 1*2*3=1+2+3 or 1*1*2*4=1+1+2+4=1*1*2*2*2=1+1+2+2+2 to generate new numbers that work . Like, 1*1*1*1*2*6=1+1+1+1+1+2+6, replace 6 with 1*2*3=1+2+3 and shuffle: 1*1*1*1*1*2*2*3=1+1+1+1+1+2+2+3

[@Mathologer]

Thanks for that. That's definitely well worth pointing out :)

[@TheOneThreeSeven]

I have not seen any of these freaky identities in the wild unfortunately, but that was neat! For the group of real numbers under the operation of addition, the exponential mapping is a homomorphism which preserves the group structure but changes the operation to multiplication; and I can't help but wonder if lie groups are hiding in the background

[@Mathologer]

Have not hear from you for a while :) 1+1+1+1+1+1+1+1+1+1+1+3+7 = 7x3x1x1x1x1x1x1x1x1x1x1x1

[@DukeBG]

I like Sophie Germain more than Twin primes. They are actually very similar conceptually to twin primes as sort of a natural pairing. Instead of +2 it's kinda x2. Of course, you cannot just multiply by a factor, you have to add a bit to get to a prime (same way with twins, they're not just the next integer, just n+1, you have to make another +1 to get to a possible prime). SG primes are most beautiful in binary. Just shift everything left and add another one in the end (and primes in binary always end in 1 anyway). Then there are SG primes of second kind, 2p-1 instead of 2p+1... Same thing in binary, but while adding another one in the end, change the one that was there to zero.

[@Vodboi]

Amazing video (I am assuming) EDIT: It was indeed a good one

[@Mathologer]

Glad you're prediction came true for you :)

[@ov8857]

I'd really like to hear your take on any mathematical relationships between brain wave states and sound phenomenon, solfeggio frequencies etc.

[@Mathologer]

Have to admit I don't know anything about all this. Any good writeup that you are aware of?

[@maze7474]

2 Observations I have: a) how would it look if we would pattern upwards (i.e. 2+1 =? 1x2)? Clearly the equation is not correct, but that's because we can't remove any further one from the left side. But what if we could? Would that correspond tot he -1? b) Also all equations seem to give a special meaning to 2 (which also visible in your p sequence, which reduces the 8 solutions down to 7). Which leads me to the question to whether there is a relation between the 7 and the 49? Are there then 7^3 solutions for 3 (special sum equals product identities)? Lovely video... now I have even more questions than I had before watching it :-) Well done!

[@Mathologer]

a) Well you can change the rules of the game and, for example, allow the numbers you are playing with to be all integers or all complex integers, or all rational numbers, or ... Depending one what you do there is definitely more and different fun to be had. E.g. since (-1)+(-1)+1+1=0 and (-1)x(-1)x1x1=0 you can splice these two blocks into an sum equals product identity to produce a longer (by 4) sum-equals-product identity. b) Also all equations seem to give a special meaning to 2. Yes, and that's mainly due to another other small number "feakishnesses". In particular, (x-1)(y-1)=N-1 and (2x-1)(2y-1)=2N-1 are special in that they automatically translate into algorithms that applies to all N. E.g. 2N-1 is automatically odd and therefore all factors are also odd. And this means that we can always solve for x and y.

[@constexprThoughts]

I loved the video, great as always!

[@Mathologer]

That's great :)

[@nefld3849]

Magnifique vidéo j’ai essayé Marty and Al to follow you but at the end its mathematical fellow ahah thanks a lots

[@seventhtenth]

5:41 electromagnetic wave functions superimpose up by multiplication (Psi) (or sum of intensity) and addition (Euler's Theorem) (or multiplying of wavevectors (momentum))... I probably remember unclearly...

[@Mathologer]

Would be great if there was something there :)

[@jorgelotr3752]

19:09 the Ns are also multiples of 3 and 3 (except the forst three, which are multiples of either 2 or 3). Coincidence?

