When it said "definitely don't pause and try this" I decided to write a program and set it to work; cause it seemed like a good stress test (and it still running almost a year later)
2:00 I tried 30 in my head a few times and gave up then pressed play. For some reason my ego said 'thousands of smart people could't solve it, my turn.'
for those that don't know, there actually is a character for cube [³] , just like there's one for squared [²] . They're Alt+0179 [U+00B3] and Alt+0178 [U+00B2] respectively
For those that don´t know, there actually is a character for cube [³] , just like there's one for squared [²]. They´re Alt Gr+[3] and Alt Gr+[2] respectively.
So all cubes (1, 8, 27 and so on) can has a parametric solution. I guess the ones with two summands x^3 and (-x)^3 are considered trivial solutions, but it is a bit weird that it was not metioned.
Shysterling28 except that's about a different number... The one where they did suggest we pause was about 30, the one where we're advised against it is about 33. Or have i missed the point you were trying to make?
I paused and tried it myself! Now my computer is in flames and space-time has been ripped in twain. Also, some demons flew out of the fissure. Like, a lot of them. Just a heads up.
9 лет назад+7
+JLConawayII So you were the one who freed Bill Cipher!
+JLConawayII Again?! You need to start being more careful. We only just finished mopping up the last of the demon-slime from your experiment last time Numberphile posted!
+Fidus V That means that there are more solutions than just the parametric ones given in the video. But Mr. Browning wanted to prove that there are infinitely many solutions and he did.
+Erik All perfect cubes are of one of the following forms: 9n-1, 9n, 9n+1 (ie they differ by at most one from a multiple of nine). You can show this by proving it for all the cubes from 0^3 to 8^3, after which the pattern repeats every nine cubes. Therefore if you add three cubes, you can get to at most three away from a multiple of nine. You can get to a number of the form 9n-3 if you add three cubes of the form 9n-1, and 9n+3 if you add three cubes of the form 9n+1 - you can also get to anything in between. However, you can clearly never get to any number of the form 9n+/-4.
+Erik Modular arithemtic solves it neatly. A cube modulo 9 can only be +1, 0 or -1 (also known as 8), and you can't reach 4 or 5 by adding three of them together.
For anyone watching this video in 2022 33 was finally cracked In 2019! "Andrew Booker, a mathematician at the University of Bristol, has finally cracked it: He discovered that (8,866,128,975,287,528)³ + (-8,778,405,442,862,239)³ + (-2,736,111,468,807,040)³ = 33"
I found it interesting that the two missing numbers are 4 and 5, then 13 (1+3)=4 and 14(1+4)=5, then 22(2+2)=4 and 23(2+3)=5, then 31(3+1)=4 and 32(3+2)=5 . Either the digits are 4 and 5 or there digits add to 4 and 5 at least up to 34 as they show.
+Furrane This trivial solution works for any number that is a simple cube itself. E.g. 27 = 3^3 +n^3 -n^3. Because it is so trivial I bet they didn't feel the need to show it.
+Furrane I'm assuming that the one shown in the video will give you *all* the solutions. Though it's true that your solution is perfectly adequate for showing that there are infinitely many solutions
+EdwardBerner Considering my equation works for every n and his works for every integer too, if you look closely at both you will come to the conclusion that both ( mine and the "m" equation ) are "subsets" of the set of all answers. Therefor they're both valid at proving his point. I do have to thank you though because I actually went back looking at his equation to figure if it was all the solutions.
+tbpotn I'm guessing that that doesn't count since it is a trivial solution for any integer of the form Z^3 (where Z is any integer). In addition, I'd guess that the equation above gets you to every solution of a^3+b^3+c^3 = 1, whereas yours for example won't get to solutions like 10^3+9^3+(-12)^3. Actually, the m equation doesn't get to every solution of the form of yours, it only gets 1^3+0^3+0^3, so neither equation gets every solution.
+Milkyway Squid no, there is no m in the integers (except 0) for which is |9m^4| = |-9m^4-3m| that is easily proven. -9m^4-3m=-3m(3m^3+1) while 9m^4=3m(3m^3) if we divide by 3m we get -(3m^3+1) and 3m^3 and their absolute values are clearly different.
This function generates infinitely many solutions to the equation at hand, but it does not generate all possible solutions. The one shown on screen does capture all, I believe.
Your work with numbers and how they allign with frequency is how i learn for i am a synesthete and within each sequence synchronisities fractalize stack and build to reveal the answer that quenches curiosity of experience
Ngl, when he said it’s suspected 30 has infinite solutions with no parameterization, I was kinda terrified. The idea of the infinite combined with the indescribable is the distillation of fear.
Looks like I am 7 years late to this. I just came across that they found solution for 33. When I think of square or cubic number, I relate them to size or volume. For example 33 cubic feet is a box 33 ft high, 33 long and 33 ft deep.
33 is congruent to 3 mod 5 implies a,b,c congruent to 1 mod 5. It also can mean -a is positive which is congruent to -1 mod 5. BTW, I liked this video so much.
