He has his own channel as well, where he posts similar stuff (it's called singingbanana, though I don't know why). It's pretty inactive, but you can binge on some of his older videos if you haven't seen them.
James Grime said it wrong though, it doesn't need to be red and blue. It needs to be neither red nor blue... so 7825 is obviously yellow, but that leads back to his question at the end of the video of what if their were 3 or more colors would it inevitably still reach a point when it no longer works (assuming a finite number of colors).
I have to do some calculus exercises... Me: "Okay, right now I'm absolutely not in the 'math-mood', let's just do other things first." *Numberphile just uploaded a video* Also me: Let's just watch this...
It is an optical illusion caused by rather specific lighting condition. It is the same effect that occasionally makes the light from a row from four lightbulbs occasionally appear to be concentrated in a row of five point sources instead.
Jan Dvořák It's not an optical illusion, i checked with a color picker. Its hue in HSL format is 159 which is pretty much green (a little bit on the turquoise side, but still objectively green).
"imagine all supercomputers in the world checking all the posibilities since the dawn of time you still wont be able to check all the posibilities ... so they used some clever mathematics to reduce the number they had to check, and it took them about 2 days" i snorted when i heard this, but this is actually great!
This problem reminded me of a project I did five or six years ago in university, which was also about finding sets of numbers without a certain structure - in our case, we studied Szemerédi's Theorem - and we were using SAT solvers to do this. Turns out that this work is from our supervisor at that time - Marijn Heule was our supervisor. And, indeed he used SAT solvers to create this proof. A little background: SAT this is the satisfiability problem, which asks "given a set of boolean variables (which can be either on or off) and a set of constraints on these variables, is there an assignment of values to these variables that satisfies these constraints?". You can translate the pythagorean triplets problem into SAT in its most basic form by creating a variable for each number from 1 to n, and creating a constraint for each pythagorean triplet stating that of the three variables corresponding to the three numbers at least one must be true (red) and at least one must be false (blue). If there is a solution for this SAT problem, then there is also a solution for the original problem, by colouring the numbers red if the variable is true, and blue if the variable is false. The main advantage of this approach is that SAT is quite a well-known problem, having quite efficient solvers. I can imagine that someone must have created an efficient SAT solver for supercomputers, or at least the Texas supercomputer, as well. This way it would be relatively simple to tap into the huge amount of computing power available in an efficient way (which is really hard if you're doing that from scratch). If I'm making it sound easy: it's not. The basics are quite easy to grasp (and can be coded in 30 lines of C), but in order to get an answer for these huge problems in a useful amount of time, you need to optimize your translation. This requires thorough knowledge of both the solver and the problem, to translate the problem in a way that is most meaningful to the solver. And finally, you need to handle all the data and include it in your proof.
I thought the reason was video coloration problems. Having experience with that very thing, that is my first guess. If someone can prove me wrong, that would be interesting also.
For the a + b = c case, there are really only 3 options, not 512 - either 1 and 2 are (say) red and 3 blue, or 1 and 2 are different colours and 3 is red or blue. The other colours are derived from there until it becomes impossible.
almightyhydra Hence why he said that it was easy to convince yourself it was true instead of simply checking every possible combination of red and blue for every space.
There are exactly 512 ways to assign the colors. Whether those colors actually follow the rules or not is up for you to decide Yes, you can reduce this space through logic. This is the kind of work mathematicians do all the time, but it’s not always so trivial to do so :)
I have synaesthesia only where numbers are concerned. As a child, I was trying to find a way to make the numbers add up in such a way that their respective colours would blend to give me the correct colour for their total. I quickly ran into the same problem you guys did.
That's interesting! I wonder if I have synaesthesia, except with feelings. I have certain emotions associated with numbers, and often, based only on how I feel about a given number, I can tell if it is prime. For example, 167. It feels sharp, and makes me feel edgy. One day, I was talking to my mom about my "synaesthesia", and threw out that number because it felt prime. Then, a few minutes later I looked it up, and sure enough! 167 is prime.
The trick to making this work is to assign numbers to colours in a balanced fashion, rather than linear. As in, red = 1 orange = 2 .....violet = 7, won't work. It would need to be arranged like the Chinese Bagua, where the opposing number was actually opposed on the spectrum. Of course, with synaesthesia, you don't get to choose the assignments so whether it works would depend on your particular individual affliction. Sorry to hear you lucked out :( That would have been really frustrating!
Colors are a three dimensional thing to our brains. This means you will always need three numbers to represent a color in a unique way. Single numbers just won't cut it.
He is just fantastic. Loves numbers. Loves math. Is completely consumed by his wonderful passion. And does his very best to spread the thrill and excitement.
I found a pattern where a^2+b^2=(b+p)^2 made pythagorean triples if p was a factor of a. (lowest factor being 2 if a is even) I noticed in the video that for 7825; the two triples that make it up one of the two follow this pattern, and the other does not.. Namely 5180^2+5865^2=(5865+1960)^2 .. obviously 1960 is not a factor of 5180 in this case whereas 625^2+7800^2=(7800+25)^2 is. It could be nothing, but.. This was the first time I've come across a case where the a pythagorean triple exists that doesn't fit the above format. (p not being a factor of a). Perhaps this is why?