[@nickyhaflinger]

So 2+2=2x2 is the start of a different infinite identity. Taking the Ackerman extension of arithmetic where exponentiation is repeated multiplication and tetration is repeated exponentiation. Using the notation [1] = + and [2] = x and [3] = ^ and so on we get the infinite identity 2[1]2=2[2]2=2[3]2=2[4]2=2[5]2... and if you use diagonalization you can extend this identity into the ordinals.

[@Mathologer]

Interesting idea but isn't (using your notation) 2[4]2=2^2^2 which is not equal to 2[1]2=2[2]2=2[3]2=4?

[@nickyhaflinger]

@@Mathologer Not at all. The pattern for the Ackermann operators is X [n] Y is Y Xes interspersed with [n-1] and [1] is just addition. So 2[4]2=2[3]2 or 2^2 =2[2]2 or 2x2 =2[1]2 =2+2 =4. 2^2^2 would be 2[4]3. I expect the reason that arrow and chain notation focus on 3's is this exact operator fix point meaning that using 1s and 2s are both disappointing.

[@agostinhooliveira5781]

Great video as always. And I noticed Euler is one of your Patreons! 😊

[@alexanderstohr4198]

some sort of results limited auto-generator might be - and also works as some sort of a proof... this: sum(1, for 1 to N-2) + 2 + N = N x 2 x 1's where sum(1, for 1 to N-2) + 2 = N or merging the 2 into the sum: sum(1, for 1 to N) = N thus the first formula is reduceable to: N + N = N x 2 (yes, it is just the basic equivalent of adding a number to itself to multiplying the number by 2.) Examples: 3 + 3 = 3 x 2 6 + 6 = 6 x 2 ;-)

[@Mathologer]

Finally got around to parsing this comment. Nice over the top proof that N+N=2N :)

[@subhranshushekharpatra7198]

Please make the video on e^ (gamma) gamma is euler's special number you were talking about ,and what is solution for continued fractions containing e^(gamma) You had already said to make one but i didn't find it Please I'm starving for it😂

[@JonatasMiguel]

I had looked into the first identity "(mod n)" for fun. It seems that only prime numbers have solutions involving x and y from 2 to n. Further, if you take the sum of all of these sums, for odd primes, you get 1 (mod n). There are also some interesting symmetries that arise when producing a table of these solutions as well, likely due to the inherently symmetrical form of these expressions. I haven't looked into the other "freaky identities" "(mod n)" yet to see if similar results arise.

[@JonatasMiguel]

Another detail that seemed unique to primes was around the number of unique sums/products for those expressions (mod n), for prime n it always seemed to be (n + 1)/2

[@JonatasMiguel]

Ah, I worded something a bit awkwardly in the first message... Only prime n seem to have solutions where *all* numbers from 2 to n are part of some solution of the expression.

[@josephyoung6749]

Wonder what the visual implications of these identities can be considering multiplication can often be represented as an area of a picture while adding can be represented as a lineament in the same picture... Could be a worthwhile novelty artistic thesis

[@Mathologer]

In the intro sequences I attempt something like this: two columns of 2 dots with a plus sign in the middle collapsing into a square made up of four points, with the plus sign turning into a times sign. Not the greatest, but the best I could think of :)

[@BrianOxleyTexan]

Why is 1 = 1 excluded? It looks like a base case of length 1. If you picture addition and multiplication as functions rather than infix operators, +(1) is just as valid as +(2, 2).

[@Mathologer]

Well, N=N for all N :) Very tempting to add the 1=1 at the top. To not do so is very much in line with not calling 1 a prime (saves you a lot of unnecessary heart ache further down the line :)

[@ShankarSivarajan]

Because you'd then have to include 2 = 2, 3 = 3, 4 = 4, and so on, and while those are all of course true, they provide no interesting insight into anything.

[@Snillocad143]

This is a marvelous topic.. Also, conjecture should only be a noun. Bring back "conject"..

