What's hiding beneath? Animating a mathemagical gem
There is a lot more to the pretty equation 10² + 11² + 12² = 13² + 14² than meets the eye. Let me show you.
00:00 Intro
00:07 Animated visual proofs
03:35 Mathologer materializes
06:31 Three puzzles
07:45 Thanks!
Notes:
The beautiful visual proof for the squares pattern is based on a note by Michael Boardman in Mathematics Magazine: tinyurl.com/2d4y7wtf
As far as I can tell, I am the first one to notice that this beautiful argument also works for those consecutive integer sums (but I am probably wrong :)
I first read about the two patterns that this video is about in the 1966 book Excursions in Number Theory by Ogilvy and Anderson (pages 91 and 92).
The article "Consecutive integers having equal sums of squares" J.S. Vidger, Mathematics Magazine, Vol. 38, No. 1 (Jan., 1965), pp. 35-42. is dedicated to finding generalisations of the sort of equations that the squares pattern is all about. Here is a particularly, nice example derived at the very end of this article: 4² + ... + 38² = 39² + ... + 48². This article is on JSTOR www.jstor.org/stable/2688015.
I first encountered the Russian painting that puzzle 2 is about in an article by Ethan Siegel about 10² + 11² + 12² = 13² + 14² and Co. tinyurl.com/y7p5k4kw Nice find :)
365 is the smallest integer that can be expressed as a sum of consecutive square in more than one way 365 = 10² + 11² + 12² = 13² + 14² (and of course 365 also happens to be the number of days in a year :) Viewer Exception2001: Knowing the result, it's fun to think about making an efficient one-page calendar where the front is a 13x13 square and the back is a 14x14 square, with each square containing a date :D
Viewer k k notes that consecutive squares also take care of leap years :) 8² + 9² + 10² + 11² = 366
Christofer Hallberg did some computer experiments and found the following beautiful equation: 4³+...+28³=30³+31³+32³+33³+34³
There are some nice families of equations involving sums of alternating sums of consecutive squares. Check out Roger Nelsen's one glance proof tinyurl.com/2xauf83u
2² - 3² + 4² = -5² + 6²
4² - 5² + 6² - 7² + 8² = -9² + 10² -11² + 12²
...
Fun fact: the top part of the logo is the top part of the last image I show in the previous video • Fibonacci = Pythagoras...
Several viewers (Exception1, Nana Macapagal, B Smith, Shay) noticed that the projected cubes pattern differences are of the form n²(n+1)²/2 = 2(1 + 2 + 3 + ... + n)².
5³ + 6³ = 7³ - 2
16³ + 17³ + 18³ = 19³ + 20³ - 18
33³ + 34³ + 35³ + 36³ = 37³ + 38³ + 39³ - 72
56³ + 57³ + 58³ + 59³ + 60³ = 61³ + 62³ + 63³ + 64³ - 200
85³ + 86³ + 87³ + 88³ + 89³ + 90³ = 91³ + 92³ + 93³ + 94³ + 95³ - 450
And that actually means that the nice visual proofs in the video do extend to these modified cubes pattern because the six slices of the cube that I show in the video actually do form the shell of a smaller cube LESS two diametrically opposed corners.
For the the 4th powers differences the formula is 4³(1 + 2 + 3 + ... + n)³ = 8n³(n+1)³
7⁴ + 8⁴ = 9⁴ - 64
22⁴ + 23⁴ + 24⁴ = 25⁴ + 26⁴ - 1728
45⁴ + 46⁴ + 47⁴ + 48⁴ = 49⁴ + 50⁴ + 51⁴ - 13824
76⁴ + 77⁴ + 78⁴ + 79⁴ + 80⁴ = 81⁴ + 82⁴ + 83⁴ + 84⁴ - 64000
There is another nice piece of 4d hypercube geometry that goes with this observation.
The new emerging pattern then breaks again with 5th powers. Here the sequence of differences starts like this: 2002, 162066, 2592552, 20002600, 101258850, .... But who knows , ... :)
The triangular numbers tn=1+2+3+...+n that feature prominently in all this arrange themselves into a nice pattern like this
t1+t2+t3=t4
t5+t6+t7+t8=t9+t10
t11+t12+...+t15=t16+t17+t18
etc.
Solving x² + (x+1)² = (x+2)² has two integer solutions. The first is 3 corresponding to 3² + 4² = 5². The second is -1 corresponding to (-1)² + 0² = 1². You also get a second solution for every other equation in the square pattern.
(-1)² + 0² = 1²
(-2)² + (-1)² + 0² = 1² + 2²
etc.
Donald Sayers and qyrghyz point out that there is a nice discussion of minimal dissections of 6³ into eight pieces that can be reassembled into a 3³ a 4³ and a 5³ in Martin Gardner's book Knotted doughnuts and other mathematical entertainments, pages 198-200. A picture of a dissection like this is shown on this wiki page on Euler's conjecture tinyurl.com/27pkbj2c Another dissection here tinyurl.com/y6c6tbj4
If you are interested in more Mathologer animations of the type shown at the beginning of this video check out Mathologer 2 and the final sections of many/most regular Mathologer videos.
T-shirt: google "super pi t-shirt"
Music: Here to fight by Roman P. and Earth, the Pale Blue Dot by Ardie Son
Enjoy!
Burkard
I was not that surprised by the square pattern, nor the linear one. But I was truly amazed when he pointed out that in the linear pattern the natural numbers appear exactly one by one 😇
1 cubed plus 2 cubed equals 3 squared. (10sq + 11sq + 12sq + 13sq + 14sq) /365 = 2 because 10 sq + 11sq + 12sq = 13sq + 14sq and 169 + 196 = 365. 3cubed + 4 cubed + 5 cubed = 6 cubed.