If you expand (b+p)² and cancel out b², you're left with a² = p(2b + p). So p (which is c - b) being a factor of a² is a necessary but not sufficient condition for a Pythagorean triple. I don't know how often it turns out that p is a factor of a, but I see quite a few examples on Wikipedia where this condition is not verified, regardless of which number you pick as a: (20, 21, 29), (33, 56, 65), (48, 55, 73), (65, 72, 97) etc.
@@yondaime500 more like a commonplace but not necessary condition. Let's move away from the cliché "necessary but not sufficient" which is tired, over-used and has become beyond vague.
The time estimate for a brute force approach is a bit off, because you don't have to check all of the possibilities. As soon as you come across a contradiction, you can stop and go on to the next set. Similarly, you only actually have to check 4 of the “512” combinations for the introductory problem to prove it never works (and indeed if you expanded the introductory problem to an arbitrary number of integers, you would never have to check more than those 4).
9:25 "doesn't tell us why". didn't he explain why? 6:46 isn't the reason it doesn't work is because it's in two Pythagorean triples? if they know this is the reason it doesn't work, then why do they need to use brute force to prove it?
Furthermore, I notice that Mr Heule has been working on Polymath 16 (the Hadwiger-Nelson problem) and just broken his own record for the smallest unit-distance graph with chromatic number 5.This follows the recent breakthrough by Aubrey de Grey, although I imagine this record may not hold for very long, considering the great progress now being made in this area.It's not so difficult to explain .. so maybe a future numberphile video please?
If there is always a number you'll reach where these equations break once you count high enough, wouldn't this calculation reflect entropy? With entropy, systems lose organisation over time. With these patterns, the number you reach before being forced into a 3-of-a-kind pattern gets higher as you add more colours. So wouldn't it be possible that this equation could end up having relevance to the concept of entropy if and when we solve it?
It’s actually kind of the opposite. If more related numbers have the same color, then that represents more structure, which is to say less entropy (The reason they are playing this specific game, avoiding 3 numbers of the same color, is that they want to see how *big* the system has to be before structure is forced to appear) Thus Ramsey Theory says that large mathematical objects often have some inherent structure, or emergent properties, which actually act against entropy! I feel like this connects to the structure of matter and the Universe in some way - Some matter/energy arrangements are just far more stable/favorable! Especially when moving towards larger scales (Yes, over long time scales, eventually all order will disappear as entropy increases. But in the meantime, it’s truly remarkable how organized and structured everything seems to be)
Then proving that, for n colors, there will always be a perfect square which can be written as the sum of n other perfect squares and that those pairs must be the same color will prove the conjecture?
It is an optical illusion caused by rather specific lighting condition. It is the same effect that occasionally makes the light from a row from four lightbulbs occasionally appear to be concentrated in a row of five point sources instead.
It's possible. The two right triangles are different. Using inverse trig functions the first one has angles 90, 85.42(approx) and 4.58(approx) The second right triangle has angles 90, 48.55(approx) and 41.45(approx) In fact I think this type of triangle seems to be rare in this case since we're using whole numbers. Otherwise we would have infinitely many such right triangles having same hypotenuse. Suppose we have a constant hypotenuse 'x' unit. The other sides are a and b. So, a=root(x^2-b2). Since x constant you can choose infinite values of b to get infinite values of a. Just make sure 0
I know this is an old post but someone else might stumble here... It's the Pythagorean theorem (using really big numbers). The theorem states that "in a right triangle a² + b² = c². Where side c is the side opposite the right angle." In other words, if there were a right angled triangle with side (a) being 3 units long, side (b) being 4 units and you were looking for side (c)... c² = a² + b² c² = 3² + 4² c² = 9 + 16 c² = 25 c = (sqrt) 25 (*sqrt 25 to remove the exponent from c) c = 5 Using this formula and some trig, so long as you know the length of 2 sides and that the triangle has a 90 degree angle, you can find the length of the remaining side and the other 2 angles. (That's what Sadman did above.) In the case of 7825 being c, there are two ways to form a right angled triangle and either option they choose of the two won't allow them to color the chart correctly to meet the other requirements and it ends their little game.
That’s saying that if you construct a triangle with sides 625-7800-7825 or 5180-5865-7825, one of the angles of the triangle (the one opposite the 7825) is a right angle.
If you extend the case of sums of first powers to 3 colours (so we're excluding monochromatic triples (a,b,a+b), where a and b are distinct), then you can go at least as far as 22, using colour classes {1,2,4,8,11,16,22}, {3,5,6,10,12,19,20,21} and {7,9,13,14,15,17,18}.
Everybody knows *42* is the base number system used by the computronium running our reality [actually, it is _Quadragesimal_ (40) plus an independent "C.R.C." on _Binary_ (2)]. -> Knowing that *7825* is part of the equation leading to the "seed" used on this _particular run_ on the multiverse will get us closer to formulate the ultimate question.
This was a fun video. I saw immediately that 1 thru 9 was impossible to not duplicate the same color. It happened for me when he showed the number 9 would be red. Bingo! That did it. Then, when he was discussing the 7825 number, I knew it was too large to ever examine every possibility. That somebody narrowed just a subset to 1 trillion was amazing, and then took 2 days to run the sets, was also thrilling to me. This sort of problem is what makes math so wonderful and inviting to me.
well this problem is also connected to fermat's last theorem as anything in the form a^n+b^n=c^n is part of the theory. If you noticed thought he addition of 1 to 9 also add the same issue as 7825. There are two sets of possible additions for 5 and 8 that break down the rules. 9 is another example of two pairs that are in opposite colors, so not matter what is chosen the coloring breaks. we might want to stick to the addition to understand this problem. multiplying is only causing us greater amount of work.
