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

The third video in a trilogy of Mathologer videos dealing with sum-equals-product identities and equations. The other two are:

Way beyond the golden ratio The power of AB=A+B (Mathologer masterclass)

Heron’s formula: What is the hidden meaning of 1 + 2 + 3 = 1 x 2 x 3 ?

Today's video is all about how 2+2=2x2 and 1+2+3=1x2x3 first extends to an infinite sequence of similar identities and then further to a whole world of sum-equals-product identities. And we chase down a new connection to the so-called Sophie Germain primes.

00:00 Intro
06:27 How many?
11:29 New from old
19:53 Sophie’s primes
27:47 Beyond Sophie
30:15 Thank you!!

Currently the following paper is the best introduction to the sum-equals-product problem circle of ideas:

Maciej Zakarczemny, On the equal sum and product problem

Very nice translation of this circle of ideas into a set of problems for the Michigan Lemma 2020 maths competition (starts with Problem 2)

Check out the Encyclopedia and integer sequences pages for
2,3,4,6,24,114,174,444 and
7,8,9,10,12,14,15,16,18,20,22,30

Some other papers to check out with links:
Michael W. Ecker, When Does a Sum of Positive Integers Equal Their Product?

M. A. Nyblom, Sophie Germain primes and the exceptional values of the equal-sum-and-product problem

Maciej Zakarczemny, Equal-Sum-Product problem II

M. A. Nyblom and C. D. Evans, An Algorithm to Solve the Equal-Sum-Product Problem

Leo Kurlandchik and Andrzej Nowicki, When the sum equals the product

Check out the list of updates in my comment that's pinned to the top of the comment section.

Music: Just jump by Ian Post
T-shirt: Forgot where I got this one from :(

Enjoy!

Burkard

[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





Check out this paper by Maciej Zakarczemny, On the equal sum and product problem (linked to in the description of this video)
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 :)


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.



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 Online Encyclopedia of integer sequences: A033178

@Frafour


@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

[michaelheeren]

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.

[StefanWeckbach]

Dear Burkard,

thanks for your very interesting math-videos, i enjoy them very much!

At the beginning of the actual video you asked for some interesting variation of the sum-equals-product problem.

Here is one, limited to a number of 2 variables a and b, both being positive *rational* numbers.

It goes like this:

Take any integer number N > 2, for example 29.

For N being odd, you have (29-1):2 possibilities to construct the identity (a times b) = a plus b.
For N being even, you have (N:2) possibilities to construct the identity (a times b) = a plus b.

Example with N = 29:

(29:11) + (29:18) = (29:11) x (29:18) (where the sign : is "divide by the following number)

As you can easily conclude, with N = 29 you have 14 different solutions to the sum-equals-product problem (with 2 variables).

The algorithm to find the suitable a and b combinations is simple, a and b must always sum up to whatever N you have choosen initially.

So in the example above, the divisors are: a = 11 and b = 18, so (a + b) = 29.

But you could also have choosen the divisors 10 and 19 and so on. As said above, for N = 29 you have a total of 14 combinations to choose from for the values of a and b (skipping the redundance of counting double each pair of sums for a + b, since the sums are equal to b + a).

[ianstopher]

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.

[julianpilbrow]

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!"

[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

[heathrobertson]

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

[sheikhalsumaiya]

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

[XODARHAZ]

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 :')

[perfredlund]

Stepping away from topic at hand. Very nice T-shirt you got there👍

[tomasebenlendr]

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)

[SaturnCanuck]

Always love your videos, a great sunday afternoon

[evanev]

always something uncanny about that ai generated yassified sophie germain

[franknijhoff]

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.

[Igor_Zdrowowicz]

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

[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.

[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.

[agostinhooliveira]

Great video as always.

And I noticed Euler is one of your Patreons! 😊

[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.

[onejumpman]

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.