Cube time! I stopped the video around 5:00 where you said those slices don't fit nicely around the cube. Well I say they fit pretty good if you just grab a couple of little unit helpers. So we get two 1^3 helper cubes to add to 5^3 + 6^3, then a 5 cube plus six 6x6 slabs plus two unit helpers in the corners we get the 7 cube putting everything together! So 2*1^3 + 5^3 + 6^3 = 7^3. So lets go further. Following from our two previous families, the central pattern for cubes looks like its gonna be 6=6*1, 18=6*(1+2), 36=6*(1+2+3), etc. So lets look at the second one. 16^3 + 17^3 + 18^3 ~= 19^3 + 20^3. We're gonna need more helper cubes, this time two sets - one for each big cube on the right. Putting stuff together we get 2*(1^3 + 2^3) + 16^3 + 17^3 + 18^3 = 19^3 + 20^3 makes perfect fit! Now it looks like our helper cubes have a pattern very similar to the central number: 6*1 2*1^3 and 6*(1+2) 2*(1^3+2^3). So now I'm gonna go right into the third cube equation with central number 36=6*(1+2+3) and say that we need 2*(1^3+2^3+3^3) helper cubes. So just jump right to the equation 2*(1^3+2^3+3^3) + 33^3 + 34^3 + 35^3 + 36^3 = 37^3 + 38^3 + 39^3, and it works! If you want to keep the Christmas tree aesthetic, we can call our helper cubes the ornaments :)
Ornaments, I love it :)
@@Mathologer 💫💫
This also extends to quads or whatever the 4th power is called. So for cubes we get this 5³ + 6³ = 7² - (2 * (1))² 16³ + 17³ + 18³ = 19³ + 20³ - (2 * (1 + 2))² 33³ + 34³ + 35³ + 36³ = 37³ + 38³ + 39³ - (2 * (1 + 2 + 3))² 56³ + 57³ + 58³ + 59³ + 60³ = 61³ + 62³ + 63³ + 64³ - (2 * (1 + 2 + 3 + 4))² And for quads, we can do this 7^4 + 8^4 = 9^4 - (4 * (1))³ 22^4 + 23^4 + 24^4 = 25^4 + 26^4 - (4 * (1 + 2))³ 45^4 + 46^4 + 47^4 + 48^4 = 49^4 + 50^4 + 51^4 - (4 * (1 + 2 + 3))³ 76^4 + 77^4 + 78^4 + 79^4 + 80^4 = 81^4 + 82^4 + 83^4 + 84^4 - (4 * (1 + 2 + 3 + 4))³ Can probably generalize that a bit more but its kind of interesting Edited, improved formatting a bit :)
Great!!!
@@tma8983 Nice, but you can get rid of all those 8/2 and make them a 4, looks nicer, and may help guide us to the solution for the pents.
For the second puzzle, my answer is 2 I already learned from the video that 10² + 11² + 12² = 13² + 14² so the numerator is just 2(13² + 14²). When (slowly) performing the addition in my head, I was surprised that 13² + 14² actually equals to 365, cancelling out the denominator and leaving 2. Knowing the result, it's fun to think about making an efficient one-page calendar where the front is a 13x13 square and the back is a 14x14 square, with each square containing a date :D
I like your calendar idea :)
I think I need to build that calendar. Just putting the julian dates into 2 squares. And as Simone Giertz said for her Every Day Calendar, the leap day is the day you should take off, sleep in and have ice cream with a movie.
Ypu can also do (12-2)^2+(12-1)+12^2+(12+1)^2+(12+2)^=5×12^2+2×1^2+2×2^2=720+10=730
@@Mathologer I used the exact same reasoning for that challenge! 10^2+11^2+12^2 = 13^2+14^2, (you said that at the beginning of the video) which I already knew was 365.
@@Gunstick Historically, the leap day was when you were getting a new calendar anyway, because it was the end of the year. (Yes, the year originally started with March. It was originally a Roman military calendar that started with the first march of the army in the spring, and stopped after ten months, because winter. January and February were added when the calendar was adapted for civilian use, and then some guy named Caesar added the leap year every fourth year rule so he could be out of the capital on long extended military campaigns, instead of sticking around to declare leap days in person. Starting the new year in January happened *much* later, and was originally a fiscal year. People wanted to do time-consuming year-end things like annual inventory during the winter because business was slow then.)
I taught woodwork for years, in good selective schools and there were always pupils that thought they could build a hollow cube out of six equal squares of plywood. My boss would actually cut them the squares and watch then trying to assemble them. I regret that it did not occur to me to tell them it was impossible because of Fermat's last theorem, that would have been cool. Martin Gardner showed a nice dissection of a 3, 4 and 5 cubes into a small number of pieces that could be assembled into the bigger cube.
"Martin Gardner showed a nice dissection of a 3, 4 and 5 cubes into a small number of pieces that could be assembled into the bigger cube." Do you still remember where you saw that? Was there any talk about what was being shown being the smallest number of pieces? Also, you may have mentioned that in previous comments, but where are you from ?
@@Mathologer Oct 1973 Scientific American, reprinted in "Knotted Doughnuts and Other Mathematical Entertainments".
@@qyrghyz Great, thank you very much for that. Just looked it up in Knotted Doughnuts. Must have seen this before ... :)
@qyrghyz Just noticed that they show a picture of such a dissection on the wiki page dedicated to Euler's conjecture :) en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture
@@Mathologer Ooh Pretty! I missed that.
For the blackboard problem, it occured to me that 10²+11²+12²+13²+14²= 5 × 12² + 2 × 1² + 2 × 2² because the double products of (12+x)² and (12-x)² cancel. Knowing 12²=144, it is straightforward that 5 × 12² = 1440/2 = 720. The remaining squares sum up to 10 and we get 730, which is 365 × 2.
Mathematicians doesn't believe in "coincidence", these are true mathematical Facts. 😌 3³+4³+5³=6³ but not possible for further case. Pls continue doing such 10-15 min videos also, Sir. 🙏🏼 Long vid are itself goldmine. ❣️❣️❣️
Okay, but do these "facts" work in other base systems?
@@arikwolf3777 yes. Nowhere do any of the equations depend on the digits of the numbers
@@arikwolf3777 This is pure arithmetic with everything on the table, not pseudo-numerologic digit sums making the hidden assumption that we work in Base ten, then making all kinds of misleading conclusions about how nine is the most special number blah blah. In REAL numerology we find mystical meaning in rigorous mathematical facts, with all mathematical assumptions as clear as possible, not in half-baked half-truths and falsehoods. Thus e.g. ten is a special number because it is both 10 = 1 + 2 + 3 + 4 and 10 = (1) + (1 + 2) + (1 + 2 + 3), not because our Homo Sapiens human bodies have ten fingers. We need to clearly distinguish cause and effect, and in numerology we (truly or falsely) claim that the causes are the eternal mathematical truths, and the effects are the dense material and dense psychological "coincidences" of our sensorial imperfect world. Thus any arguments for Base ten "coincidences" being relevant for Homo Sapiens humans must be tied to that the number of our fingers/digits equals ten being an EFFECT that is CAUSED by the mathematical and mystical properties of the number ten AS SUCH, and the providentially chosen place of our human evolution in the great non-random Divine scheme of manifestation. Or something similar to that. You cannot just claim "there are only ten digits, how could there be more digits?" or "the human (body, psyche and mind together) is the measure of everything".