"...this kind of proof does not increase our understanding of why..." I would soften this statement, the tricks used to reduce the number of possibilities to check can reveal some interesting structures of the problem.
It does strike me as a potentially useful step in understanding why this is the case in that having a brute force example may help mathematicians identify patterns that hadn't been tangible before.
...but it did tell us something about why it failed at that particular number, James, it failed because it's in two sets of Pythagorean triples. Which will be where they fail for more colours as well, except that for three colours the number's gotta be in three sets of Pythagorean triples in order for it to break, and so on. And surely that must be relatively easy to find and check, at least for small numbers of colours
These things are interesting, but what I would like to see in these videos is more explanation of why it matters.I know he talks about not knowing why it is true, but I want to know why it matters at all. Why does it matter if you can or can't color in numbers a certain color in a formula?
That's probably outside the scope of mathematics. For example, another similar problem is the graph coloring theorem. This actually is important in practice when you draw country maps, because it says that you will never need more than four colors (I think there's an older Numberphile video about this specifically). Maybe there's some similar application for this, but that's not the concern of mathematicians.
That is not something mathematicians worry about. Physicists and engineers care about the use of these things. But first comes the discovery, then the application.
I think the question of why it matters isn't a question for mathematics, it's a personal question. Because clearly it matters to him much more than it does to me, and maybe to you. He has a passion for math and for numbers, and passion is one of those things that the more you try to justify it to someone else, the muddier it gets. That isn't to say you can't share your passion with someone. But the more intense your interest in a field, the more alone you'll be. I think the reason this matters to this guy and to the people who did the research is because they like numbers and think this kind of thing is fun. With $100 at stake they're not in it for the money! Cheers!
I'm reminded of the following quote: Reporter: Mr Mallory, why do you want to climb Mt Everest? George Mallory: Because it is there. Mathematicians are like George Mallory. They don't care about whether doing something is useful, only whether it can be done at all, and how.
Not only is 7825 the sum of two distinct combinations of two perfect squares, but oddly enough, it's the difference of two perfect fourth powers: 13^4 - 12^4 = 28561 - 20736 = 7825
@ 10:00 Do you mean it is conjectured that if we have more than 2 colors the property will hold until 7824? Or do you mean it will always hold up to SOME n>=7824?
This vid popped on my recommendations and I'm first time watching this channel. Mby not the biggest fan of math but I absolutely love how passionate and happy this guy is when explaining all this :D
So there are no solutions for a+b+c when the total amount of numbers is 9, but it is possible for a²+b²=c²? And in that case also for 10,11, 12, ... 7824?
The 1-9 “a+b=c” problem is EASILY disproved simply because ever number can be used to equal 9, meaning that you would need a third color to at the very least be 9 in order to prevent a match with a, b and c.
Or maybe not. I haven't looked at the paper, but I suspect (given the size of the problem) that they stopped at the first "failure"; what if there are infinite others? Or demonstrably finite others?
dlevi67 The first failure is the only failure. Finding another failure would mean that you have to skip 7825. What if it's actually part of another triplet, in the same way that 5 is part of two triplets? What would you do at this point?
Tumbolisu The first failure is sufficient to declare failure over N. You (or as far as I know anyone) haven't demonstrated that there is/isn't a partition of the numbers _above_ 7825 (or above an arbitrary number greater than 7825). The number of triples that 7825 belongs to is neither here nor there, though it's clearly got to be more than one.
Mitchell Steinder I never said otherwise; quite the opposite. All I'm saying is that 7825 by itself may not be particularly useful on the way to an analytical solution, just like 2, as the first prime, isn't particularly useful to generalise about primes.
I love that James is still a guest on this channel. I think he was in the very first Numberphile video. That was like 5 years ago, wasn't it? Holy mackerel.
Isn't the reasoning easy? 5180 and 5865 also appear in two pythagorean triples (so each is already restricted to having to be opposing colours). So is this the lowest pythagorean triple where all three member appear in two pythagorean triples? So would a three colour number set break down at the lowest pythagorean triple where each member appears in 3 pythagorean triples? etc. ie. at any point we're all three terms of the pythagorean triple are restricted to a single colour, it breaks down because of the domino effect of these restrictions (through the other pythagorean triples they are a part of) on all other numbers within the number set which, in some cases (as denoted by the coloured numbers in the published chart), were already restricted to a single colour.
this is a trinary issue. since you are using 3 color possibilities, we can call them 0,1,2 in this proof we are trying to find connecting trinary values that do not repeat within different functions of base 10. this is unavoidable for ANY series if you get large enough. the reasoning comes down to combinatorial limits in 3 dimensions. we are trying to find the limit scales of platonic sets that scale without partial numbers (only whole). try doing these variants in a renderer like blender and you will find these same issues cropping up.
It turns out 9 is the smallest number for which the "a+b=c" coloring described at the start of the video is impossible. Here's one that works for 1 through 8: Color 1, 2, 4, and 8 red; color 3, 5, 6, and 7 blue.