576, 865 and 1512 but only when stored as 32 bit integer. 576^3 = 191,102,976 865^3 = 647,214,625 1512^3 = 3,456,649,728 sum = 4,294,967,329 But, 32 bit goes up to 4,294,967,296. After it wraps around, the leftover is... 33 (credit to David King)
This is so interesting. I'm currently in the process of relearning programming and now I'm thinking about creating a 3 cubes number crunching program. It would be actually pretty ridiculously simple on my end......just have to leave the program running for enough system processes to go through all the permutations in a set limit.
+Legendententertainment Yes that works, though is not one of the parametric solutions. I guess there are solutions that don't fit the parametric family, or maybe more than one parametric family.
7:05 if we choose m very big for this solution, then (1+9m^3)^3 + (9m^4)*3 is approximately equal to (9m^4+3m)^3. That is a near-miss solution for Fermat's last theorem.
As of 2019-03-09, Tim Browning has found a sum-of-cubes representation for 33; namely, 8866128975287528^3+(-8778405442862239)^3+(-2736111468807040)^3. That makes this video outdated.
Fun fact: everytime the number is + 9 and the number next to it meaning 4, 5 13, 14 22, 23 31, 32 40, 41 49, 50 58, 59 67, 68 76, 77 85, 86 94, 95 103, 104 112, 113 121, 122 130, 131 etc.
Yes, and that also implies that if there is a proof that for numbers of the form 9k+4 or 9k+5 there are no solutions, it means that no number of the form 9k+4 or 9k+5 is a perfect cube, which is not something that I for one would have guessed off the top of my head.
84 years ago I paused. I'm just now writing down the final digit of my life's work, which covers every square inch of slate, paper, and whiteboard on earth. Please copy it down before I die, which is in 13 minutes.
Solve this: 1 1 1=6 2 2 2=6 3 3 3=6 4 4 4=6 5 5 5=6 7 7 7=6 8 8 8=6 9 9 9=6 Just fill in math symbols (Plus minus times devided roots factorial brackets...but no integers or extra numbers e.g. Squares or cubes)
Draw a row of squares. From the center square form a column of squares. Now you have a giant plus shape made of squares. Start labeling the square by going from the center square out to the right. Label the squares as you go with 1 cubed, 2 cubed, 3 cubed, 4 cubed, etc. Leave the center square empty and work your up from the center. Label them the same as when you were going to the right. Go back to the center square and move to the left labeling all the negative cubes. Also label the negative cubes going down from the center square. Add more box so that instead of a plus you end up with a cartesian plane with some of the square filled in. The plus shape we constructed divided the plane into the four quadrants that cartesian planes tend to be divided into. Treat this like a times table only instead of multiplication you add cubes. For example the square when 3 cubed and 5 cubed meet should be show the sum of the two cubes. Fill in the four quadrants extend this however far you like and you start to generate the set of numbers that can be represented as the sum of two cubes, both positive and negative to clarify. With the first step done compile a list of sums of c cubed plus 33. Compared this list with your bijections and look for matches. If there is an answer that would be the way to find it.
Regarding the proof that the sum of cubes can never be of the form 9k+4 or 9k+5: You can show that for any integer n it holds that n³ is of the form 9k-1,9k or 9k+1. Just check the cases 0, .. , 8 and then use the equivalence a³ mod 9 = (a mod 9)³ mod 9 to see that it's true for all integers. Now it's clear that three of those terms can't be combined to get 9k+4 or 9k+5.
You should make a video about 4,8,15,16,23,42 - the very famous sequence of numbers from LOST. See does it have ANYTHING interesting about it, mathematically. Is there even a pattern? Numberphile are definitely the best guys to research this.
What's with all the comments not realizing you can't use decimals? They're so proud to have found a simple solution, but don't think that a computer could have?
The criminal numbers can be added to ultimately the number "9". 4+5=9, 13+14=27...7+2=9m, 22+23=45...4+5=9, 31+32=63...6+3=9. Perhaps the number 9 is manifesting on the linear series not as a particle, but as a wave viewed from another perspective. Puzzling how there are seven integers between 5 and13, 14 and 22 and 23 and 31.
I noticed that there is another (simple) solution for 1 outside the parametric solution. 1^3-1^3+1^3. So could there be more than one parametric equation for 1? ------- sorry answered my own question. this ones kind of simple and works for any perfect cube! 1^3+x^3-x^3
He didn’t explain why the numbers that aren’t on the list ( 4,5,13,14,22,23,31,32 ) can’t be written as; a^3 + b^3 + c^3. Can someone who knows why explain please. I really can’t sleep because of this.