There are also mathematically rigorous "exceptional" patterns that are only finite, not infinite (as long as we stay in certain "worlds", like finite dimensional space, locally Euclidean/positive definite metrics, loops being associative (i.e. groups) or Moufang, space being Euclidean or Elliptic not Hyperbolic etc). One such finite, exceptional pattern is the E series of crystallographic root systems, Lie algebras and Lie groups. Note that if we watch the diagrams of the affine ~E series root systems, then ~E_8 has only a 1-fold diagram symmetry (i.e. no symmetry), ~E_7 has only a 2-fold diagram symmetry, ~E_6 has "only" a 3-fold diagram symmetry, ~E_5 = ~D_5 has only a 4-fold diagram symmetry, ~E_4 = ~A_4 has only a 5-fold diagram symmetry, ~E_3 = ~A_1 + ~A_2 has only a 6-fold diagram symmetry. To be rigorous, the symmetry groups of the affine E series root systems are (going from ~E_8 to ~E_3): Triv = Cyc1 = Sym1, Cyc2 = Sym2, Sym3 = Dih3, Dih4, Dih5, Sym2 × Sym3 = Dih6. Of cardinalities 1, 2, 6, 8, 10, 12. So it seems that an order 2 mirror symmetry is "missing" in the diagram automorphism group of the two largest root systems of the affine ~E series. This is to be expected though because the Dih series connection to regular polygons break down for digons/dihedra and monogons/dihedra if we only count allowed permutations of vertices rather than permutations of edges and faces (seeing polygon dihedra as sitting inside spherical space S2 and dihedral groups as sitting inside SO(3)). So it seems that when it comes to diagram symmetries of affine root systems, the ~E series behaves almost inverted to the ~A series who have Dih_(n+1) symmetry for ~A_n (except for ~A_1 which has Cyc2 symmetry, again expected), while the ~D series is more or less constant with the notable exception of ~D_4 having Sym4 spacial 4-fold diagram symmetry, while all the other ~D_n for n >= 2 have Dih4 diagram symmetry.
Oh, and one more thing: God geometrizes. 😜
Awesome video. From the animation to the soundtrack to the explanation. Just amazing.
That was a wonderfully powerful intro. So beautiful and fully deserving of the absolutely epic music.
This channel is mostly above my head but the quality of education here never ceases to amaze me
Glad you enjoy it!
Agreed, I like to take some of the simpler things into 7 to 12 grade math classes just to show them that math can be fun.
And thank you so much for getting another video out before Christmas. Especially one with things to ponder over the break!
In principle I've got another short one ready and so who knows, there might even be yet another video before the end of the year :)
@@Mathologer … the gift that keeps on giving :)
Surprisingly fun for such a short video! Also, as a big fan of empty operations, I was quite happy to see the "0 =" and "0^2 =" at the tops of the trees :)
I think you are the first one to agree with me in this respect (and say so :)
I thought this vid was going to be more than 30 minutes long, but it's still nice to have an occasional short video from time to time! :) As for the puzzles: 5³ + 6³ ≈ 7³ (just off by two which just so happens to be the cube root of the next number 8) The answer to the second puzzle is 2 since 10² + 11² + 12² = 13² + 14² = 365 (I knew it was 365 earlier since when mentioning this equation some people like to point out that there are about 365 days in a year!) Funnily enough the sequence in the third puzzle goes 5, 6, 6.893..., 7.80557... which seem to round up to 5, 6, 7, and 8. Now I'm curious to know if this rounding pattern continues or if it's just a coincidence . . .
Cannot tell you how much I enjoyed making this short video. The long ones are real killers to make :) Also your 365 remark is spot on. People do find it remarkable that the result is the number of days in a year although that is really just a coincidence :)
the next pattern does in fact round up to 9 and i expect it to continue for several more terms but eventually diverge. it appears to simply be getting further away from the """expected"" value
The rounding pattern continues forever. Indeed, 3^n+...+(n+2)^n < (n+3)^n for all n>3. Moreover, solving 3^n+...+(n+2)^n = x^n, we find that x-(n+2) converges to 1-ln(e-1)=0.458... Fact 1: We have 3^n+...+(n-k)^n < (n-k+1)^n for all integers k >= 0 and n >= k+3. Integrating x^n from 3 to n-k+1, we find that 3^n+...+(n-k)^n < 1/(n+1) * (n-k+1)^(n+1). The result follows, because n-k+1 = 0 and n >= k+3, we have the following. (n-k+1)^n+(n-k+2)^n+...+(n+2)^n < 3^n+...+(n+2)^n < 2(n-k+1)^n+(n-k+2)^n+...+(n+2)^n By Fact 2, dividing everything by (n+2)^n, the left hand side converges to e^(-k-1)+e^(-k)+...+1, and the right hand side converges to 2e^(-k-1)+e^(-k)+...+1. Letting k go to infinity, we find that (3^n+...+(n+2)^n) / (n+2)^n converges to 1+e^(-1)+e^(-2)+... = e/(e-1). Fact 4: For any sequence of positive real numbers a_n converging to a positive real value c, the value n * ((a_n)^(1/n) - 1) converges to ln(c). For any ε>0, we have the following for n large enough. (1 + ln(c - 2ε)/n)^n < c-ε < a_n < c+ε < (1 + ln(c + 2ε)/n)^n ln(c - 2ε) < n * ((a_n)^(1/n) - 1) < ln(c + 2ε) Fact 5: The value (3^n+...+(n+2)^n)^(1/n) - (n+2) converges to 1-ln(e-1). Define a_n = (3^n+...+(n+2)^n) / (n+2)^n, such that (3^n+...+(n+2)^n)^(1/n) - (n+2) = (n+2) * ((a_n)^(1/n) - 1). By Fact 3, the value a_n converges to e/(e-1), so (a_n)^(1/n) - 1 converges to 0. By Fact 4, the value n * ((a_n)^(1/n) - 1) converges to ln(e/(e-1)) = 1-ln(e-1).
If you don't already know that 10² + 11² + 12² = 13² + 14² then you can still calculate it quickly by noticing that it's approximately 5 x 12² = 720, plus an "error term" of (13-12)(13+12) - (12-11)(12+11) + (14-12)(14+12) - (12-10)(12+10) = (13+12-12-11) + 2*(14+12-12-10) = 2+8 = 10. In fact, in general, (a-2)² + (a-1)² + a² + (a+1)² + (a+2)² = 5a² + 10.