There's another triple involving 7825 4695^2+6260^2=7825^2 It's an extension of sequence of 3^2+4^2=5^2 i.e.. 6^2+8^2=10^2 9^2+12^2=15^2 12^2+16^2=20^2 Since 7825 = 5*5*313 We can have (3*(5*313))^2+(4*(5*313))^2=(5*(5*313))^2 i.e 4695^2+6260^2=7825^2
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
I love these classic Numberphile videos where James Grime talks about one particular number
So do we.
make a table of video of every real number like periodic video
Can we add the sequence of numbers that numberphile has covered in the order they made the videos to the integer sequence database?
@@numberphile I only watch videos where James appear.
@@revenevan11 numberphile numbers
Probably my favorite guy on Numberphile.
He has his own channel as well, where he posts similar stuff (it's called singingbanana, though I don't know why). It's pretty inactive, but you can binge on some of his older videos if you haven't seen them.
i dont think i could ever like anyone more than i like cliff stoll
James Grime and Matt Parker are my favorites!
"Singing banana" captures his personality very well.
I think they are all very likable - and with very different personalities - but there is something very special about James :)
I can feel 7825. When I go somewhere where everyone is having a good time, everything stops when I get there.
100th like I know you don't care.
OMG... that is hilarious!!!
Especially when the address says 78125 and you are lost
@@brendawilliams8062 bad take
@@thorodinson6649 look Thor grab 1001035 and let’s trash the place
Interesting how 7, 8, 2, and 5 were the digits for 7,825 which also gave you the original issue in a+b=c.
*_lol blue and red makes purple, so 7825 is purple, silly mathematicians_*
I know!
James Grime said it wrong though, it doesn't need to be red and blue. It needs to be neither red nor blue... so 7825 is obviously yellow, but that leads back to his question at the end of the video of what if their were 3 or more colors would it inevitably still reach a point when it no longer works (assuming a finite number of colors).
with my experience of mixing multiple colors I bet you 7825 will be brown
7825 is purple? I didn't know a number could be comfortable!
And don't you wind up about powder. The powder is harmless.
Thijs van Dijk
I mean, it's a sort of splitting-the-difference-between-purple-and-brown sort of color.
I have to do some calculus exercises...
Me: "Okay, right now I'm absolutely not in the 'math-mood', let's just do other things first."
*Numberphile just uploaded a video*
Also me: Let's just watch this...
what is the derivative of an inverse function
depends
Maybe use that (f^(-1))' (y) = 1/f'(x) as long as f'(x) =/= 0 and assuming that f is differentiable? :)
pentix
I think you mean f'(x)≠0
Thanks, obviously it shouldn't have a pole at x=0, therefore f'(x) ≠0 :)
I always get excited when James grime is on numberphile
That blue pen appears rather green
It is an optical illusion caused by rather specific lighting condition. It is the same effect that occasionally makes the light from a row from four lightbulbs occasionally appear to be concentrated in a row of five point sources instead.
Parker pen
Oh no its another controversy
Jan Dvořák
It's not an optical illusion, i checked with a color picker. Its hue in HSL format is 159 which is pretty much green (a little bit on the turquoise side, but still objectively green).
Can't help but picture Picard: "There ... are ... four .... lights !!!"
Ramsay theory : The formula to find lamb sauce.
Expressed by the notation LS=f*ck
* Ramsey
Badhbhchadh no. It is ramsay
Its bland.
If you don't understand it, you're a sack of yankee dankee doodle shite.
Absolutely loves James Grime, always a pleasure to see him featured on Numberphile again.
Fine example of a Parker Grid at 2:31 there James
Will Price
Ah, yes. The Parker Grid, the ideal space for containing a Parker Square.
Except he already knew he was going to fail.
"imagine all supercomputers in the world checking all the posibilities since the dawn of time you still wont be able to check all the posibilities ... so they used some clever mathematics to reduce the number they had to check, and it took them about 2 days" i snorted when i heard this, but this is actually great!
Whenever there are videos with James in it, I never hesitate to see it at that instant.
I bet I could do this for a^3 + b^3 = c^3
Clingfilm Productions that would be very interesting to see XD! But will your proof fit in the margin of a book?
I can do it with one colour.
that's a joke about Fermat last theorem.
Clingfilm Productions Indeed, o man full of wiles. I'll raise you one and bet I can do a^4 + b^4 = c^4 in about the same time as you do the cubes.
I might be able to do for both a^3 + b^3 = c^3 and a^4 + b^4 = c^4 at the same time, although you need to give me some time.
James Grime appears in my subscription box: instant view.
That’s what we like to hear.
This is probably the least click-baity title I've ever seen. That's something I really like about this channel.
Always enthusiastic, always interesting. So glad James Grime does these.
This problem reminded me of a project I did five or six years ago in university, which was also about finding sets of numbers without a certain structure - in our case, we studied Szemerédi's Theorem - and we were using SAT solvers to do this. Turns out that this work is from our supervisor at that time - Marijn Heule was our supervisor. And, indeed he used SAT solvers to create this proof.
A little background: SAT this is the satisfiability problem, which asks "given a set of boolean variables (which can be either on or off) and a set of constraints on these variables, is there an assignment of values to these variables that satisfies these constraints?". You can translate the pythagorean triplets problem into SAT in its most basic form by creating a variable for each number from 1 to n, and creating a constraint for each pythagorean triplet stating that of the three variables corresponding to the three numbers at least one must be true (red) and at least one must be false (blue). If there is a solution for this SAT problem, then there is also a solution for the original problem, by colouring the numbers red if the variable is true, and blue if the variable is false.