[@HakoHak]

19:12 One is not a prime ?? 1 ?! The unit building block of all others ? What can be primer than literally what's under the symbol 1 or I which builds all the other numbers, even primes ?

[@piyushchoudhury3709]

Hello sir you videos has been a delight, recently i have studying perron Frobenius Theorem ,Its proof has beautiful concepts behind it and ton of applications

[@Mathologer]

Yes, very beautiful and powerful mathematics :)

[@spiderjuice9874]

7:22 "... between the smallest length 2, ..." Hold on: if we look at the triangular chart (6:33) then it is clear to see that the sum equals product identity of length 1 is possible, namely 1 = 1 (and also x = x, but when we use 1 it fits the pattern nicely). Yes, it's trivial.

[@Mathologer]

Very tempting to add the 1=1 at the top. To not do so is very much in line with not calling 1 a prime (saves you a lot of unnecessary heart ache further down the line :)

[@spiderjuice9874]

@@Mathologer Yes, that's the other side. Oh, decisions, decisions!

[@adamcionoob3912]

In the beginning of the list of these primes I found a chain: 2, 2*2+1=5, 5*2+1=11, 11*2+1=23, 23*2+1=47. I wonder if there are more and longer chains among the Sophie Germain's primes.

[@Mathologer]

Well, spotted. Yes there are. Google "Cunningham chain"

[@solarwonder]

it occurs to me that the 2+2=2x2 and 1+2+3=1x2x3 correspond to triangular graphs. there is also the next triangular graph, 10, in the case of 1+1+2+4=1x1x2x4. considering the rich properties of pascal's triangle, i'd wager there is at least one more way to find a pattern that produces a closed form, involving triangular graphs. this might be derived in reverse from relevant identities that involve infinite sums and infinite products.

[@nanamacapagal8342]

CHALLENGES! 9:12 Find all identities of length 10. (1 * 8) + 2 + 10 = 10 * 2 * (1 * 8) (1 * 8) + 4 + 4 = 4 * 4 * (1 ^ 8) Brute force checking 3 non-padding-1s didn't work, and it's easy to prove that checking any more than that is impossible. 9:22 Find a length N that has more than 10 identities. Just look at the graph from earlier, there's plenty of such N plotted above y = 10 ;) On a more serious note, consider the base case that all but two of the values are padding 1's. The equation is now (N-2) + A + B = AB. Rearrange a little bit... AB - A - B + 1 = N - 1 (A - 1)(B - 1) = N - 1 A and B cannot be 1 so A-1 and B-1 are both positive. We are now looking for an N-1 that has more than 10 distinct pairs of factors, or essentially more than 20 factors. N-1 = 576 has 21 factors as 11 pairs so N = 577 has at least 11 distinct identities. BONUS: 11:24 Do you enjoy being... you know... Not with a yassified AI-generated Sophie Germain staring at me, no thanks. But the journey itself is fine 22:38 The Hyper Sophie Primes It's possible to generalize this even further. If we have a bunch of 2s (T of them) and then the last 2 terms to worry about are A and B, then we have the following rule: (2^T * A - 1)(2^T * B - 1) = 2^T * (N + T - 2) + 1 For T = 0, 1 we get the familiar prime and Sophie Germain prime conditions: (A-1)(B-1) = N-1 (2A-1)(2B-1) = 2N-1 But following that come the following conditions: (4A-1)(4B-1) = 4N + 1 (8A-1)(8B-1) = 8N + 9 (16A-1)(16B-1) = 16N + 33 (32A-1)(32B-1) = 32N + 97 etc. Every single one of those up to when T = N-2 must work. All of those RHS terms must be prime.