Quick proof why numbers of the form (9k + 4) or (9k + 5) can't be written as the sum of 3 cubes: (9k)^3 = (9^3 * k^3) (remainder 0 when divided by 9) (9k+1)^3 = (9^3 * k^3) + 3(9^2 * k^2) + 3(9k) + 1(remainder 1 when divided by 9) (9k+2)^3 = (9^3 * k^3) + 3(9^2 * k^2)(2) + 3(9k)(2^2) + 2^3(remainder 8 when divided by 9) ... If you look carefully, you can see that only the last term matters when you are trying to find the remainder when divided by 9, because all other terms are multiplied by 9, so we can stop expanding the whole expression and just look at the last term, which is always the second number in the parentheses cubed according to binomial theorem. Doing that, we get: (9k)^3 / 9 --> remainder = 0^3 % 9 = 0 (9k + 1)^3 / 9 --> remainder = 1^3 % 9 = 1 (9k + 2)^3 / 9 --> remainder = 2^3 % 9 = 8 (9k + 3)^3 / 9 --> remainder = 3^3 % 9 = 27 % 9 = 0 (9k + 4)^3 / 9 --> remainder = 4^3 % 9 = 64 % 9 = 1 (9k + 5)^3 / 9 --> remainder = 5^3 % 9 = 125 % 9 = 8 (9k + 6)^3 / 9 --> remainder = 6^3 % 9 = 216 % 9 = 0 (9k + 7)^3 / 9 --> remainder = 7^3 % 9 = 343 % 9 = 1 (9k + 8)^3 / 9 --> remainder = 8^3 % 9 = 512 % 9 = 8 You can see that the only remainders we get are: {0,1,8} When we add up the cubes, you add up the remainders to get the remainder of the resulting sum when divided by 9. However, there's no way to add up 3 copies of 0, 1, and 8 to get either 4 or 5(which is the case because all of the numbers only have a maximum distance of 1 from the closest multiple of 9, and both 4 and 5 have a distance of 4 from the closest multiple of 9, and you can't get to 4 with a max of 3 copies of 1). This means that you can't sum 3 cubes to get numbers of the form (9k + 4) or (9k + 5), completing the proof. You can do a more elegant proof of this with modular arithmetic, but I wanted to keep it algebraic in case you didn't know how that number system works.
The Spegee thank you for the proof I really appreciate it and btw I know modular arithmetic can you proove it that way too if it won’t be to much work for you.
Aditya Yergolkar I once walked into a casino, went to the roulette table and put $5 on 33. Won and then left. I see this as EVIDENCE the number is meant for me
Proof of the 9k+4, 9k+5 property can be obtained by bruteforcing all modulo classes for a,b,c mod 9 (that makes 9^3 computations... which can be reduced by symmetry but stil). Python takes something sub-secondish for that computation.
If you think about it 30 = 1, theoretically any integer is equal to 1. So 1^3 + 1^3 + 1^3 + 1^3 all the way to 30 will equal to 30. Another way is, 9^3 + 1^3 + 1^3 + 1^3 = 30
According to wolfram Alpha there is no solution for 33. Wolfram Alpha gives me the result "False" for "solve 33 = a^3+b^3+c^3 for a, b, c over the integers".
Instead of "9xK+4" or "9xK+5" you can also say the sum of the digits is 4 or 5 (if it's more than 9 you take the sum of the digits of that result until you end up with one digit).
Proof is quite easy for (4, 5) (mod 9). Any integer cubed is has a remainder of either 0, 1 or 8 from 9 (can check by bruteforce). Then, the sum of 3 cubic numbers (mod 9) can be either 0, 1, 2, 3, 6, 7 or 8 (bruteforce - checking all combinations of 0 1 and 8). So, the sum of 3 cubic numbers cannot have a remainder of 4 or 5 (mod 9).
A way easier parametric solution for 1 would be 1^3 + m^3 + (-m)^3, which should work for any number if you use the number's cube root as the first number in the equation...what am I getting wrong here?...
A theorem about this equation: If x; y; z are integers such that x^3 + y^3 + z^3 = 33 then there exist two of x; y; z are congruent to each other modulo 9. Suppose that x ≡ y (mod 9). Then (x − y)(x + y)(x + 4y)(x + 3y) is divisible by 99.
Sorry, I paused and tried and got 30 (after 3 hours). I tries 33 and worked through the night to no avail.... they I click play, and smack myself on the head. *sigh*
And now 42 has been found as well! By the same guy who got 33.
(-80538738812075974)^3+80435758145817515^3+12602123297335631^3 = 42
Anyone can show the 3d graph of founded numbers ?is it possible ?3d representation of funded numbers in 3d space ?
Is that the answer to life, the universe and everything?
@@Siddoable no
I didn’t expect 42 to have an answer, but rather that 42 is an answer to anything
@@Siddoable yes, but the question to life the universe and everything, we’ll that’s a bit tricky.
"Don't pause and try it yourself"
The best advice you can give, knowing what type of people watches your videos.
Kind of like "kids don't try this at home".
Turns out someone did that
"But try it with 42 please"
@@munjee2 yeah a guy from numberphile himself.
Honestly I paused just before that message has wasted 15 minutes of my life trying to figure it out ,and DAMN I WOULD NOT HAVE BEEN ABLE TO DO IT
When it said "definitely don't pause and try this" I decided to write a program and set it to work; cause it seemed like a good stress test (and it still running almost a year later)
14ercooper any update on progress?
Progress: Zero
14ercooper did you use a binary searching technique? or just linear?
14ercooper still going or did you shut it down?
How much power are you throwing at it? How many threads? What kind of hardware? How many iterations have you already completed?
2:00 I tried 30 in my head a few times and gave up then pressed play.
For some reason my ego said 'thousands of smart people could't solve it, my turn.'