The main advantage of this approach is that SAT is quite a well-known problem, having quite efficient solvers. I can imagine that someone must have created an efficient SAT solver for supercomputers, or at least the Texas supercomputer, as well. This way it would be relatively simple to tap into the huge amount of computing power available in an efficient way (which is really hard if you're doing that from scratch).
If I'm making it sound easy: it's not. The basics are quite easy to grasp (and can be coded in 30 lines of C), but in order to get an answer for these huge problems in a useful amount of time, you need to optimize your translation. This requires thorough knowledge of both the solver and the problem, to translate the problem in a way that is most meaningful to the solver. And finally, you need to handle all the data and include it in your proof.
Cool! I just learned how to come up with pythagorean triples. One of the side affects of watching numberphile: you might accidently learn something.
...and those two days of supercomputation probably cost more than 100$ :)
That's what grant money is for.
I think the check itself is worth more than 100$, so its even.
The university pays for the electricity.
WE pay for the electricity.
FOR SCIENCE!
it doesn't look blue because it's on brown paper.
If you draw on it with a yellow marker it would look orange.
but thats a story for a different video
I thought the reason was video coloration problems. Having experience with that very thing, that is my first guess. If someone can prove me wrong, that would be interesting also.
that makes sense, neutral browns are basically dark shades of yellow
It’s looks green
For the a + b = c case, there are really only 3 options, not 512 - either 1 and 2 are (say) red and 3 blue, or 1 and 2 are different colours and 3 is red or blue. The other colours are derived from there until it becomes impossible.
almightyhydra Hence why he said that it was easy to convince yourself it was true instead of simply checking every possible combination of red and blue for every space.
Four options, actually, because if 1 and 2 are the same colour, 4 can be the same colour as either 1 or 3.
There are exactly 512 ways to assign the colors. Whether those colors actually follow the rules or not is up for you to decide
Yes, you can reduce this space through logic. This is the kind of work mathematicians do all the time, but it’s not always so trivial to do so :)
5:30 Actually Ramsays theory allows you to solve for the location of the lamb sauce .
Where you referring your joke to the famous chef gordon Ramsay??.
@@andrjsjan4231 what does it look like
Michael Sowierszenko like your mom opssss by the way I didn’t understand your joke?
I laughed so hard when he so calmly said "so they reduced the number of scenarios they had to check to about 3 Trillion" XDD
I have synaesthesia only where numbers are concerned. As a child, I was trying to find a way to make the numbers add up in such a way that their respective colours would blend to give me the correct colour for their total. I quickly ran into the same problem you guys did.
Interesting!
That's interesting! I wonder if I have synaesthesia, except with feelings. I have certain emotions associated with numbers, and often, based only on how I feel about a given number, I can tell if it is prime. For example, 167. It feels sharp, and makes me feel edgy. One day, I was talking to my mom about my "synaesthesia", and threw out that number because it felt prime. Then, a few minutes later I looked it up, and sure enough! 167 is prime.
The trick to making this work is to assign numbers to colours in a balanced fashion, rather than linear. As in, red = 1 orange = 2 .....violet = 7, won't work. It would need to be arranged like the Chinese Bagua, where the opposing number was actually opposed on the spectrum.
Of course, with synaesthesia, you don't get to choose the assignments so whether it works would depend on your particular individual affliction. Sorry to hear you lucked out :( That would have been really frustrating!
Cube the Squid I def have some version of this as well lol
Colors are a three dimensional thing to our brains. This means you will always need three numbers to represent a color in a unique way. Single numbers just won't cut it.
He is just fantastic. Loves numbers. Loves math. Is completely consumed by his wonderful passion. And does his very best to spread the thrill and excitement.
"The number 7825 must be both red and blue, which it can't"
Superposition: "Bonjour"
well actually, 7825 must be none of red and blue.
Actually, it has to be neither or red or blue; if it’s either color, there is a trip,e of one color.
It is in the superposition |R>|B>|R'>|B'>!
(did I get the notation right?)
These videos have no expiration date. Enthusiasm for mathematics is everlasting.
I almost never understand this channel, but I love it anyway
how he is amazed about what he's talking about, it's contagious :)
this problem reminds me of the 17 sudoku clues problem
I don’t know what this man is saying but his excitement sells me every time.
If everyone had a teacher/professor as excited about Math as James then everyone would be a mathematician.
07:45 7825 can be both red and blue at the same time if you put it in a box! (Schrödinger joke)
I found a pattern where a^2+b^2=(b+p)^2 made pythagorean triples if p was a factor of a. (lowest factor being 2 if a is even)
I noticed in the video that for 7825; the two triples that make it up one of the two follow this pattern, and the other does not.. Namely 5180^2+5865^2=(5865+1960)^2 .. obviously 1960 is not a factor of 5180 in this case whereas 625^2+7800^2=(7800+25)^2 is.
It could be nothing, but.. This was the first time I've come across a case where the a pythagorean triple exists that doesn't fit the above format. (p not being a factor of a). Perhaps this is why?
Um, 7, 24, 25?