[@Mathologer]

That's great! Just at the end you gotto be a bit careful. For (2A-1)(2B-1) = 2N-1, the number 2N-1 is odd and therefore also every one of its factors is odd, which means we can always solve for integers A and B. However, this is no longer always possible for the higher equations. E.g. the factors of a number of the form 4N+1 are not necessarily of the form 4A-1 :)

[@nanamacapagal8342]

@Mathologer noted, thanks for the info

[@measurezero0102]

Probably because I'm a novice but proving that no integers satisfy the identity for length 10 past 3 non padding isn't that easy for me. Could you provide the proof

[@Tletna]

There is no problem with using 1's. The problem is your broke the pattern of using unique integer (so using each integer only once per side). If we follow that rule there is there a four integer line after the 1,2, 3 line?

[@Mathologer]

Good point. However, the answer to your question is 'No' :)

[@ratneshsudha]

@mathologer Following the same logic, a number N with just one product=sum identity also has the property that either 3N+1 is prime or (3X-1)x(3Y-1)=3N+1 has no integer solutions different from N (besides the properties of N-1 and 2N-1 being primes). Does this make sense?

[@shoam2103]

23:00 I was wondering this as it seemed like the next logical step!

[@Mathologer]

(x-1)(y-1)=N-1 and (2x-1)(2y-1)=2N-1 are special in that they automatically translate into an algorithm that applies to all N. E.g. 2N-1 is automatically odd and therefore all factors are also odd. And this means that we can always solve for x and y. But you are on the right track in terms of additional insights. Check out this paper by Maciej Zakarczemny, On the equal sum and product problem (linked to in the description of this video) www.iam.fmph.uniba.sk/amuc/ojs/index.php/amuc/article/view/1662 Theorem Theorem 4.3. If there is only one sum-equals-product identity for length n and n > 8, then all of the following conditions hold: 1) n − 1 is a Sophie Germain prime number, 2) all divisors of 3n + 1 are congruent to 1 modulo 3, 3) all divisors of 4n + 1 are congruent to 1 modulo 4, 4) all divisors of 4n + 5 are congruent to 1 modulo 4, 5) all divisors of 6n + 7 are congruent to 1 modulo 6, 6) 8n + 9 has no divisors congruent to 7 modulo 8, 7) 8n + 17 has no divisors congruent to 7 modulo 8, 8) 8n + 41 has no divisors congruent to 7 modulo 8, 9) 10n + 31 has no divisors congruent to 9 modulo 10, 10) 12n + 25 has no divisors congruent to 11 modulo 12, 11) 12n + 37 has no divisors congruent to 11 modulo 12, 12) 12n + 49 has no divisors congruent to 11 modulo 12, 13) 27n + 109 has no divisors congruent to 26 modulo 27, 14) 30n + 151 has no divisors congruent to 29 modulo 30.

[@ratneshsudha]

​​@@Mathologer❤ Thanks for taking the time to explain.

[@gregorymorse8423]

The question I keep asking now is does the pattern N-1, 2N-1 continue onto either 3N-1 or 4N-1, I suspect either it continues by integer multiples or powers of 2. Because that would place great restrictions on the numbers of each frequency which would be basically have to be unsatisfiable after a certain point on the number line.

[@Mathologer]

Have a look at the comment by @nanamacapagal8342 just following yours (and my response)

[@djsmeguk]

23:00 is anyone else curious about what happens with (mx-1)(my-1)=mn-1 for other m than 1 and 2?

[@Mathologer]

(x-1)(y-1)=N-1 and (2x-1)(2y-1)=2N-1 are special in that they automatically translate into an algorithm that applies to all N. E.g. 2N-1 is automatically odd and therefore all factors of 2N-1 are also odd. And this means that we can always solve for x and y. Check out this paper by Maciej Zakarczemny, On the equal sum and product problem (linked to in the description of this video) www.iam.fmph.uniba.sk/amuc/ojs/index.php/amuc/article/view/1662 for some other insights gained from other patterns in the list of sum-equals-product identities Theorem 4.3. If there is only one sum-equals-product identity for length n and n > 8, then all of the following conditions hold: 1) n − 1 is a Sophie Germain prime number, 2) all divisors of 3n + 1 are congruent to 1 modulo 3, 3) all divisors of 4n + 1 are congruent to 1 modulo 4, 4) all divisors of 4n + 5 are congruent to 1 modulo 4, 5) all divisors of 6n + 7 are congruent to 1 modulo 6, 6) 8n + 9 has no divisors congruent to 7 modulo 8, 7) 8n + 17 has no divisors congruent to 7 modulo 8, 8) 8n + 41 has no divisors congruent to 7 modulo 8, 9) 10n + 31 has no divisors congruent to 9 modulo 10, 10) 12n + 25 has no divisors congruent to 11 modulo 12, 11) 12n + 37 has no divisors congruent to 11 modulo 12, 12) 12n + 49 has no divisors congruent to 11 modulo 12, 13) 27n + 109 has no divisors congruent to 26 modulo 27, 14) 30n + 151 has no divisors congruent to 29 modulo 30.