@@SmileyMPV n=39 gives sum(k^n,k=3..n+2)^(1/n) = 41.4992... ≈ 41 < 42
So for that first puzzle... I think my mind just got blown I started with 5 since I was basing on the fact that the first equation of both patterns end and start with 3 (1 + 2 = 3 and 3² + 4² = 5²), so why not do the same for the cube pattern. What I ended up is that 5³ + 6³ = 341, which is almost close to 7³ = 343. I was curious enough, so I tried to expand this by utilizing the same trick from the previous patterns and used 6*(1 + 2) for the next pattern and got 16³ + 17³ + 18³ ≈ 19³ + 20³ (with the difference being 18) Expanding this pattern leaves me with a list of differences and... I don't know about you, but (2, 18, 72, 200, etc.) just screams "I have a pattern"... and it does! Each difference is just twice a triangular number *squared*, and knowing that just blew my goddamn mind... because THAT'S LITERALLY HOW I STARTED working on this pattern. For each sum that uses 6(1 + 2 + ... + n) cubed as a starting point, there is a difference of 2(1 + 2 + ... + n)². How cool is that!? In honor to document this amazing pattern, here's my christmas tree for the cube pattern, with the difference added in. You can think of them like they're ornaments or something xD (the formatting might only work with monitors) 5³ + 6³ = 7³ *- 2* 16³ + 17³ + 18³ = 19³ + 20³ *- 18* 33³ + 34³ + 35³ + 36³ = 37³ + 38³ + 39³ *- 72* 56³ + 57³ + 58³ + 59³ + 60³ = 61³ + 62³ + 63³ + 64³ *- 200* 85³ + 86³ + 87³ + 88³ + 89³ + 90³ = 91³ + 92³ + 93³ + 94³ + 95³ *- 450* ... and here's another one with the sum expanded, just to see the beauty :D 5³ + [6(1)]³ = 7³ - 2(1)² 16³ + 17³ + [6(1 + 2)]³ = 19³ + 20³ - 2(1 + 2)² 33³ + 34³ + 35³ + [6(1 + 2 + 3)]³ = 37³ + 38³ + 39³ - 2(1 + 2 + 3)² 56³ + 57³ + 58³ + 59³ + [6(1 + 2 + 3 + 4)]³ = 61³ + 62³ + 63³ + 64³ - 2(1 + 2 + 3 + 4)² 85³ + 86³ + 87³ + 88³ + 89³ + [6(1 + 2 + 3 + 4 + 5)]³ = 91³ + 92³ + 93³ + 94³ + 95³ - 2(1 + 2 + 3 + 4 + 5)² ...
ALSO just to add, because (1 + 2 + ... + n)² is just 1³ + 2³ + ... + n³, we can replace each squared triangular sum with that so we can get rid of the pesky squares and get a purely cubic pattern :D
Very nice :)
@@Exception-mk3xh that cubic pattern does indeed have a geometric interpretation. You just need to write it like this: 5³ + 6³ + 2(1³) = 7³ for it to make sense After wrapping the slices of the cube around the 6 faces (3 around a corner) you have just enough room for the correcting cubes to fit in on the 2 opposite corners
Well done, was just about to try to figure this out and then realised it might have been done down here
Patterns. Animation. Understanding. One follows the other. Explanation in its purest form without danger of distraction. This is educational genius.
What a treat! Thank you Mathologer, happy holidays!
Es bewahrheitet sich immer wieder: Die wirkliche Kunst ist es, Schwieriges so einfach zu vermitteln, dass es aussieht als wäre es selbstverständlich. Top gemacht!! Vielen Dank 🙂
This was fascinating! I enjoyed the shorter form for a change, and I look forward to the next deep dive.
Glad you enjoyed it!
Noticed another "super pattern" for the squares. Looking at the base of the last term of one row, you can derive the base of the first term of the next row. Multiply the base by (n+1)/n, where n = the row number. For example, the base of the last term of the 1st row is 5, and 5 * 2 / 1 = 10, which is the base of the first term of the second row. 14 * 3/2 = 21, 27 * 4/3 = 36, etc...
Every other triangular number
@@DitDede specifically, it's every even triangular number! So it can be described as r(2r+1), where r is the row number. It also looks like the last term is the odd triangular number - 1 or (2r+1)(r-1)
you have blown my mind once again. absolutely loved this video, and the animation was amazing! amazing work!
Glad you liked it!
I once noticed the square pattern had triangular numbers somewhat at its heart. Happy to have stumbled upon this video!! The pattern involving the squares seems like another extension of the Pythagoras Theorem just like De Gua’s Theorem.
These visual proofs are so comforting to watch :)
5^3 + 6^3 = 341, which is short of 7^3 by just two. I love that because you'd already mentioned trying to wrap a cube in square slices, I immediately imagined such a wrapping where two opposite corners were missing (though without checking I'm not sure that could really be done without further slicing the slices).
Actually, that almost wrapping that you mention there extends to a generalized cubes pattern. I am saying something about this in the description of this video. Check it out :)
@@Mathologer I just read it - thank you! I also love the hypothetical calendar arrangements. I once designed a calendar based on the Twelve Days of Christmas song, which features tetrahedral numbers - 364 is a tetrahedral number, and is actually the total number of gifts given in the song. I made Dec 25th a special "non calendar" day, and then had a series of months whose lengths were the first 12 triangular numbers, summing 364. A fairly silly calendar, but good fun. I have a spreadsheet that lets me generate it for any given year :) This actually came about because I was writing a novel that featured some magic based on the song... but it never got published (yet).
For the list which begins with 3^2+4^2=5^2. You had a short cut (formula) at the item before the equal sign. I have a short cut (formula) to jump from the last item of the previous line to the first item of the next line. 1st line ends with 5 2nd line begins with 10. Because 5 (last of previous line)/1(number of items on the right side)*2(number of items on the left side) 2nd line ends with 14 3rd line begins with 21. Because 14 (last of previous line)/2(number of items on the right side)*2(number of items on the left side).
I love the no comment animation at the beginning. Well paced to be able to capture the sequence. Maybe do some more of this conzent. Not for shorts though as there's not enpugh time.
Great video! Another way I noticed of describing why that (1st power) integer sums pattern works is that the left hand side always begins with a square number n^2, and apart from that term, there are n terms on each side, with each of the terms on the right side each having a term on the left side which it is exactly n larger than. For example, with 9+10+11+12 = 13+14+15, the terms 13, 14, and 15 each have a term on the left hand side they are 3 larger than (10, 11, and 12) making the difference between those be 3 times 3, which equals the square number 9 on the left.
Hello 💪I remember you made a short about this
@@jasimmathsandphysics Hey, yeah I've mentioned each of these identities in my videos before, but never noticed this connection between the two of them until seeing this Mathologer video :)
Hi combo class, I enjoy your videos. I know the pattern. Look at the other comment I made on this video.
@@ComboClass Hi, Domotro! In which video did you talk about the linear pattern?
@@wyattstevens8574 It was one of the shorts on my bonus channel @Domotro , it hasn't been in a full episode yet but may appear when it fits
Video still on before start ads and i already love it. That image next to 'skip ad' is 👌
Finally some music and math together! Wonderful work done mathologer!
