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

Welcome to another Mathologer video. You all  know those two freaky identities over there, right? And for most people they are just that:  pretty freaks without any deeper meaning… Unless of course, you are a Mathologer regular. 

Because if you are, you would have found out in two earlier videos that these identities  and their underlying equations, those two there lead very interesting secret lives. Heron’s  formula, the heptagon golden ratio, and all that jazz. But what even most Mathologer regulars  won’t know yet is that these two identities up there are just the start of a very interesting  infinite sequence of similar identities.

Really? Yes, and in fact, there is a whole world of  sum-equals-product monsters out there. Let’s go and explore :)Okay, infinite sequence, hmm,  so what’s next (point) over there?

:) Hmm. Well, obviously, the next equation at the bottom is  this one here. If you’ve not done so already, definitely check out those two earlier videos for  a lot of amazing and beautiful maths :)Anyway, what even most Mathologer regulars won’t know yet  is that these two identities up there are just the start of a very interesting infinite sequence  of similar identities.

Really? Yes, and in fact, there is a whole world of sum-equals-product  monsters out there. Let’s go and explore :)Okay, infinite sequence, so what’s next?

:) Hmm. Well,  obviously, the next equation at the bottom is this one here. x plus y plus z plus u equals  x times y times z times u :) And, at the top, we are looking for a solution of this new  equation in positive integers.

So, to start with, we want four positive integers on both sides  of the equal sign at the top. Can we spot a pattern? Well, if we shuffle things around like  this, as I did in the thumbnail of this video, things are starting to look very SUM-Metrical. 

Okay, okay, bad one :)2, 3. What comes next? 4, of course :)Then2, 2, what’s next?

Another 2,  maybe? Now. 1.

What’s next? Well, let’s first add and multiply what we’ve got so far. 2+4 that’s  64 times 2 that’s 8.

Now, if you’d have to bet your life on what number comes after 1 in this  set-up, you’d probably go for 1, right? Okay, so let’s try 1. On the right, multiplying by  1the product actually does not change.

Aha! Can you see what’s happening here? On the other  hand, adding the corresponding 1 on the left ups the sum by 1.

7 and 8, almost equal. Obvious how  to finish, right? Just chuck in another 1 :)1 on the right.

Nothing changes. Add 1 on the leftTada  :) Equality. What’s next?

Well, After 4 comes 5. So replace 4 by 5. That makes the sum on the left  go up by 1.

On the right, replacing 4 by 5 ups the product by 2. And now we know what to do. Just pad  with another 1:)And it’s clear sailing from here, forever and ever after.

2, 3, 4, 5, 6, and so on,  in the middle. Then2, 2, 2, 2, 2 and so on. And pad out the rest with 1s such that in every  identity the number of terms on the left and right (click) is equal to the largest number in  the middle.

There (point) three terms and 3 in the middle. 4 terms and 4 in the middle, and so on. Two  remarks: One: The whole padding with 1s business may come across a little bit like cheating but if  you want to play this game with positive integers, you really don’t have a choice.

Because,  if you also insist on not involving any 1s, then 2+2=2x2 is the only identity that works.  Second remark: Taking this padding-with-1s-idea one step further, it’s clear that there are many  more of these sum-equals-product identities. Just start with a random assortment of positive  integers, say 2 5 7.

Doesn’t matter which, you can even have some numbers repeat. Like for  example start with the numbers 2 5 5 7. Calculate their sum and product.

Then, as in this example,  in general, the product will be greater than the sum. Subtract the sum from the product. Then  if you pad out the left and right sides with that difference number of 1s you get one of our  identities.

So, in this case add 331 1s on the left and multiply the same number of 1s on the  right and you get equality. Super easy. In fact, the first time I stumbled across all this I  thought: Okay, cute but in the end pretty trivial, right?

But let’s slow down :) After all, even  if something is easy to understand, it may still lead to deep and beautiful mathematics. Right? You  can explain to a 10-year old what a prime number is.

But at the same time some of the greatest  mathematical theorems, and some of the most important and hardest problems are about prime  numbers. As the Mount Everest example, the Riemann hypothesis is all about primes :)And, it turns out  that the world of sum-equals-product identities is also full of pretty surprising prime number  marvels and mysteries. Maybe not the Riemann hypothesis, but still great stuff :) In the rest  of this video, I’d like to tell you about one particularly nice mathematical marvel involving  the mysterious Sophie Germain primes.