[@RealQinnMalloryu4]

I like this video already i will check other video's after this video

[@ulychun]

2x2=3+1 in the context of Lie groups representations. This can be linked to the categorical concept of colimits in a monoidal category. (I start to sound like a category theorist 😂)

[@mysterion9686]

21:10 nice job, editor.😉

[@Mathologer]

:)

[@rachidsadou4178]

hi I really like your adventures in the world of mathematics, I'm waiting for your videos and I hope one day to make videos on the proof of the last theorem of Fermat for n=3 (euler) and for (p=2q+1) sophie germain

[@Mathologer]

Great :) Did you already watch this video from a couple of years ago? kzbin.info/www/bejne/d4CQiGiXeq9mjck

[@rachidsadou4178]

@Mathologer of course and I followed the demonstration for n=4 with the infinite descent of fermat

[@Mathologer]

Great :) Explaining the n=3 case nicely is definitely trickier than the n=4 case.

[@denisdaly1708]

Terrance Howard needs to watch these brilliant videos

[@Simon_Jakle__almost_real_name]

Does 2+2 = 2x2 also being "2 to the power of two" bring anything three dimensional, like on the axxis x, y and z? like in some "identity tree" or something?

[@Mathologer]

I sort of tried to capture this visually in the little animation superimposed on the intro section, two bars of 2 dots each separated by a plus sign standing for 2+2, merging into a square consisting of 4 dots and the plus turning into a x :)

[@gadxxxx]

I wish there was a book in English (not French) about Sophie Germain's life. Can't find one on Amazon.

[@Mathologer]

Prime Mystery: The Life and Mathematics of Sophie Germain by Dora Musielak Sophie's Diary: A Mathematical Novel by Dora Musielak Sophie Germain: An Essay in the History of the Theory of Elasticity” by L.L. Bucciarelli and N. Dworsky Have not read any of these books. If you do end up checking them out please report back on whether they are any good :)

[@gadxxxx]

@@Mathologer Thanks. OK, I'll look into it soon.

[@doodlebug1820]

I feel like the integer box problem must be related . The diagonals, which are sums of squares of sides, must have a relationship with the volume, which is a product of sides. Here we have a bunch of n dimensional boxes whose volume equals the length of their edges.

[@Mathologer]

Nice insight :)

[@alre9766]

With negative number, infinitely-long equalities can be made 0+1+(-2)+(-3)+4+5+(-6)+(-7)+8+9+(-10)+(-11)+12=0x1x(-2)x(-3)x4x5x(-6)x(-7)x8x9x(-10)x(-11)x12

[@Mathologer]

What you show there is still of finite length. True infinite equalities with integers are actually not possible apart from 0+0+0+...= 0x0x0x... Well if you are happy with infinity=infinity then a lot of things are possible :)

[@shivamshukla123]

Make a video on Dual numbers and split complex number

[@mienzillaz]

Since when we dropped the remote. Best thing that my daughter noticed this😅

[@BartDooper]