I have a degree in maths and physics and I am working in the field so math is all I do. Yet you manage to keep finding this amazing beautiful things in such simple ideas like Pythagorean triplets and circles which I thought for a long time have nothing to them. Absolutely amazing and beautiful thank you
Some years ago, I noticed the pattern of the first entries: (2n-1)^n + 2n^n = (2n+1)^n. Though; of course, knowing Fermat’s last theorem, I knew the pattern wouldn’t hold exactly, from cubes onward; but still, I was pleased to find that the near-miss -solutions seem to go on forever and ever after. You just need to round off the left sides of the equations to the nearest whole numbers, like so: [(2n-1)^n + 2n^n] = (2n+1)^n. Sadly, my friend and I couldn’t prove this conjecture; maybe, because we didn’t think to consider other entries, and we tried to find an algebraic proof, because of Mathematical rigor.
This video calmed me down with all its magic glory
Vielen Dank für das tolle Video, es ist in höchstem Maße faszinierend. Es löst Glücksgefühle aus. Unglaublich...
Wouldn't mind these shorter videos coming more often between masterclasses. Merry Christmas professor.
Great animation, very good explanation
Thanks! This is the kind of videos where, when you want to click LIKE at the end, you realize you already did it earlier because it was so good.
Glad that this video worked so well for you :)
I tried tiling triangles with small triangles. Every row has 2 more small triangles than the row above it, so that the total number of small triangles is 1, 4, 9, 16, 25 etc. (I.e. square numbers.) You can then slice such a tiled triangle so that it envelops another tiled triangle. For example, 9^2 + 12^2 = 15^2. The slices look pretty bad though, and not intuitive at all. 😀
Main thing is that you are having fun :) And definitely a worthwhile thing to try :)
This was beautiful to watch 👍🏻
For the second problem: the bogdanov-belski painting. I already knew it was 2. -) But yeah following the video: It's 2x(10²+11²+12²). And 100+121+144=365, goes indeed fast. (memorized) - ) 13²,14² (never memorized those for some reason): I tend to do 10x16+9 and 10x18+16, which is also still fast. (explanation: 13² = (13-3)x(13+3)+3²). So for instance if I have to square 62: 60x64+4=3844, goes decently fast. It's somewhat similar I think to what was described in the video: 7² = 6x8+1 = 5x9+4 = 4x10+9 = 3x11+16 = 2x12+25 = 1x13+36. Note that the added numbers are all squares. Each time you take a column of a squared number and you turn it into a row, you'll lose a square.
ditto, but for 10²+11²+12² I had a quick calculation: 10²+11²+12²=(11-1)²+11²+(11+1)²=3x11²+2=3x121+2=365
10^2+11^2+12^2 = 13^2+14^2 was stated at the very beginning!
Woah! I just got done animating a classic visual proof of one of these identities. I’ll tag you when it eventually posts and link to this one. Thanks!
Great :)
@@Mathologer here it is if you’re interested : kzhead.info/sun/daarpKyaiZWmi6s/bejne.html
Truly spectacular and elegant.
Easier to grasp, still enjoyable. If you split your usual format into a series of semi-standalone videos of this format.. you could become a youtube GIANT
In the mental arithmatic painting you can use the squre rearangement trick to turn 10²+11²+12² into 13²+14² which turns the expresion into 2×(13²+14²)/365 Which is 2×365/365. There is another square rearangment trick you can use here: X²+x+(x+1)=(x+1)² So the top of the fraction can be written as 10²+11²+12²+13²+14² =10²+11²+12²+13²+13²+13+14. =10²+11²+12²+2×(12²+12+13)+13+14 And so on... Eventually we are going to get 5×10²+4×10 +(3+4)×11+(2+3)×12+(1+2)×13+1×14 Which is equal to 540+77+60+39+14 = 730 = 2×365.
Great :)
Fantastic video!!! I'm wondering which software do you use to create those brilliant animations?
Exquisit selection for the intro song. 😎Added it to my Spotify liked songs.
Early Christmas gift, thank you!
Animation of rearranging of squares to show that 1+2=3 just blows my mind
For people who like to see the algebra version of these patterns, if we say that the largest term to the left of the equals sign is n², then the pattern with squares looks like [ ∑ (n-i)² from k=0 to k ] = [ ∑ (n+i)² from i=1 to k ], where k is the number of terms on the right (or 1 less than the number of terms on the left). In fact, solving that equation gives n = 2k(k+1), or n = 4(1+2+⋯+k), so the only way to make that pattern with is with 4(1+2+⋯+k), as revealed at 1:21. The pattern without squares is [ ∑ (m-i) from k=0 to k ] = [ ∑ (m+i) from i=1 to k ] with m being the last term on the left, and this is exactly equivalent to m = k(k+1) = 2(1+2+⋯+k).
The solution to the math problem in the painting is 2. 10²+11²+12² → 100+121+144 = 365 (denominator) Since we know 13²+14² is the same quantity, we're effectively doubling the numerator to get 2/1.
Outstanding animation, doctor! 👍👍👍👍 , yeah, same Timelord math trick indeed 🤓🖖🤓🖖
Can you sometime do a geometric proof like these that shows why: 1^3 = 1 2^3 = 3 + 5 = 8 3^3 = 7 + 9 + 11 = 27 4^3 = 13 + 15 + 17 + 19 = 64 Etc
Yes. In fact, I've got a video for this pretty much ready to go :)
@@Mathologer Fantastic. This has been bugging me ever since I watched your Moessner's Miracle video. I thought for sure it was going to show up there somehow but it never did. Maybe it's in there somewhere still but I couldn't see it. Looking forward to this video.
The sum of connective odds, starting from 1, is a square (there are nice visual pros for that, btw) 1, 1+3, 1+3+5 etc The sequence you ask about is based on the differences of squares 1,9-1,36-9,100-36 ... numbers are pretty :-)
@@Mathologer This is equivalent to ∑(n³) = (∑n²)² right?
@@yinq5384 If you mean the sum of integers up to a point (you said of squares) is the square root of the sum of cubes up to the same point, I think you're right.
Adoro este canal!
For the cube sequence, presumably the 1st line would be: 5^3 + 6^3 = 7^3 which isn't quite correct, but is very close (left hand side = 341, right hand side = 343)!
Exactly, the universe is definitely teasing us here with a near miss :)
Last problem: Modulo 10, x⁴ can only be 0,1,6 or 5. The four monomes yield 1+6+5+6 =8(mod10), hence there is no integer solution.
I nice samller video. I always enjoy watching your videos on Sunday with a cup of coffee. Relaxing.
I needed a break from these long videos :)
@@Mathologer And just in time for Christmas
I think the answer to the problem on the board is 1 and 364/365. I was adding 100, 121, and 144 when I realized that makes 365, so the problem is just 1 + (169 + 196)/365 Edit: Apparently it's 2. I guess I forgot to completely add a number.