„What are you doing for a living“
„I am trying to calculate 33“
I dont want to break your 33 likes but I did it just to prove something
@@yourneighbour5738 oh! you're something
for those that don't know, there actually is a character for cube [³] , just like there's one for squared [²] . They're Alt+0179 [U+00B3] and Alt+0178 [U+00B2] respectively
I use alt gr + 2/3
^x is much simpler.
For those that don´t know, there actually is a character for cube [³] , just like there's one for squared [²]. They´re Alt Gr+[3] and Alt Gr+[2] respectively.
Also ¹,⁴,⁵,⁶,⁷,⁸,⁹ and ⁰
¹²³⁴⁵⁶⁷⁸⁹⁰
Great video. I love unsolved math problems that are easy to state and I had never heard of this one, really neat.
MindYourDecisions I love your vids
Fresh Tall Walker! Do you get your inspiration from Numberphile videos?
MindYourDecisions I
Sebastian Elytron, *presh
But 30 can also = to
10 cubed + -3 cubed + 3 cubed
There's a much simpler proof that 1 has infinite solutions.
1 = 1^3 + x^3 + (-x)^3
+Landon Kryger Cool :)
+Landon Kryger Same for any cube number!
y^3 = y^3 + x^3 + (-x)^3
So all cubes (1, 8, 27 and so on) can has a parametric solution. I guess the ones with two summands x^3 and (-x)^3 are considered trivial solutions, but it is a bit weird that it was not metioned.
+Landon Kryger simpler but not exhaustive
+Landon Kryger But with an infinite array of numbers, isn't there infinite solutions for all solvable ones?
I have a different set of solutions for 1: 1^3 + n^3 - n^3
Ivo van der Hoeven exactly, but following how it must be a^3 + b^3 + c^3 wouldnt the solution set be (1)^3 + (n)^3 + (-n)^3
even more generally, for any n = k^3, n = k^3 + m^3 + (-m)^3
NERDS!!! Haha!!! 😂
But in the video they're showing as a solution for 53 this 3^3+3^3+-1^3
I was screaming this inside my head the whole time
"Don't pause and try this yourself."
Oh, now you tell me. Thanks.
Shysterling28 you saved me. I read your comment right before that part
Shysterling28 except that's about a different number... The one where they did suggest we pause was about 30, the one where we're advised against it is about 33. Or have i missed the point you were trying to make?
@@zachyepez8279 lol wtf
He has a very pleasing voice.
it's the British accent mate :D
It's more than that, thought the British accent doesn't detract! I'm thinking more of his timbre.
+Brandon Shaffer Sounds like someone has a man-crush...
+Brandon Shaffer Dat thumbnail though...
+phoeNYkx Oxbridge accent :P
I paused and tried it myself! Now my computer is in flames and space-time has been ripped in twain. Also, some demons flew out of the fissure. Like, a lot of them. Just a heads up.
+JLConawayII So you were the one who freed Bill Cipher!
+JLConawayII Again?! You need to start being more careful. We only just finished mopping up the last of the demon-slime from your experiment last time Numberphile posted!
+JLConawayII meh. Not bad. but you can do better. Try opening the matrix next time
+JLConawayII Again!
+JLConawayII
69 upvotes
1:47 I paused. At least my computer checked every combination between -500 and 500. ^^
+ritwik sinha What do you mean?
+Fidus V That means that there are more solutions than just the parametric ones given in the video. But Mr. Browning wanted to prove that there are infinitely many solutions and he did.
I see.
+Piotr Matysiak 1 = 1^3 + n^3 - n^3 there you go i proved the infinite solutions without having to write a wierd equation.
+DorFuchs Dass man DorFuchs hier auch mal sieht :D
0:33 The number 20 appears twice lol
The Cracked Problem with 0:33
Ammar Faraz 0.33*
Ammar Faraz 11 months later you still have 20 likes! This whole comment thread is odd. 😁
Dammit Ammar ... your comment just went to 34, so i'll temporarily withdraw my own like to make it perfect again. For however long it might last. ;-??
@@Ammar34567 io
Back in the day when there were no solutions for 33😂
What’s the solution then?
@@fenrez3293 check the vid about 42 because 33 has been done
"Their great crime...."
@@fenrez3293 read the description
No math here
Teased with a proof you don't show! Arrgh!
+Erik All perfect cubes are of one of the following forms: 9n-1, 9n, 9n+1 (ie they differ by at most one from a multiple of nine). You can show this by proving it for all the cubes from 0^3 to 8^3, after which the pattern repeats every nine cubes.
Therefore if you add three cubes, you can get to at most three away from a multiple of nine. You can get to a number of the form 9n-3 if you add three cubes of the form 9n-1, and 9n+3 if you add three cubes of the form 9n+1 - you can also get to anything in between. However, you can clearly never get to any number of the form 9n+/-4.
+Erik Modular arithemtic solves it neatly. A cube modulo 9 can only be +1, 0 or -1 (also known as 8), and you can't reach 4 or 5 by adding three of them together.
+MasterHigure Ninja'd ;p
Alex Potts MasterHigure Thank you both =) Great explanations - this makes perfect sense.
+Alex Potts Your comment wasn't there when I wrote mine...