Hereby another interesting geometric triangle where the horizontal fractions can be factorized from the outer to the inner of the triangle. The outcome of that factor is the location and the number in the Pascal's triangle. 1/1 2/2 2/2 3/3 x 2/1 x 3/3 4/4 x 3/1 3/1 x 4/4 5/5 x 4/1 x 3/2 x 4/1 x 5/5 6/6 x 5/1 x 4/2 4/2 x 5/1 x 6/6 7/7 x 6/1 x 5/2 x 4/3 x 5/2 x 6/1 x 7/7 8/8 x 7/1 x 6/2 x 5/3 5/3 x 6/2 x 7/1 x 8/8 Where 1 is the unit of length: 1/1, 2/2, 3/3, 4/4, .. The length is going up 1 unit: 2/1, 3/1 , 4/1 for each row.. The "fibonacci triangle" variant of the Pascal's triangle (the number is the sum of the 2 diagonal numbers above that number) is : 1 1 2 1 1 3 2 1 1 5 3 2 1 1 8 5 3 2 1 1 can also be seen as answers to the fractions, factorized from right to left (in this excample below): 1/1 1 2/1 1/1 1 3/2 2/1 1/1 1 5/3 3/2 2/1 1/1 1 8/5 5/3 3/2 2/1 1/1 1 So ,1,2,3, Cheers to the catalan numbered sencorship networks.

[@yinq5384]

Wonderful video as always! 9:20 Length 10: Case 1: 8 copies of 1, a and b (2≤a≤b), 8+a+b=ab, (a-1)(b-1)=9, (a,b)=(2,10) or (4,4). Case 2: 7 copies of 1, a, b and c (2≤a≤b≤c), 7+a+b+c=abc. c is not an integer when a=b=2. Thus ab≥6 and c≥3, then 7+3c≥7+a+b+c=abc≥6c, contradiction! Case 3: 6 copies of 1, a, b, c and d (2≤a≤b≤c≤d), 6+4d≥6+a+b+c+d=abcd≥8d, contradiction! By the same argument as in case 3, we know there are no solutions with fewer than 6 copies of 1. More than 10 identities: Consider the equation (a-1)(b-1)=2³·3²·5 i.e. 359+a+b=ab where 1≤a≤b, there are (3+1)(2+1)(1+1)/2=12 integer solutions. Length 361 is one solution.

[@Mathologer]

Very good!

[@gwalla]

Not a fan of the AI-generated animations. Aside from the ethical questions regarding AI, they just look weirdly squishy and disconnected.

[@major__kong]

It's called the uncanny valley.

[@robertunderwood1011]

I like to see Sophie as if she were alive

[@davidvilla9581]

Not only does 2+2 = 2x2, they are also equal to 2^2, 2^^2, 2^^^2, etc.

[@Manoj_b]

I had a question in Newtown gregory method we get a constant sequence taking the differences but what if we get a periodic sequence at last sequence taking the differences. Imagine like we get like an infinite periodic sequence instead of infinite constant sequence

[@atmosphericSkull]

Does "almost prime" mean that they're either prime powers or square-free?

[@Mathologer]

Check out this paper by Maciej Zakarczemny, On the equal sum and product problem (linked to in the description of this video) www.iam.fmph.uniba.sk/amuc/ojs/index.php/amuc/article/view/1662 Theorem 5.1. says If there are exactly two sum-equals-product identities of length n , then one of the following two conditions is true: 1. n - 1 is a prime and 2n - 1 \in { p, p^2, p^3, pq } , 2. 2n - 1 is a prime and n - 1 \in { p, p^2, p^3, pq } , where p, q denote prime numbers.

[@mananself]

There are special properties about N-1 and 2N-1, but how about 3N-1. Any properties related to 3, and if not, why not?