These were discovered 200 years ago by – you guessed it :) –  the mathematician Sophie Germain. She discovered these primes while working on Fermat’s last  theorem one of the most famous open mathematical problems at the time. But before I get started,  I’d like to take this opportunity to pick all your brains to dig for mathematical treasure :)In your  mathematical journeys, did you ever encounter any of these sum-equals-product identities or  equations in surprising settings or contexts?

For example, my friend Marty recently told me  about this paper by Stanford mathematician Brian White in Inventiones Mathematicae,one of the most  prestigious maths journals. Turns out that amidst a lot of very high-powered mathematics, 2+2=2x2  ends up being the ingredient that makes everything come together. Really quite wonderful.

Anyway,  if you know of anything like this for any of the sum-equals-product identities and equations please  let the rest of us know in the comments. Okay, now let’s have a close look at the world  of sum-equals-product identities. Let’s call the number of terms on one side of a  sum-equals-product identity the length of the identity.

So, 2+2=2x2 has length 2  and 1+2+3=1x2x3 has length 3. It’s then natural to ask how many sum-equals-product  identities there are of a given length. Well, it’s natural for us mathematicians to ask this  :) It’s what we do :) So, here straightaway is a plot of these numbers of identities up to  length 10,000 taken from a really nice paper by the Polish mathematician Zah-kar-CHEM-nee that  I link to in the description of this video.

What a weird looking plot :) Definitely not the nice  curve that you may be expecting to see here, but a standard plot regardless, consisting of  one dot each for the possible lengths between the smallest length 2 and length 10,000. Right,  take any length. .

then the height of the dot above it is the number of sum-equals-product  identities of that length. in this case around 25. Clearly the dots are all over the place. 

Meaning that the number of sum-equals-product identities for different lengths are all over  the place. The general trend is a logarithmic growth. But the trend is not very trendy, is  it?

As we increase the length, we come across larger and larger numbers of identities (click)  but, at the same time, really small numbers keep popping up. (click) Let’s have a closer look at  what happens for the smallest lengths. Alright, we know that for every length there is one  of our basic sum-equals-product identities over there.

In fact, for length 2 there is only  this one sum-equals-product identity. That is, the equation X + Y = Y x X only has one solution  in positive integers. The same is true for the next equation.

The only solution (up to shuffling  of the terms) corresponds to our second basic sum-equals-product identity. Same for length  4. Then 5.

But now it turns out that for length 5 we also have this second sum-equals-product  identity and also this one hereBut then that’s it. For length 5 there is a total of three identities,  those three (point)And now that we’ve got going, there should be lots of identities of length  6, right? Nope.

For length 6 there is just one again. And, after this, things continue up to  length 9 as follows. Here are two little challenges for you.

First, can you find all the identities  of length 10? Also, can you find a length for which there are more than 10 identities?  Leave your answers in the comments :)Okay, so what’s a natural question to ask here? 

Who cares? No, no, no, no. Natural or not, that question is taboo around here :) Seriously,  what might we ask about these lengths?

Well, how about “For which lengths is there only one basic  sum-equals-product identity, the basic identity is the only one? ” Right? Natural question? 

Well, so far we’ve seen four such lengths2, 3, 4, and 6. Are there more? Yep, 24.

It turns out  there is also only the one sum-equals-product identity of length 24. And then, we also have  114174and 444What’s next? Should I leave you to hunt for the next one?

Maybe not :) In fact,  mathematicians have used computers to check all lengths up to 10 billion and have not discovered  another length with only the basic solution. So, maybe that’s it, and the list ends with 444. But  we don't know.

Checking all the way to 10 billion is very suggestive, but ten billion is still a  long long way from infinity :)How could we settle this question? Well, we could leave the computer  running :) But that will take an infinite amount of time, and we’re probably a little too impatient  for that. So, instead, how might we think our way to a solution?

Well, a good starting point would  be to figure out what all these special lengths have in common. Right? Okay, so have a look at  those numbers up there.

Is there any pattern, any common trait that sticks out to you? Hmm.  Lots of 4s?