If you expand (b+p)² and cancel out b², you're left with a² = p(2b + p). So p (which is c - b) being a factor of a² is a necessary but not sufficient condition for a Pythagorean triple. I don't know how often it turns out that p is a factor of a, but I see quite a few examples on Wikipedia where this condition is not verified, regardless of which number you pick as a: (20, 21, 29), (33, 56, 65), (48, 55, 73), (65, 72, 97) etc.
@@yondaime500 more like a commonplace but not necessary condition. Let's move away from the cliché "necessary but not sufficient" which is tired, over-used and has become beyond vague.
7:52 - That's okay, you just have to defragment it and that will make the red go away. Or am I misunderstanding the problem?
The time estimate for a brute force approach is a bit off, because you don't have to check all of the possibilities. As soon as you come across a contradiction, you can stop and go on to the next set. Similarly, you only actually have to check 4 of the “512” combinations for the introductory problem to prove it never works (and indeed if you expanded the introductory problem to an arbitrary number of integers, you would never have to check more than those 4).
Number: **exists**
Numberphile: Do you have a problem?
Aaaaahh! I've missed seeing James!! 🤗🤗
9:25 "doesn't tell us why". didn't he explain why? 6:46 isn't the reason it doesn't work is because it's in two Pythagorean triples? if they know this is the reason it doesn't work, then why do they need to use brute force to prove it?
Great video, love that we're discussing a specific number
Ramsey theory has always fascinated me. Thank you so much for this video!
Furthermore, I notice that Mr Heule has been working on Polymath 16 (the Hadwiger-Nelson problem) and just broken his own record for the smallest unit-distance graph with chromatic number 5.This follows the recent breakthrough by Aubrey de Grey, although I imagine this record may not hold for very long, considering the great progress now being made in this area.It's not so difficult to explain .. so maybe a future numberphile video please?
If there is always a number you'll reach where these equations break once you count high enough, wouldn't this calculation reflect entropy? With entropy, systems lose organisation over time. With these patterns, the number you reach before being forced into a 3-of-a-kind pattern gets higher as you add more colours. So wouldn't it be possible that this equation could end up having relevance to the concept of entropy if and when we solve it?
It’s actually kind of the opposite. If more related numbers have the same color, then that represents more structure, which is to say less entropy
(The reason they are playing this specific game, avoiding 3 numbers of the same color, is that they want to see how *big* the system has to be before structure is forced to appear)
Thus Ramsey Theory says that large mathematical objects often have some inherent structure, or emergent properties, which actually act against entropy!
I feel like this connects to the structure of matter and the Universe in some way - Some matter/energy arrangements are just far more stable/favorable! Especially when moving towards larger scales
(Yes, over long time scales, eventually all order will disappear as entropy increases. But in the meantime, it’s truly remarkable how organized and structured everything seems to be)
Then proving that, for n colors, there will always be a perfect square which can be written as the sum of n other perfect squares and that those pairs must be the same color will prove the conjecture?
That would probably be ridiculous to prove, though
Where have you been James?! My favorite numberphile guy! Glad to see you back! J
That blue pen didn't look very blue to me
Agree ahahahah
probably the yellow-ish paper mixing to make it look green!
Nelson Goodman says hi to you
WanderingRandomer perfect amount of blue in the video
It is an optical illusion caused by rather specific lighting condition.
It is the same effect that occasionally makes the light from a row from
four lightbulbs occasionally appear to be concentrated in a row of five
point sources instead.
Love watching your channel! It’s the never ending pursuit of beauty!
*625² + 7800² = 7825²*
*5180² + 5865² = 7825²*
WTH, how will we construct a right triangle with this?
It's possible. The two right triangles are different. Using inverse trig functions the first one has angles 90, 85.42(approx) and 4.58(approx)
The second right triangle has angles 90, 48.55(approx) and 41.45(approx)
In fact I think this type of triangle seems to be rare in this case since we're using whole numbers. Otherwise we would have infinitely many such right triangles having same hypotenuse. Suppose we have a constant hypotenuse 'x' unit. The other sides are a and b. So, a=root(x^2-b2). Since x constant you can choose infinite values of b to get infinite values of a. Just make sure 0
I know this is an old post but someone else might stumble here...
It's the Pythagorean theorem (using really big numbers).
The theorem states that "in a right triangle a² + b² = c². Where side c is the side opposite the right angle."
In other words, if there were a right angled triangle with side (a) being 3 units long, side (b) being 4 units and you were looking for side (c)...
c² = a² + b²
c² = 3² + 4²
c² = 9 + 16
c² = 25
c = (sqrt) 25 (*sqrt 25 to remove the exponent from c)
c = 5
Using this formula and some trig, so long as you know the length of 2 sides and that the triangle has a 90 degree angle, you can find the length of the remaining side and the other 2 angles. (That's what Sadman did above.)
In the case of 7825 being c, there are two ways to form a right angled triangle and either option they choose of the two won't allow them to color the chart correctly to meet the other requirements and it ends their little game.
@@adraedin Thanks a lot. That's very helpful
That’s saying that if you construct a triangle with sides 625-7800-7825 or 5180-5865-7825, one of the angles of the triangle (the one opposite the 7825) is a right angle.
If you extend the case of sums of first powers to 3 colours (so we're excluding monochromatic triples (a,b,a+b), where a and b are distinct), then you can go at least as far as 22, using colour classes {1,2,4,8,11,16,22}, {3,5,6,10,12,19,20,21} and {7,9,13,14,15,17,18}.