[@Mathologer]

(x-1)(y-1)=N-1 and (2x-1)(2y-1)=2N-1 are special in that they automatically translate into an algorithm that applies to all N. E.g. 2N-1 is automatically odd and therefore all factors are also odd. And this means that we can always solve for x and y. Check out this paper by Maciej Zakarczemny, On the equal sum and product problem (linked to in the description of this video) www.iam.fmph.uniba.sk/amuc/ojs/index.php/amuc/article/view/1662 Have a look at Theorem 4.3. for what's known beyond what I talk about in this video :) If there is only one sum-equals-product identity for length n and n > 8, then all of the following conditions hold: 1) n − 1 is a Sophie Germain prime number, 2) all divisors of 3n + 1 are congruent to 1 modulo 3, 3) all divisors of 4n + 1 are congruent to 1 modulo 4, 4) all divisors of 4n + 5 are congruent to 1 modulo 4, 5) all divisors of 6n + 7 are congruent to 1 modulo 6, 6) 8n + 9 has no divisors congruent to 7 modulo 8, 7) 8n + 17 has no divisors congruent to 7 modulo 8, 8) 8n + 41 has no divisors congruent to 7 modulo 8, 9) 10n + 31 has no divisors congruent to 9 modulo 10, 10) 12n + 25 has no divisors congruent to 11 modulo 12, 11) 12n + 37 has no divisors congruent to 11 modulo 12, 12) 12n + 49 has no divisors congruent to 11 modulo 12, 13) 27n + 109 has no divisors congruent to 26 modulo 27, 14) 30n + 151 has no divisors congruent to 29 modulo 30.

[@DeathSugar]

Wonder if it could be smashed with twin primes conjecture.

[@philkeyouz2157]

@28'50 did you see the kaprekar constant !!!

[@Mathologer]

What a coincidence. I did not notice :)

[@johanjotun1647]

I wish i would have kept up on math in school, i had no idea that 1 wasn't a prime.

[@crimsonvale7337]

It’s more of a convention thing, there are many statements about primes (fundamental theorem of arithmetic) that if going off “primes are numbers that are only divisible by 1 and itself” would have to exclude 1. Really, so long as you get the ideas going on this is just dotting your i and crossing your t.

[@Man20045]

If f(x) is uniformly continuous on a set S1 & S2 is a proper subset of S1 then is f(x) uniformly continuous on S2? If f is uniformly continuous on S1 & f is uniformly continuous on S2 then is f uniformly continuous on S1 union S2? Please someone help.

[@oelarnes]

Who is depicted in the thumbnail? Shouldn’t it be Sophie Germain?

[@Tumbolisu]

I'm pretty sure it's an AI-generated image and therefore completely made up.

[@oelarnes]

@@Tumbolisu I know, just trying to make the point a different way. I don't think they'd treat Newton that way.

[@Mathologer]

For this video I played around with generating some AI animated photos of Sophie Germain using www.myheritage.com/deep-nostalgia as well as with the ChatGPT generated rendering of Sophie Germain in the thumbnail. Here I fed the AI some of the existing pictures of Sophie Germain and asked it to play artist to come up with a picture of her in the same style used in this real picture of Leibniz www.alamy.com/gottfried-von-leibniz-image5071937.html). Lots of fun and quite eerie :)

[@NepTunez-ff9bp]

I thought he was going to iterate to 3n-1, and then show the general formula, and then explain how that proves that there are finitely many lengths with only 1 identity. (No need for infinitely running computers!)

[@Mathologer]

(x-1)(y-1)=N-1 and (2x-1)(2y-1)=2N-1 are special in that they automatically translate into an algorithm that applies to all N. E.g. 2N-1 is automatically odd and therefore all factors are also odd. And this means that we can always solve for x and y. For what comes next in this respect, check out this paper by Maciej Zakarczemny, On the equal sum and product problem (linked to in the description of this video) www.iam.fmph.uniba.sk/amuc/ojs/index.php/amuc/article/view/1662 Theorem Theorem 4.3. If there is only one sum-equals-product identity for length n and n > 8, then all of the following conditions hold: 1) n − 1 is a Sophie Germain prime number, 2) all divisors of 3n + 1 are congruent to 1 modulo 3, 3) all divisors of 4n + 1 are congruent to 1 modulo 4, 4) all divisors of 4n + 5 are congruent to 1 modulo 4, 5) all divisors of 6n + 7 are congruent to 1 modulo 6, 6) 8n + 9 has no divisors congruent to 7 modulo 8, 7) 8n + 17 has no divisors congruent to 7 modulo 8, 8) 8n + 41 has no divisors congruent to 7 modulo 8, 9) 10n + 31 has no divisors congruent to 9 modulo 10, 10) 12n + 25 has no divisors congruent to 11 modulo 12, 11) 12n + 37 has no divisors congruent to 11 modulo 12, 12) 12n + 49 has no divisors congruent to 11 modulo 12, 13) 27n + 109 has no divisors congruent to 26 modulo 27, 14) 30n + 151 has no divisors congruent to 29 modulo 30.