I found a generalization that works for all degrees, not just 1 and 2! It's not as clean, but it's still fun! Theorem: For every d ≥ 1, there is a unique degree-d polynomial P (up to scaling) and a unique constant c such that for every n, P(x - n) + ... + P(x) = P(x + 1) + ... + P(x + n), where x = c·(1 + 2 + ... + n). Furthermore, we must have c = 2d, and P(0) = 0. The patterns shown in the video are the special cases d = 1 and d = 2, for which the polynomial P is quite simple: x and x^2. Here are the first few: - For d = 1, P(x) = x - For d = 2, P(x) = x^2 - For d = 3, P(x) = 6x^3 + x^2 - For d = 4, P(x) = 32x^4 + 16x^3 + x^2 - For d = 5, P(x) = 4500x^5 + 4500x^4 + 930x^3 - 19x^2 - For d = 6, P(x) = 46656x^6 + 77760x^5 + 33696x^4 + 1392x^3 - 511x^2 - ... (Since they're unique up to scaling, I've scaled them up so that the coefficients are integers in lowest terms.) And here's the "Christmas tree" formed for d = 3: (6·0^3+0^2) = (6·5^3+5^2) + (6·6^3+6^2) = (6·7^3+7^2) (6·16^3+16^2) + (6·17^3+17^2) + (6·18^3+18^2) = (6·19^3+19^2) + (6·20^3+20^2) (6·33^3+33^2) + (6·34^3+34^2) + (6·35^3+35^2) + (6·36^3+36^2) = (6·37^3+37^2) + (6·38^3+38^2) + (6·39^3+39^2) .... It's not as pretty, but it works!! I would be very interested to know if there's a geometric interpretation to 6n^3 + n^2, or even the higher-degree ones!
That also looks like a nice idea. Will also ponder a bit more once I got a bit of time. Maybe also check out another kind of generalization that I am writing about in the description of this video :)
@@Mathologer Thanks! And yeah I've seen the various possible generalizations in the description and they're quite interesting!
By the way, there are various ways to find the polynomials P, but the one I used (and implemented in a program) make use of the following slight extensions: Lemma 1: For every d ≥ 1, every polynomial Q with degree < d, there is a unique monic degree-d polynomial P such that for every n, P(x - n) + ... + P(x) = P(x + 1) + ... + P(x + n) + Q(x), where x = 2d·(1 + 2 + ... + n). Lemma 2: For every d ≥ 1, every polynomial Q with degree ≤ d and every constant c ≠ 2d, there is a unique polynomial P with degree ≤ d such that for every n, P(x - n) + ... + P(x) = P(x + 1) + ... + P(x + n) + Q(x), where x = c·(1 + 2 + ... + n). These immediately imply the theorem, and these in turn can be proven with straightforward recursion/induction, though you'll have to use the fact that the sum of odd powers is a polynomial in 1 + 2 + ... + n = n(n+1)/2, which has a nice proof using central differences and symmetry; it also shows the fact that the sum of even powers is (n+1/2) times a polynomial in n(n+1)/2. (Use central differences instead of forward or backward differences for the cleanest proof :D) So I use these repeatedly to compute the coefficients one by one, starting with the larger terms first. In Lemma 2, the coefficient of x^d in P(x) is p_d := q_d/(c^(d-1) (c - 2d)) where q_d is the coefficient of x^d in Q(x). Then we simply replace P(x) by P'(x) := P(x) - p_d x^d to get a new equation of the form P'(x - n) + ... + P'(x) = P'(x + 1) + ... + P'(x + d) + Q'(x) with a smaller degree, so we can use Lemma 2 recursively. (That's basically also how the proof works.)
Here's the meat of my (Python) code: ``` def get_P(d, c, Q): if Q.deg > d: raise Exception("No answer") # base case if d < 0: return Q # compute the leading coefficient if c == 2*d: if Q[d] != 0: raise Exception("No answer") P_d = 1 # P is monic else: P_d = Q[d] / (c**(d-1) * (c - 2*d)) # compute the new Q by evaluating at several points and then interpolating nQ = Poly.interpolate(*( (s1(n), Q(s1(n)) + P_d * sum((c*s1(n) + i)**d for i in range(1, n+1)) - P_d * sum((c*s1(n) - i)**d for i in range(0, n+1))) for n in range(d+3))) # recursively solve for the remaining coefficients return P_d*x**d + get_P(d - 1, c, nQ) ```
simply magnificent!
1# puzzle: I guess there is a turnaround point at n=3 and the numbers go up from there? Was that the pattern or did I miss something? 2. Well I didn't have a specific tactic to handle this, I just did 100+121+144 in my mind and got 365. Then I've done 169+196 and got 365 again. So, result is 2. 3. Well, you got me there. Thought there was a pattern because 3³+4³+5³=6³, so I thought the next one would be 7⁴ but no it isn't. Big Mathologer always reminding us of the law of small numbers, great! Keep up the wonderful work, and happy holidays!
Looks like you really enjoyed this one :)
@@Mathologer yes I did. Well not just this one, the other ones were very good too!
Wonderful. Very nice. Sometimes a question arises. There are numbers, or is it a way to perceive information. And as it is correct, the number of combinations, or the frequency of occurrence. Great performance. Thank you.
2. I mentally wrote the numerator as 10^2 + (10+1)^2 + (10+2)^2 + (10+4)^2, and applied the squared sum rule. So I have 5 tens, that's 500 in total. Next I have 1^2 + 2^2 + 3^2 + 4^2, this adds up to 30, and finally 2×10×(1+2+3+4), which is 200. So In total I have 500 + 30 + 200 = 730 = 2×365 It is surprising because 365 just happens to be the number of days in a year.
Full marks :) That's it. Of course, it's even quicker if you apply 10² + 11² + 12² = 13² + 14² Then you just have to calculate one side of the equation.
I did (12-2)^2+(12-1)^2+12^2+(12+1)^2+(12+2)^2=5*12^2+2*(1^2+2^2)=5*146
Oh right. Clever solving. 13² + 14² = 169 + 196 = 170 + 195 = 365 This simplifies with the denominator, and you're left with 2 from the other addend.
I first tried using the sum of squares formula but that would have taken more time than just calculating it. I have memorized the squares up to 16^2 very well and the addition wasn't bad, once I got 10²+11²+12² = 365 I knew what the other one would be
I found the pattern. Each of those equations start with an nth triangular number, where n is an even whole number, and 1 is the first triangular number. Each equation has 1 more term on each aide than the previous one. For example: 21 is the 6th triangular number, so if my conjecture is true, this is also true: 24 27 Σ n² = Σ n² n=21 n=25 And since 36 is the 8th triangular number, we see another one of those equations: 40 44 Σ n² = Σ n² n=36 n=41 If you don’t know what the weird symbol Σ means, you should learn it by the time you take algebra 2.