Lol, paused right before the "don't try this yourself" text XD
+MMMIK13 Me too, haha.
+MMMIK13 Same.
+MMMIK13 likewise! haha
+MMMIK13 20 years later...
:D
ITR Read it as "do stop and try it for yourself" day there for a few seconds, read my screen again and the realised.
I always wondered if I should have done a maths degree instead of a computing degree. This video has set my mind at rest.
both! :)
For anyone watching this video in 2022 33 was finally cracked In 2019!
"Andrew Booker, a mathematician at the University of Bristol, has finally cracked it: He discovered that (8,866,128,975,287,528)³ + (-8,778,405,442,862,239)³ + (-2,736,111,468,807,040)³ = 33"
wow it reached 10^15 number
I found it interesting that the two missing numbers are 4 and 5, then 13 (1+3)=4 and 14(1+4)=5, then 22(2+2)=4 and 23(2+3)=5, then 31(3+1)=4 and 32(3+2)=5 . Either the digits are 4 and 5 or there digits add to 4 and 5 at least up to 34 as they show.
In fact, this will always be true. A integer of the form 9k + 4 will always sum to 4, and an integer of the form 9k + 5 will always sum to 5.
1 = 1^3 + n^3 - n^3 for whatever integer n. No need for that complex "m" equation ^^
+Furrane Ikr xD
+Furrane This trivial solution works for any number that is a simple cube itself. E.g. 27 = 3^3 +n^3 -n^3. Because it is so trivial I bet they didn't feel the need to show it.
+Furrane I'm assuming that the one shown in the video will give you *all* the solutions.
Though it's true that your solution is perfectly adequate for showing that there are infinitely many solutions
+EdwardBerner Considering my equation works for every n and his works for every integer too, if you look closely at both you will come to the conclusion that both ( mine and the "m" equation ) are "subsets" of the set of all answers. Therefor they're both valid at proving his point. I do have to thank you though because I actually went back looking at his equation to figure if it was all the solutions.
+JesterIF They shown 1+0+0 which is a travial solution to my already so called trivial equation :p
Makes me think of narcissistic numbers. I.e.... 153 = 1^3 + 5^3 + 3^3
why do we do so difficult with 1? wouldn't 1^3 + (-k)^3 + k^3 work for any integer k?
+danai50 his example in the video was 1^3 + 0^3 + 0^3 so clearly that doesn't matter
+tbpotn I'm guessing that that doesn't count since it is a trivial solution for any integer of the form Z^3 (where Z is any integer). In addition, I'd guess that the equation above gets you to every solution of a^3+b^3+c^3 = 1, whereas yours for example won't get to solutions like 10^3+9^3+(-12)^3.
Actually, the m equation doesn't get to every solution of the form of yours, it only gets 1^3+0^3+0^3, so neither equation gets every solution.
+Milkyway Squid no, there is no m in the integers (except 0) for which is |9m^4| = |-9m^4-3m|
that is easily proven.
-9m^4-3m=-3m(3m^3+1) while 9m^4=3m(3m^3) if we divide by 3m we get -(3m^3+1) and 3m^3 and their absolute values are clearly different.
+Jakkal115 Just because it is trivial, doesn't reduce its validity to show 1 has infinite solutions. Another case wouldn't be more powerful.
This function generates infinitely many solutions to the equation at hand, but it does not generate all possible solutions. The one shown on screen does capture all, I believe.
Your work with numbers and how they allign with frequency is how i learn for i am a synesthete and within each sequence synchronisities fractalize stack and build to reveal the answer that quenches curiosity of experience
Ngl, when he said it’s suspected 30 has infinite solutions with no parameterization, I was kinda terrified. The idea of the infinite combined with the indescribable is the distillation of fear.
Looks like Chris went into Mathematics after surviving Until Dawn.
+Yojiro That's exactly what I thought! ^^
+Yojiro
That's what i thought all along. He looks identical to Chris.
Me too!
Yojiro i was waiting for this comment
I've come up with an elegant solution to this problem, but this comment box is too small to contain it.
Sure you did ;)
Schizophrenic Enthusiast
Whoops, I forgot it again. That's life.
+Scott Wallace Careful now, this just might be your last theorem.
Asad Mirza
Hehe, don't worry, I let better minds theorize for me. In my work I don't need anything more advanced than logarithms.
+Scott Wallace I see what u did there ;)
The most impressive thing in this video, is his massive collection of Yellow Springer books.
His what?
Looks like I am 7 years late to this. I just came across that they found solution for 33. When I think of square or cubic number, I relate them to size or volume. For example 33 cubic feet is a box 33 ft high, 33 long and 33 ft deep.
33 is congruent to 3 mod 5 implies a,b,c congruent to 1 mod 5. It also can mean -a is positive which is congruent to -1 mod 5. BTW, I liked this video so much.
Someone should give the 33 problem to any job interview candidates that they don't like.
If trump figures out the numbers he gets to be president
I'm sure Maxine Water and Nancy Pelosi can figure it out.
30 + 3 = 33
Was it that hard
Ieah Leen
Lol yes
Well this won't work anymore if he had Google
Now we got 33 and 42!
We got 33 in March 2019
Well the advice not to pause came on screen only after pressing resume...