I missed you Professor Grime 💕
I love how the start is basically tic-tac-toe with colors instead of x's and o's.
What, a new Internet Comment Etiquette AND a new Numberphile video at the same time? is it Christmas already??''
But it's not even July!
Aleatorio can’t wait for the collaboration
stylo big money salvia here bouncing on my boy's number?
imagine the venn diagram of the fans of both channels
Bush did 9/11
(_)(_)::::::::::::::::::D~~~~~~~~~~~@tedcruz~~~~~
Everybody knows *42* is the base number system used by the computronium running our reality [actually, it is _Quadragesimal_ (40) plus an independent "C.R.C." on _Binary_ (2)].
-> Knowing that *7825* is part of the equation leading to the "seed" used on this _particular run_ on the multiverse will get us closer to formulate the ultimate question.
Shout out to the nail and gear in the background!
The Mighty Nail and Gear.
@@numberphile I wonder where it is now
One of the most fascinating videos you've ever uploaded ! Great one !
"It can't be red and blue at the same time"
Violet: bonjour
Did you mean: *Purple*
Always happy to see James on numberphile😄
8:52 3.6 x 10^2355, for the record.
This was a fun video. I saw immediately that 1 thru 9 was impossible to not duplicate the same color. It happened for me when he showed the number 9 would be red. Bingo! That did it.
Then, when he was discussing the 7825 number, I knew it was too large to ever examine every possibility. That somebody narrowed just a subset to 1 trillion was amazing, and then took 2 days to run the sets, was also thrilling to me. This sort of problem is what makes math so wonderful and inviting to me.
Who else is here just because it has James Grime in it?
well this problem is also connected to fermat's last theorem as anything in the form a^n+b^n=c^n is part of the theory. If you noticed thought he addition of 1 to 9 also add the same issue as 7825. There are two sets of possible additions for 5 and 8 that break down the rules. 9 is another example of two pairs that are in opposite colors, so not matter what is chosen the coloring breaks. we might want to stick to the addition to understand this problem. multiplying is only causing us greater amount of work.
7825 should be either:
1. In a superposition of red and blue
2. Purple or violet
"...this kind of proof does not increase our understanding of why..." I would soften this statement, the tricks used to reduce the number of possibilities to check can reveal some interesting structures of the problem.
It does strike me as a potentially useful step in understanding why this is the case in that having a brute force example may help mathematicians identify patterns that hadn't been tangible before.
one color->5
two colors ->7825
Not a very useful pattern (though I understand what you mean to say )
We'll that's what they do lot of times
mnikhk what ?
4:32 omg I tried to find this for such a long time thank you!
Great video, though I'd mention that the Babylonians were making lists of Pythagorean triples long before the Greeks!
...but it did tell us something about why it failed at that particular number, James, it failed because it's in two sets of Pythagorean triples. Which will be where they fail for more colours as well, except that for three colours the number's gotta be in three sets of Pythagorean triples in order for it to break, and so on. And surely that must be relatively easy to find and check, at least for small numbers of colours
These things are interesting, but what I would like to see in these videos is more explanation of why it matters.I know he talks about not knowing why it is true, but I want to know why it matters at all. Why does it matter if you can or can't color in numbers a certain color in a formula?
That's probably outside the scope of mathematics. For example, another similar problem is the graph coloring theorem. This actually is important in practice when you draw country maps, because it says that you will never need more than four colors (I think there's an older Numberphile video about this specifically).
Maybe there's some similar application for this, but that's not the concern of mathematicians.
That is not something mathematicians worry about. Physicists and engineers care about the use of these things. But first comes the discovery, then the application.
I think the question of why it matters isn't a question for mathematics, it's a personal question. Because clearly it matters to him much more than it does to me, and maybe to you. He has a passion for math and for numbers, and passion is one of those things that the more you try to justify it to someone else, the muddier it gets. That isn't to say you can't share your passion with someone. But the more intense your interest in a field, the more alone you'll be.
I think the reason this matters to this guy and to the people who did the research is because they like numbers and think this kind of thing is fun. With $100 at stake they're not in it for the money!
Cheers!
It's hard to tell what's going to be useful in the future..
I'm reminded of the following quote:
Reporter: Mr Mallory, why do you want to climb Mt Everest?
George Mallory: Because it is there.
Mathematicians are like George Mallory. They don't care about whether doing something is useful, only whether it can be done at all, and how.
Not only is 7825 the sum of two distinct combinations of two perfect squares, but oddly enough, it's the difference of two perfect fourth powers:
13^4 - 12^4 = 28561 - 20736 = 7825
I would've loved to have you as my math teacher.
@ 10:00 Do you mean it is conjectured that if we have more than 2 colors the property will hold until 7824?
Or do you mean it will always hold up to SOME n>=7824?
I wonder if this has something to do with the built-in limitations in DNA.
I wonder how you were dropped as a kid SAD!
New Kid WTF is wrong with you
That's a neat idea, maybe there's something to that
This vid popped on my recommendations and I'm first time watching this channel. Mby not the biggest fan of math but I absolutely love how passionate and happy this guy is when explaining all this :D
"I didn't do it. I failed."
So this is a Grime Square™?
I’d love to live at that squareat nr. 7825.