I'm still impressed with the great gift. The connections presented were known to me (some I invented myself). But there are people who see the soul of a stone, thank you, amazing :). And now the task is simple :). There is a plane for which the parameter is set - time (you can use the definition of Newton, you can Einstein, you can geometrically). And now imagine that the topology of the plane is a Möbius strip. How, mathematically, you can match clocks on two surfaces. Thanks:).
Also noticed that in the case of the squared sequence the last number from the previous term and the first number from the current term, are exactly 3, 5, 7, 9, ... apart from each other, which follows the difference of the squares! 0² = 0 3² + 4² =5 ², ----------------------------------->3 - 0 = 3 10² + 11² + 12² = 13² + 14², ---------> 10 - 5 = 5 21² + 22² + 23² + 24² = 25² + 26² + 27², ---------> 21 - 14 = 7 Pretty cool
Looks like a Christmas present: wunderbar!
7:26 since 10^2 + 11^2 + 12^2 = 13^2 + 14^2 we can reduce the equation to 2(13^2 + 14^2) we can then calculate what 13^2 + 14^2 is : ( 169 + 196 ) which gives you... 365 ! Answer ends up being 2(365)/365, cancel out the 365 and you get 2 as the final answer. ( trick to do 169 + 196, just do 169 + 200 and then subtract 4 )
2nd puzzle is easy. (2*365)/365. And yes, I also noticed those patterns and extended them to 0. Tried to look for other ones with different exponents (after I learned of Fermat last theorem, I looked for non-positive and non-integer exponents), but to no avail. I always liked that first pattern doesn't have a "hole" - so it uses all numbers. I always wondered if the unused numbers in 2nd pattern can be used for something. I guess I will sit down with my pen and paper later this evening to check if there isn't some interesting there. Even if not as beautiful.
While swimming I discovered the difference in consecutive natural number cubes is 6 x triangle number + 1. 2^3 - 1^3 = 8 - 1 = 6*1 + 1 3^3 - 2^3 = 27 - 8 = 6*3 + 1 4^3 - 3^3 = 64 - 27 = 6*6 + 1 Proof is that (x+1)^3 - x^3 = 3x^2 + 3x + 1 Factor 6x out of the first two terms. 6(x)(x+1)/2 + 1. x(x+1)/2 is a triangle number for natural number x. I thought it was cool. Embedded in triangle numbers are square numbers. Embedded in cube numbers are triangle numbers. 6 triangles to cover each face + 1 more to complete the cube.
Yes, nice insight :) There is actually a nice geometrical interpretation of all this. Maybe have a look at the first couple of minutes of my video on Moessner's miracle where I show how to interpret the difference of two cubes of two consecutive numbers as a certain hexagon. That a hexagon can be divided into 6 equilateral triangles then translates into you insight :)
As soon as I learned multiplication and square numbers in school, I figured out that b^2 = a^2 + a + b when a +1 = b, so also knwon as (n+1)^2 = n^2 + n + n + 1. And then I learned that (a|5)^2 = (a * (a+1))|25
Haha that school picture had me laughing, despair, deep thinking and one kid visper the answer
dang ! you're a mighty force of bringing Enlightenment in this Age of Confusion !
I bought a few packs of Rogers sugar cubes to have fun with watching this video. Happy New Year!
For the second puzzle, I did not recognize it as some people did so I did the calculation in my head. The sum is of the form: (x-2)^2+(x-1)^x+x^2+(x+1)^2+(x+2)^2 it can be seen that if you expand the squares, all the terms linear in x cancel out, so you're left with 5x^2+2(1^2)+2(2^2) which simplifies to 5x^2+10, which, for x=12, evaluates to 730 which is twice 365.
Problem 1: If the pattern in 1 and 2 dimensions continued, the first 3D example would have used 6 in the bridging term. (6 is the number of faces on a cube). Thus the first identity would have been 5³ + 6³ = 7³. Unfortunately, the RHS is short by 2. Problem 2: Since 10² + 11² + 12² = 13² + 14² = 365, the calculation on the blackboard yields 2. Problem 3: By direct calculation, 3³ + 4³ + 5³ = 6³ (surprise!) and 3⁴ + 4⁴ + 5⁴ + 6⁴= (6.8934...)⁴ (not so pretty). Thanks for sharing another shiny nugget from your mine of mathematical gems.
Also check out the description of this video for some extra fun :)
(1) If you consider triangular, square, pentagonal, hexagonal, ... numbers, do you get an infinite pattern of patterns! (2) Note the starting numbers of each row. On the tower with the sums of consecutive integers, the starting numbers are squares. On the tower with the sums of squares, they're alternating triangular numbers: 2n(2n+1)/2
Strange but true! Didn't know about the infinite pattern of first powers & squares, but I did stumble onto the single equation with cubes many years ago (3³ + 4³ + 5³ = 6³). Tried to find some general rule to extend it, but was never successful. As for the attempt to extend your 1st- and 2nd-power patterns to cubes, the obvious try almost works, for the first line: 5³ + 6³ = 341 = 7³ - 2 Fred
"5³ + 6³ = 341 = 7³ - 2" There is actually something super nice hiding just around the corner. If you are interested have a look at the description of this video.
Me too about the cubes which we studied in school where it was phrased like something like what will be the side/radius length of a cube/sphere made by melting cubes/spheres of side length/radii 3,4 and 5 units.
Puzzle 1: (x+2)^3=(x+1)^3+(x+0)^3, difference is (x^3-3x^2-9x-7) Puzzle 2: This is simple, we know that 10^2+11^2+12^2=13^2+14^2, so we simply evaluate 10^2+11^2+12^2 then multiply it by 2, which gives us 365*2, but we don't have to evaluate that since now we have 365*2/365=2. Puzzle 3: 3^3+4^3+5^3=6^3, this is sorta well known as well. However, the follow-up doesn't have a real non-negative integer solution.
In David Wells's (The Penguin) Dictionary of Curious and Interesting Numbers, in the entry for triangular numbers, there is an analogous pattern: T1 + T2 + T3 = T4; T5 + T6 + T7 + T8 = T9 + T10; T11 + T12 + T13 + T14 + T15 = T16 + T17 + T18, and so on, attributed to M. N. Khatri. I don't understand everything about how this happens, but the fact that the sum of consecutive triangular numbers is a square (whose base is the larger index of the two triangular numbers), plus the fact that the difference of consecutive triangular numbers is the index of the larger triangular number, seems to be at the heart of it.
Yes, I've also seen a visual proof of this in one of the one glance proofs book by Roger Nelsen. I think it was the third book.