576, 865 and 1512 but only when stored as 32 bit integer.
576^3 = 191,102,976
865^3 = 647,214,625
1512^3 = 3,456,649,728
sum = 4,294,967,329
But, 32 bit goes up to 4,294,967,296. After it wraps around, the leftover is... 33
(credit to David King)
This is so interesting. I'm currently in the process of relearning programming and now I'm thinking about creating a 3 cubes number crunching program. It would be actually pretty ridiculously simple on my end......just have to leave the program running for enough system processes to go through all the permutations in a set limit.
Or 1=1³+(-1)³+1³. Is there a reason they didn't mention this?
I was thinking this myself, it also doesn't follow the rule of (1+ 9m³)³ + (9m^4)³ + (-9m^4 - 3m)³
What about 1^3+ 1^3-1^3
That was my first thought and it seems to fit the bill.
That makes 2.
+Legendententertainment Yes that works, though is not one of the parametric solutions. I guess there are solutions that don't fit the parametric family, or maybe more than one parametric family.
+Vicvic W No it doesn't.
+Reece Robin He showed it wrong, it should be 1^3+ (1^3-1^3) which is 1 + 0 = 0
This was really fascinating, thank you guys. Is there any particular reason that the exceptions seem to occur in pairs?
7:05 if we choose m very big for this solution, then (1+9m^3)^3 + (9m^4)*3 is approximately equal to (9m^4+3m)^3. That is a near-miss solution for Fermat's last theorem.
For one there is another solution : 1^3 + (-1)^3 + 1^3
And I don't think it can be reached by using the "m" formula
The sound of that marker pen is killing me.
too much friction
Survival of the fittest... apparently doesn't include you
@@not2tired Come out on the streets , it won't suit you as well keyboard warrior🤣🤣🤣
how come sonic is 15 and hedgehogs only live 2-5 years?
he's a furry
And he bathes in baby unicorn blood twice a month
fot the same reason he is blue and walks on two legs
drugs
Simple. He moves so fast that time slows down for him relative to what we experience.
33 is solved now.. aaahhhh .. Came here after the latest numberphile video 😂
Numbers that are not 4 or 5 mod 9(do not give 4 or 5 as a remainder when divided by 9) is the sequence at the beginining.
I love this channel,numbers are facinating.Even more with all the weird properties you come up with.
I tried it for 10 minutes before he got to the don't pause part...
As of 2019-03-09, Tim Browning has found a sum-of-cubes representation for 33; namely, 8866128975287528^3+(-8778405442862239)^3+(-2736111468807040)^3. That makes this video outdated.
I found 3³+2³+(-1.25992105)³ to equal 33 in a few seconds...
33rd?
33rd dislike
33rd degree?
Fun fact: everytime the number is + 9 and the number next to it
meaning
4, 5
13, 14
22, 23
31, 32
40, 41
49, 50
58, 59
67, 68
76, 77
85, 86
94, 95
103, 104
112, 113
121, 122
130, 131
etc.
a^3 + b^3 + c^3 = 1 For example if a = 1, b = -1, c = 1, this is valid for all a=1 and b=-c :) Just another infinite series :)
Wouldn't any number that is a cube have infinitely many solutions? Like, if the number is n^3, the formula would be:
n^3=n^3 + x^3 - x^3
Yes, and that also implies that if there is a proof that for numbers of the form 9k+4 or 9k+5 there are no solutions, it means that no number of the form 9k+4 or 9k+5 is a perfect cube, which is not something that I for one would have guessed off the top of my head.
He's got some nice figurines of Tintin and Captain Haddock on his bookshelf :D
84 years ago I paused. I'm just now writing down the final digit of my life's work, which covers every square inch of slate, paper, and whiteboard on earth. Please copy it down before I die, which is in 13 minutes.
ThisMightBeMe
Ok
I have a lot of pieces of paper and they don't have your number on it. So obviously you didn't write it down on every paper on Earth.
@@medexamtoolscom i mean it was 2 years ago, it has since been transfered to a computer and the papers have been cleared.
Solve this:
1 1 1=6
2 2 2=6
3 3 3=6
4 4 4=6
5 5 5=6
7 7 7=6
8 8 8=6
9 9 9=6
Just fill in math symbols (Plus minus times devided roots factorial brackets...but no integers or extra numbers e.g. Squares or cubes)
Draw a row of squares. From the center square form a column of squares. Now you have a giant plus shape made of squares. Start labeling the square by going from the center square out to the right. Label the squares as you go with 1 cubed, 2 cubed, 3 cubed, 4 cubed, etc. Leave the center square empty and work your up from the center. Label them the same as when you were going to the right. Go back to the center square and move to the left labeling all the negative cubes. Also label the negative cubes going down from the center square. Add more box so that instead of a plus you end up with a cartesian plane with some of the square filled in. The plus shape we constructed divided the plane into the four quadrants that cartesian planes tend to be divided into. Treat this like a times table only instead of multiplication you add cubes. For example the square when 3 cubed and 5 cubed meet should be show the sum of the two cubes. Fill in the four quadrants extend this however far you like and you start to generate the set of numbers that can be represented as the sum of two cubes, both positive and negative to clarify. With the first step done compile a list of sums of c cubed plus 33. Compared this list with your bijections and look for matches. If there is an answer that would be the way to find it.
what about 1= 1^3+(-1)^3+1^3? it doesn't seem to fit that parametric equation
That can also be parameterized for all m. (1)^3 +(-m)^3 +(m)^3
Of course there are infinitely many ways to write 1 as a sum of three cubes:
x^3 + (-x^3) + 1^3
ive been seeing 33 everywhere and sure enough this video pops up under recommended.