So there are no solutions for a+b+c when the total amount of numbers is 9, but it is possible for a²+b²=c²? And in that case also for 10,11, 12, ... 7824?
Even tho I'm a tenth grader, and sometimes I don't decipher completely what numberphile's guests are explaining, but I still love this chanel😍.
The 1-9 “a+b=c” problem is EASILY disproved simply because ever number can be used to equal 9, meaning that you would need a third color to at the very least be 9 in order to prevent a match with a, b and c.
Wait so is 7825 the first time there is a c in two Pythagorean tripled that have to be of opposite colors
Spencer Schmidt -- I'm confused as well. Yes for 7824, no for 7825. But what about lower (and higher?) no's?
I'm a C shell user so consequently not the best person to ask, but I'm assuming it's more of a C++ thing. I feel old. Again
I noticed that the difference between 7800 and 7825 was 25, and that the multiple of 7800 squared was 25 to the fourth, squared. Why is that?
Can you talk about other patterns showing in the image? I see a lot of vertical lines.
That's just a consequence of the size of the grid they used
Related trivia: Ramsey Theory you mentioned and a vaguely similar problem about graph colouring is where we got the gigantic Graham's Number from.
Gotta love how briskly and casually he goes through the method of creating a Pythagorean triple. That could be its own whole video!
I love Dr. Grime's enthusiasm.
whats the name of the generalized version of this problem
Maybe what you are looking for is Ramsey Theory?
2:50
He turns into Michael Rosen for a split second when saying "twelve".
We may not understand why it's true, but knowing that 7825 has this property is surely a stepping stone towards finding an analytical proof?
Or maybe not. I haven't looked at the paper, but I suspect (given the size of the problem) that they stopped at the first "failure"; what if there are infinite others? Or demonstrably finite others?
dlevi67
The first failure is the only failure. Finding another failure would mean that you have to skip 7825. What if it's actually part of another triplet, in the same way that 5 is part of two triplets? What would you do at this point?
Tumbolisu The first failure is sufficient to declare failure over N. You (or as far as I know anyone) haven't demonstrated that there is/isn't a partition of the numbers _above_ 7825 (or above an arbitrary number greater than 7825).
The number of triples that 7825 belongs to is neither here nor there, though it's clearly got to be more than one.
dlevi67 I think you entirely missed the point. Analytical proofs are useful
Mitchell Steinder I never said otherwise; quite the opposite. All I'm saying is that 7825 by itself may not be particularly useful on the way to an analytical solution, just like 2, as the first prime, isn't particularly useful to generalise about primes.
I love that James is still a guest on this channel. I think he was in the very first Numberphile video. That was like 5 years ago, wasn't it? Holy mackerel.
CGP gray print could not be unnoticed
I had to scroll way too far to find this comment
I'm confused. He says there's no explanation yet, but didn't he give it? It's cuz 7825 occurs in two Pythagorean triples. What am I missing?
Hey Brady, please resume making videos on the wordsoftheworld channel! Bring it back to its past glory
Isn't the reasoning easy? 5180 and 5865 also appear in two pythagorean triples (so each is already restricted to having to be opposing colours). So is this the lowest pythagorean triple where all three member appear in two pythagorean triples?
So would a three colour number set break down at the lowest pythagorean triple where each member appears in 3 pythagorean triples? etc.
ie. at any point we're all three terms of the pythagorean triple are restricted to a single colour, it breaks down because of the domino effect of these restrictions (through the other pythagorean triples they are a part of) on all other numbers within the number set which, in some cases (as denoted by the coloured numbers in the published chart), were already restricted to a single colour.
Am I colorblind or is that blue green??
I think it's halfway between Battery charged blue and Deep green-cyan turquoise.
blue on yellowish brown paper appears green
this is a trinary issue.
since you are using 3 color possibilities, we can call them 0,1,2
in this proof we are trying to find connecting trinary values that do not repeat within different functions of base 10.
this is unavoidable for ANY series if you get large enough. the reasoning comes down to combinatorial limits in 3 dimensions.
we are trying to find the limit scales of platonic sets that scale without partial numbers (only whole).
try doing these variants in a renderer like blender and you will find these same issues cropping up.
It turns out 9 is the smallest number for which the "a+b=c" coloring described at the start of the video is impossible. Here's one that works for 1 through 8: Color 1, 2, 4, and 8 red; color 3, 5, 6, and 7 blue.
James' "blue" color doesn't look so "blue" to me... Is it my screen or is it really green?
Necko Agic it's more blue when the ink is wet.. I think it's the brown paper that makes it look green?
Necko Agic Yanny or Laurel?
For me the actual grid switches between blue and green
So of cause it didn't work!
Necko Agic it's because it's blue ink on brown paper, distorts the outcome.
There's another triple involving 7825
4695^2+6260^2=7825^2
It's an extension of sequence of
3^2+4^2=5^2
i.e..
6^2+8^2=10^2
9^2+12^2=15^2
12^2+16^2=20^2
Since 7825 = 5*5*313
We can have
(3*(5*313))^2+(4*(5*313))^2=(5*(5*313))^2
i.e
4695^2+6260^2=7825^2
it says 7825 views! o.0
Now I know that 7825 is the traitor with the Pythagorean Triple that is going to the ruin the red and blue integer set so thanks Grime.
And how can we use these results?
It will definitely be used in the future, but I'm really curious about how!