Yes, but sadly that’s not it :)Now, I could just tell you the surprising answer.  But that’s not the Mathologer way :) Instead, I will do what I often do, I’ll kidnap you, and  take you on a journey of discovery, which leads naturally to this answer. Let me know in the  comments whether you enjoyed being kidnapped.

Okay, let’s start with something easy. Let’s  figure out why, for length 2, there is only one positive integer solution of this equation,  namely 2+2=2x2. How would you get started on this mission?

Easy! Solve for one of the variables,  say Y and hope for the best :) Algebra autopilot on. Okay, just one Y left and it would now  be easy to finish solving for Y.

However, at this point an experienced algebra torturer  may notice something else, an opportunity to simultaneously isolate both variables :) Have  a look. Take minus 1 on the rightand add 1 to cancel out the -1. Right, -1+1, cancels out to  0, so nothing’s changed.

Except…Can you see it? ThereThat’s nice isn't it? We managed to rewrite  our initial equationin a second supersymmetric way.

In particular, this means that both  equations have exactly the same solutions. And, so, what are those solutions? Well, easy.

Since  X and Y are supposed to be integers, so are the numbers X-1 and Y-1. But then what this says is  that the product of the orange integer and the green integer equals 1. Of course this means that  either both orange and green are equal to 1 or both are equal to -1.

Right? 1 times 1 is 1 and -1  times -1 is also 1, and that’s all. Now, if both numbers are equal to 1.

Then, well, what? X-1 is 1.  So X=2 and so is Y.

That’s our 2+2=2x2 solution. On the other hand, if both orange and green are  -1then we have X-1 = -1 and so X = 0 and so is Y. 0+0=0x0 that’s also correct but we decided,  well rather I made an executive decision for all of us :) that we’d only be interested in solutions  with positive integers.

Alright, that was definitely a slick way to show that we have only  one solution for length 2. But, more importantly, all this mucking around leaves us just one step  short of another crucial insight. Okay, again, this nicely symmetric equation is equivalent  to our initial sum-equals-product equation.

But here’s another trick. Just shuffling everything in  the bottom equation to the left of the equal sign, we obtain a third equivalent equation. Does  the quantity on there left ring a bell?

That difference between a product and a sum? Right,  if you start with two positive integers X and Y, then this difference tells you how many 1s you  need to pad these two numbers with to get one of our identities. In the case of X and Y both  being equal to 2, this difference is 0.

But now let’s consider any old X and Y. Then the  difference between the product and the sum is some other number, let’s call it D. Then the exact  same algebraic acrobatics as earlier give us this equivalent equation at the top.

Easy for you to  nut out the details. And of course you should. No freeloading here :) Anyway, let’s check using  an example: X=2 and Y=7.

What’s that difference? 2 times 7, that’s 14 minus 2+7,…, 9, 14-9 is 5. Okay  that means we need to pad 2 and 7 with five 1s.

And, actually, if we do that we just get one  of our basic identities. That one here. Now what about at the top?

X-1, that’s 2-1. 2-1 is 1. andY-1,  that’s 7-1 that’s 61 times 6 is 6Works.

That 6 is the difference 5 from before plus 1, 5+1 is 6.  Now here comes the nifty bit. 6 is also 2 times 3.

But, then, if X-1 is 2then X is 3. And if Y-1 is  3, then Y is 4. But, now we can use these new X and Y, to get another difference of 5 at the bottom. 

3 times 4 is 12 minus 3+4, 7, 12-7 equals 5. But now that means that if we pad out 3 and 4 with  five 1s, we obtain a second sum-equals-product identity of the exactly the same length as the  basic identity we started with. Can you see where I am going with this?

Well, what we just  discovered boils down to a nice way of creating new sum-equals-product identities from our basic  ones, of the same lengths. Let me spell out this streamlined way of creating new sum-equals-product  identities using a couple of examples. Okay?

Well, let’s start with the basic sum-equals-product  identity that we just encountered, 2 and 7 padded out with 5 1sAlright, 7 minus 1, that’s the 6. 6 is  2 times 32 plus 1 is 3 and 3 plus 1 is 4And that’s our new sum-equals-product identity of length 7  once again. Now let’s start with the basic identity of length 5.

5 minus 1 is 44 is 2 times 22 plus  1 is 3And there is the new sum-equals-product identity of the same length as the basic one  we started with. Does this always work? Well, here is what quite a few of you will already  be waiting for by now.