Same here.
Me too.
33
I see it everywhere, as a matter of fact the time is 22:33 now
Daffeh Tutorials cool tnx
6:36
9^3 + 8^3 + (-6)^3 = 1 is a mistake. The correct equality would be 9^3 + (-8)^3 + (-6)^3 = 1.
omg that dry sharpie is killing me
Regarding the proof that the sum of cubes can never be of the form 9k+4 or 9k+5:
You can show that for any integer n it holds that n³ is of the form 9k-1,9k or 9k+1.
Just check the cases 0, .. , 8 and then use the equivalence a³ mod 9 = (a mod 9)³ mod 9 to see that it's true for all integers.
Now it's clear that three of those terms can't be combined to get 9k+4 or 9k+5.
You should make a video about 4,8,15,16,23,42 - the very famous sequence of numbers from LOST. See does it have ANYTHING interesting about it, mathematically. Is there even a pattern? Numberphile are definitely the best guys to research this.
Oh. Thank you!
What's with all the comments not realizing you can't use decimals? They're so proud to have found a simple solution, but don't think that a computer could have?
In fact, 1^3+n^3+(-n)^3=1 for all n. So there is another formula for infinitely solutions.
The criminal numbers can be added to ultimately the number "9". 4+5=9, 13+14=27...7+2=9m, 22+23=45...4+5=9, 31+32=63...6+3=9. Perhaps the number 9 is manifesting on the linear series not as a particle, but as a wave viewed from another perspective. Puzzling how there are seven integers between 5 and13, 14 and 22 and 23 and 31.
I noticed that there is another (simple) solution for 1 outside the parametric solution.
1^3-1^3+1^3. So could there be more than one parametric equation for 1?
-------
sorry answered my own question. this ones kind of simple and works for any perfect cube!
1^3+x^3-x^3
This is what immediately occurred to me, and I was wondering what is wrong with it? Nobody is answering. What's going on, it looks obvious.
Perhaps it's just too obvious for them to talk about.
He didn’t explain why the numbers that aren’t on the list ( 4,5,13,14,22,23,31,32 ) can’t be written as; a^3 + b^3 + c^3. Can someone who knows why explain please. I really can’t sleep because of this.
Quick proof why numbers of the form (9k + 4) or (9k + 5) can't be written as the sum of 3 cubes:
(9k)^3 = (9^3 * k^3) (remainder 0 when divided by 9)
(9k+1)^3 = (9^3 * k^3) + 3(9^2 * k^2) + 3(9k) + 1(remainder 1 when divided by 9)
(9k+2)^3 = (9^3 * k^3) + 3(9^2 * k^2)(2) + 3(9k)(2^2) + 2^3(remainder 8 when divided by 9)
...
If you look carefully, you can see that only the last term matters when you are trying to find the remainder when divided by 9, because all other terms are multiplied by 9, so we can stop expanding the whole expression and just look at the last term, which is always the second number in the parentheses cubed according to binomial theorem. Doing that, we get:
(9k)^3 / 9 --> remainder = 0^3 % 9 = 0
(9k + 1)^3 / 9 --> remainder = 1^3 % 9 = 1
(9k + 2)^3 / 9 --> remainder = 2^3 % 9 = 8
(9k + 3)^3 / 9 --> remainder = 3^3 % 9 = 27 % 9 = 0
(9k + 4)^3 / 9 --> remainder = 4^3 % 9 = 64 % 9 = 1
(9k + 5)^3 / 9 --> remainder = 5^3 % 9 = 125 % 9 = 8
(9k + 6)^3 / 9 --> remainder = 6^3 % 9 = 216 % 9 = 0
(9k + 7)^3 / 9 --> remainder = 7^3 % 9 = 343 % 9 = 1
(9k + 8)^3 / 9 --> remainder = 8^3 % 9 = 512 % 9 = 8
You can see that the only remainders we get are: {0,1,8}
When we add up the cubes, you add up the remainders to get the remainder of the resulting sum when divided by 9.
However, there's no way to add up 3 copies of 0, 1, and 8 to get either 4 or 5(which is the case because all of the numbers only have a maximum distance of 1 from the closest multiple of 9, and both 4 and 5 have a distance of 4 from the closest multiple of 9, and you can't get to 4 with a max of 3 copies of 1).
This means that you can't sum 3 cubes to get numbers of the form (9k + 4) or (9k + 5), completing the proof.
You can do a more elegant proof of this with modular arithmetic, but I wanted to keep it algebraic in case you didn't know how that number system works.
The Spegee thank you for the proof I really appreciate it and btw I know modular arithmetic can you proove it that way too if it won’t be to much work for you.