[@whozz]

2^4 = 4^2 is my favorite

[@williamtomlinson85]

Wait until Terrence Howard finds out about this.

[@50oje]

Terence Tao?

[@Mathologer]

What do you think is going to happen? :)

[@cliffwroberts]

@@Mathologer it's going to shake the very foundations of Terrology. You can't do this to him! He's hanging onto reality by a thread as it is.

[@williamtomlinson85]

​@@Mathologer what would be great is if his followers would come watch this instead. How do we reach those people?

[@Mathologer]

Terrology, now there is a term I had not heard before. Scary stuff :)

[@danadnauseam]

What did attempts to generalize the mn-1 approach lead to?

[@Mathologer]

(x-1)(y-1)=N-1 and (2x-1)(2y-1)=2N-1 are special in that they automatically translate into an algorithm that applies to all N. E.g. 2N-1 is automatically odd and therefore all factors are also odd. And this means that we can always solve for x and y. Check out this paper by Maciej Zakarczemny, On the equal sum and product problem (linked to in the description of this video) www.iam.fmph.uniba.sk/amuc/ojs/index.php/amuc/article/view/1662 Theorem 4.3. says that if there is only one sum-equals-product identity for length n and n > 8, then all of the following conditions hold: 1) n − 1 is a Sophie Germain prime number, 2) all divisors of 3n + 1 are congruent to 1 modulo 3, 3) all divisors of 4n + 1 are congruent to 1 modulo 4, 4) all divisors of 4n + 5 are congruent to 1 modulo 4, 5) all divisors of 6n + 7 are congruent to 1 modulo 6, 6) 8n + 9 has no divisors congruent to 7 modulo 8, 7) 8n + 17 has no divisors congruent to 7 modulo 8, 8) 8n + 41 has no divisors congruent to 7 modulo 8, 9) 10n + 31 has no divisors congruent to 9 modulo 10, 10) 12n + 25 has no divisors congruent to 11 modulo 12, 11) 12n + 37 has no divisors congruent to 11 modulo 12, 12) 12n + 49 has no divisors congruent to 11 modulo 12, 13) 27n + 109 has no divisors congruent to 26 modulo 27, 14) 30n + 151 has no divisors congruent to 29 modulo 30.

[@Windprinc3]

What, if any, relationship does this have with the following facts about the quadratic equation a*x^2+b*x+c=0? If X and Y are the roots, then Sum of the roots (X+Y)= -b/a Product of the roots (XY)= c/a Then XY - (X+Y)= c/a - (-b/a)= (c+b)/a 🤔

[@Mathologer]

Nothing comes to mind. Still an interesting observation :)

[@teh_kaczuch]

what's funny is that the pattern continues backward, with one term on left and right too! (1 = 1)

[@Mathologer]

Yes, tempting but I resisted the temptation :) Not only 1=1 but N=N for all N. And just like declaring 1 not to be a prime it makes sense to not include the 1=1 case. In this way we save ourselves a lot of heart aches later on in the piece when we talk about these identities in general. Also with 1=1 there is really no sum/product in sight.

[@catastrophe3049]

Love from Bhaarat ( India) 🇮🇳

[@joels7605]

Terrence Howard says you forgot one: 1+1 = 1*1

[@Mathologer]

Is this person a known fool?