6 minus 1 is 5But now, since 5 is prime, the only way to write 5 as a  product of positive integers is 1 times 5Adding 1 to both 1 and 5, gives…2 and 6. Which after  padding with 1s only gets us back to where we started. In general, for a given length N our  construction will give no new sum-equals-product identities if N-1 is a prime number and at least  one new sum-equals-product identity if N-1 is not a prime number.

And that tells us something very  interesting about our puzzle. Remember, our puzzle was to try to determine, without using a computer  for infinity hours, for which special lengths N there exists just the one basic sum-equals-product  identity. And this is what we just discovered: any of those special numbers minus 1 must be a  prime number.

How neat is that? :) Let’s check. Those are the known special lengthsAll minus  1.

1, 2, 3, 5, 23, etc. Works, those are all prime numbers. Well, that 1, the first of the  numbers, that’s not a prime, but close enough, right?

1 used to be a prime and is at least still  an honorary prime :) Nothing goes into it except 1 and itself, which is what matters here. Anyway,  what we’ve discovered is that if there is another one of these special lengths, does not matter how  large, then that special length minus 1 must be a prime number. So, we only need to look for special  lengths among those numbers that diminished by 1 are prime numbers.

Hmm, that’s all definitely  some pretty prime number action here :) Okay, but still there are a lot of prime numbers, and  so there must be something extra special about those particular primes up there. To track down  that extra speciality :) let’s have another close look at our line-up of small cases. All clear so  far?

Great? Again, over there is our list of small cases. What I’ve hightlighted in this list are  the first three in the infinite family of special identities that our streamlined way of making new  sum-equals-product identities from the basic ones produces.

Remember, this infinite family and the  streamlined way of producing it resulted in our neat prime number insight. Another close look at  our list of small cases, yields a second infinite family of identities. This second infinite family  and the rule underlying it will lead us to another really neat prime insight about our special  lengths.

Ready for this? Okay, let’s go. Alright, here are the first two identities in our second  family.

Hmm…What’s the rule here? How do you go from length 5 at the top to 2,2,2 in the  yellow box or from length 8 to 2, 2, 3 in the blue box? Well we’ve not got much to work with so  far.

So let me also show you the next couple of identities in this family. 222, 223, 224. So we’re  all guessing that 225 is next, right?

:). Well, let’s see…233 not 225 :) In fact, if you look at  a couple more examples, the only obvious pattern is the fact that we always get a 2 and something,  and something else. But, since that one 2 is common to all these new identities, maybe the better  question to ask is how does the 5turn into 2 2.

And how does the 8turn into 2 3? Well, actually,  it’s not hard to figure out what the general rule is, now that we know what we are gunning for  :)Right, starting with some N, we want to find X and Y up there such that both sum-equals-product  identities have length N. And how do we do that?

Easy, do some algebra torturing :) First,  the number of 1s in the basic sum-equals-product identity at the top is what? Well, the length is N  and only the 2 and the N are not 1s. And so there are N-2 1s.

Right? On the other hand, the number of  1s in the second sum-equals-product identity is, as usual, the product of 2, X and Y minus the sum  of 2, X and Y. Next insight: There is one more 1 in the first identity than in the second.

Have  a look. One more 1 at the top. So adding 1 to the second expression makes both expressions equal.

Now  simplifying on autopilot and isolating the X and Y as previously recasts this equation in this  supersymmetric form. At this stage the algebra involved should be a piece of cake for you and  so I won’t bother with the details. For comparison the corresponding equation for our first rule is  this.

So exactly the same except for the 2s. Now, remember, the first equation translates into this  rule for generating new special sum-equals-product identities: subtract 1 from NWrite N-1 as a  product. Add 1 to both factors to get the new X and Y.

Similarly, for our new second rule we multiply  N by 2 and minus 1Write 2N-1 as a productAnd, finally, add 1 to each factor and divide by  2 to get the new X and Y. Right? :) And here is the new rule in practice:2 times N-1, so 2  times 5 is 10 minus 1 is 9write as a product, 9 is 3 times 3add 1 to each factor and then divide  by 2.

3 plus 1 is 4, divided by 2 is 2. throw in that common 2. E voila, here is our new special  sum-equals-product identity, exactly as long as the basic one we started with.