It's because they are numbers that are 9n+4 and 9n+5
@@spegee5332 legendary YT comment. What a guy
He just solved it.
The problem can restated as cubed(a) = cubed(b) + cubed(c) + N. and look for N=33, or N=-33.
He wrote the 1 solution out to prove that EACH number could be different, other than having 1 as the first integer each time.
Let me know when you crack it, 33 is my lucky number
+brod2man If you try to solve it, it might turn out to be your unlucky number...
Samuel Estenlund
Not if I DIE while trying, then I'll truly know it was meant to be with me forever.
brod2man well, if you do you may think so, but the rest of us will just think of it as your unlucky number :P
+brod2man 33 is both my lucky and favorite number too.
Aditya Yergolkar
I once walked into a casino, went to the roulette table and put $5 on 33. Won and then left. I see this as EVIDENCE the number is meant for me
It's been over 4 years and I still have it paused. I feel like I'm getting close
The number 33 has significant occult meaning as well. Along with this 33 is a very interesting number.
Ssssh numberphile doesn't want you to know that.
Proof of the 9k+4, 9k+5 property can be obtained by bruteforcing all modulo classes for a,b,c mod 9 (that makes 9^3 computations... which can be reduced by symmetry but stil). Python takes something sub-secondish for that computation.
Oops. 20 appears twice in the brownpaper graphic. I'm pedantic that way. -_^ I seriously love Numberphile. Keep on!
The first solution for 1 that I thought of was 1 = 1^3 +1^3 + (-1)^3.
1 = 1^3 + (-1)^3+1^3
I thought that would be obvious!
1:22 there are two twenty's on the list.
If you think about it 30 = 1, theoretically any integer is equal to 1. So 1^3 + 1^3 + 1^3 + 1^3 all the way to 30 will equal to 30. Another way is, 9^3 + 1^3 + 1^3 + 1^3 = 30
According to wolfram Alpha there is no solution for 33. Wolfram Alpha gives me the result "False" for "solve 33 = a^3+b^3+c^3 for a, b, c over the integers".
This is why I wish that I had a real crazy mathematician as a math teacher.
his 9's are horrible
+condenador The way he writes some of his numbers is very strange.
+condenador Yet, he knows way more math than you or me. Go figure.
he definitely does.
+condenador yeah those lefties nines written from the bottom kind of freak me out :P
+codediporpal Don't forget to mention he's left handed
Trying to unlock time and space are we ?
ha they dont have enough time left to do that
Or enough brown paper...
Instead of "9xK+4" or "9xK+5" you can also say the sum of the digits is 4 or 5 (if it's more than 9 you take the sum of the digits of that result until you end up with one digit).
I cracked it!!!
It's easy
let a = +3, then a3 = 27
let b = 6th root of 36, then the 6th root of 36 cubed = 6
then c = 0
then 27 + 6 + 0 = 33
AAAANNNNNDDDDDDD.....The problem is finally cracked!
I paused it to try it, like a fool
I find interesting that every k(3^2)+2^2 and k(3^2)-2^2 has no solution
Proof is quite easy for (4, 5) (mod 9). Any integer cubed is has a remainder of either 0, 1 or 8 from 9 (can check by bruteforce). Then, the sum of 3 cubic numbers (mod 9) can be either 0, 1, 2, 3, 6, 7 or 8 (bruteforce - checking all combinations of 0 1 and 8). So, the sum of 3 cubic numbers cannot have a remainder of 4 or 5 (mod 9).
1 can also be written as 1 cubed + (-1 cubed) + 1 cubed
I have the solution for 33, but its too big to write down in this comment.
Lynus Garrett Kho made me laugh. Nice one😀
I wanted to upvote this comment but it has 33 likes.
Lynus Garrett Kho You just made my day, maybe hundreds of years later that solution will be discovered
It's me Fade you only can use who,e numbers not decimals
that's irrational
Is the proof allready uploaded?
nice. I just found my birthyear with some puzzling in Excel.... (26)^3 + (-25)^3 + (3)^3 = 1978
I was born in 2003.2003-5=1998 which is 999*2 so there is no solution :(
For my birthyear there is also no solution for sum of 3 cubes.
A way easier parametric solution for 1 would be 1^3 + m^3 + (-m)^3, which should work for any number if you use the number's cube root as the first number in the equation...what am I getting wrong here?...
A theorem about this equation: If x; y; z are integers such that x^3 + y^3 + z^3 = 33 then there exist two of x; y; z
are congruent to each other modulo 9. Suppose that x ≡ y (mod 9). Then
(x − y)(x + y)(x + 4y)(x + 3y) is divisible by 99.
Well this is now solved!
Sorry, I paused and tried and got 30 (after 3 hours).
I tries 33 and worked through the night to no avail.... they I click play, and smack myself on the head.
*sigh*
First of all stop lying. Of course you won't be trying to find a solution to that problem all night.
33 is the answer to the universe... not 42 ???
*Life, the Universe, and Everything
Coincidentally, 42 is the other number below 100 which is left.
Please do a video about even numbers (0, 2, 4, 6, etc)....they are fascinating!
I learn so much Holy Math from Numberpile...thank you.. I put your stuff up many times..Ahman