Another example2  times 6 is 12, minus 1 is 11A prime number and we are stuck. Next,2 times 7 is 14, minus 1 is  13stuck again. 2 times 8 is 16 minus 1 is 15aha, 3 times 5+1 divided by 2, both of them.

3+1 is 4  divided by 2 is 2 and 5 plus 1 is 6 divided by 2 is 3… 2 3join in the common 2 in front. And  done. :)Okay, where does this leave us in terms of characterising our special lengths?

Well, we  already know that if N is a special length, then N minus 1 must be prime. But now with our new rule,  we also know that, in addition, twice N minus 1 is prime as well. Right?

There, all those new numbers  3, 5, 7, 11, 47 etc. are also prime. And, so, if we ever find another one of those special lengths,  just times it by 2 and minus 1, and you get another prime.

Super interesting and surprising,  don’t you think? :) Well, it sounds unlikely, doesn't it? What are the chances that both N-1 and  2N-1 are prime numbers?

Or, phrased differently, what are the chances that, given a prime number P,  twice P plus 1 is also a prime number? As I already mentioned at the beginning of this video, it was  the mathematician Sophie Germain (click) who was the first to ponder this question around  1800 while trying to prove Fermat’s last theorem. Because of this, a prime number P such  that 2P+1 is also a prime is now called a Sophie Germain prime.

One of Sophie Germain’s major claim  to fame is that using her primes she was the first to make significant progress towards proving  Fermat’s last theorem for infinitely many prime exponents. Prior to her research people only  considered individual exponents. If you have not heard of Sophie Germain, definitely check  out the remarkable story of her mathematical life.

Completely self-taught, had to pretend  that she was a man to enter maths competitions, the whole bit. Okay, history lesson done. Back to  work.

So we know that every special length minus 1 is a Sophie Germain prime, right? What about the  other way around? Is every Sophie Germain prime also a special length minus 1?

The answer to this  question is … Sadly, or not sadly, depending what you are cheering for, there are a lot more Sophie  Germain primes :) As it happens, Sophie Germain primes are nowhere near as rare as one might  expect. To illustrate, here are all the Sophie Germain primes up to 113. Computer searches have  produced ridiculously large Sophie Germain primes which also happen to be useful in cryptography. 

In fact, the people in the know believe that there may well be infinitely many Sophie Germain  primes. Apart from the two special infinite families of sum-equals product identities that  led to our prime number insights there are of course many other infinitely families one can  investigate and milk for further insights into our special lengths. This is in fact possible,  however nothing quite as neat as what I’ve talked about today has been discovered.

Anyway, check out  the linked papers in the description of this video for more details. Ooookay, what other interesting  open sum-equals-product results and questions are there? Let’s finish up with that.

First off, if I  had to bet my life whether or not those numbers up there are the only special lengths, with just  one sum-equals-product identity each, I’d go for “Yes”, those are the only ones :) But, then, what  about the lengths that correspond to exactly two special sum-equals-product identities? A couple  of the small lengths that we considered earlier were of that type. There, at the bottom, there  are exactly two identities each for lengths 7, 8 and 9.

Turns out that the list of known lengths  of this type is also fairly short. To be precise we know of 49 such lengths, the largest known  being 6324There are also some primes hiding in these numbers. Turns out that if there are  exactly two special identities of length N, then the number N-1 or the number 2N-1 or both are  prime.

For the first 13 in this list (click) here are the corresponding N-1s and 2N-1s. And here  are the primes among these N-1s and 2N-1s. That’s a pretty cute result, too, don’t you think?

:)  Now it’s possible to show that that even if one of these N-1s or 2N-1s is not a prime, such a  non-prime is nevertheless always close to being a prime number in a certain sense. What does it  mean to be close to being a prime? To find out, check out the details in the description of this  video.

By now you will probably not be surprised to hear that all sum-equals-product identity experts  also suspect that those 49 lengths are the only ones corresponding to exactly 2 identities.  In fact, it’s reasonable to conjecture that for any fixed number of identities there are  only finitely many lengths corresponding to that number. All very intriguing and lots  of open interesting problems remain to be tackled in this area.

Anyway, that’s it for  today. Don’t forget to let the rest of us know of any interesting occurrences of our special  identities in the wild.