Ah! Playing around with the triangular-numbers patterns a little more, using the facts I mentioned above, I see that they correspond exactly (bijectively!) to doubled versions of each square pattern. They're just the square patterns in the mirror/double-verse: 6^2 + 8^2 = 10^2; 20^2 + 22^2 + 24^2 = 26^2 + 28^2; and so on. And yet they walk through all the triangular numbers in order without missing a beat. Neat.
I LOVE THE SOUNDTRACK 🥰
Details in the description of this video :)
combining both animations mentally, I've presumed that by doubling the "central" number n×(n+1) in imaginary squares pattern for the linear pattern, in addition to "left" and "right" we could furnish "top" and "bottom" as we did in the squared pattern and... it's true! 1² + 2² × 2 = 3² 4² + 5² + 6² × 2 = 7² + 8² 9² + 10² + 11² + 12² × 2 = 13² + 14² + 15² 16² + 17² + 18² + 19² + 20² × 2 = 21² + 22² + 23² + 24² ... What's next? :)
I'm a bit of a math nerd, but the best part of the video is still the music! It's so good!
Isn't it :)
Yes, a nice find. Actually both of them: the one I use in the main part of the video and the one I use for the Thank you section at the end of the video :)
Good job, you successfully teased me into grabbing a calculator for 3^4+4^4+5^4+6^4. Also, I will now remember 7^4 until I die.
Hah! Mission accomplished then :)
The near miss in the cubes, I believe starts with Ramanujan's taxi cab number 1729 = 9^3+10^3 = 12^3+1. The board problem was easy due to the insight you revealed, answer 2 3^3+4^3+5^3 = 6^3, for the 4th powers we infact have a prime times 2 😀
It's a "near hit." It is an almost hit. It is not a "near hit." The sums did, in fact, miss equaling other.
Concerning the linear pattern, put the sequence into a table, the perfect squares on the left. Then, you get a beautiful Ulam-like pattern with the primes in the diagonals and in the columns; I especially like the diagonal 5-11-19-29-41-(55)-71-89…. with the differences 6-8-10-12-(14)-16-18… and the second one, 3-7-13-(21)-31-43… So there is something very deep in this about primes in arithmetic progressions as well. Moreover, there is also a hint in the pattern about additive and multiplicative properties of the integers, because just by lining up the additive properties of the integers, you automatically get them ordered by means of the perfect squares. Amazing (how my math teachers missed this revelation)!
Second puzzle: short answer is no, since I'm not good at mental arithmetic under pressure. But armed with a few factoids and nerves of steel, I could solve the problem in about 30 using the following strategy: first, what's (n-1)^2 plus (n+1)^2 ? We see that the middle terms in the resulting trinomials cancel out, and the end terms add together, giving 2*(n^2 +1^2). Same pattern for n-2. We can write the series in the numerator as (12-2)^2 + (12-1)^2+ ...+ (12+2)^2. Using the above observation we get 5*12^2 plus 10 (=2*2^2 + 2*1^2) in the numerator. In the denominator, we know 365 is also divisible by 5. If you happen to know the other factor is 73(I didn't) = 72+1 then you're home free, otherwise 360=12*30 is pretty common knowledge, so add another 5 to it and you get 5*(6*12) +5 in the denominator. Now, 2*72=144 (gross!) is also pretty common knowledge, so we have twice( 72 + 5) divided by 72+5 =2. I freely admit to having worked this out with an electronic calculator, ahead of time, in about 5 seconds. When you do that, you get 365 for the first three terms in the numerator and another 365 for the two remaining terms, thus proving, by brute force, the original Mathologer proposition.
6:50 Two other nice patterns in the left-hand terms (ignoring the powers): the top one is the square numbers, and the second one is alternate triangular numbers :)
I've looked at 7^x + 8^x = 9^x 22^x + 23^x + 24^x = 25^x + 26^x ... and found that interestingly enough, while it isn't 4 exactly, it appears to be approaching 4 as we continue down the line, with its limit being 4.
7^3 + 8^3 = 9^3 Because the pattern is 2^1, 2^2, 2^3 for each pattern
Actually, I'd say the first equation should be 5^3 + 6^3 = 7^3 :)
@@Mathologer well, you're still two short 😀
I noticed a couple of little patterns not directly mentioned. In the first sequence, the first terms follow the sequence 1**2, 2**2, 3**2, ... In the second sequence, the last term on the left is consecutive multiples of four squared. These observations make it trivial to figure out the Nth row of the patterns in your head.
At first, I looked at a preview to this video and I thought about how does "10² + 11² + 12² = 13² + 14²". And my math intuition was like: _0² + _1² + _2² = _3² + _4² _1 + _4 = _9 + _6 The operation with dividing numbers in groups and forming a border around is a wonderful discovery!
For the pattern of squares first shown at 1:48, in each step n the number of integers skipped to get to the next step n+1 equals 2n + 2. For the equivalent, slightly more complicated pattern of cubes wonderfully sorted out by Exception2001 in a must-read comment, it is 4n + 4. But it isn't n + 1 for the linear pattern. Instead, no integers are skipped. What the hell, cosmos?
I am looking at the super-pattern for the squares... Doesn't look so nice, but 5*(2/1)=10, 14*(3/2)=21, 27*(4/3)=36... Next would be 44*(5/4), so 55²+56²+57²+58²+59²+60²=61²+62²+63²+64²+65². Next one would start at 65*(6/5), or 78.
Good idea to look for the super pattern there too :)
Why can't you start the pattern a^3+b^3=c^3+d^3 or a^3+b^3+c^3=d^3+e^3 or similar?
I am not going to stop you. However, the natural choices for these numbers suggested by everything we did in this video do not work :)
The missing numbers in the tree of sums of squares have a nice pattern to: 1*1 , 2*1 ; 2*3 ... 3*3 ; 5*3 ... 5*4 ; 7*4 ... 7*5 ; 9*5 ... 9*6 etc.
1:44 The last number of one series and the first number of the next are in type: ab, a(b+1)
I remember finding a photo on google once about this. I showed my teacher it. The proof was very simple. It’s really interesting how there are an infinite number of ways you can construct these types of numbers. I am looking forward to watching this on your Channel. I don’t usually comment before watching a video. So I might have some questions.
The beautiful visual proof for the squares pattern is based on a note by Michael Boardman in Mathematics Magazine. Here is link to that note: tinyurl.com/2d4y7wtf
In fact, there are many unnoticed number theory patterns. For example, when I was memorizing squares, I realized that 12^2 = 144 and 21^2 = 441. The mirrored numbers had mirrored squares! In fact, the same thing happens with 13 and 31! But nothing after that. However, this is just a consequence of addition carrying. In higher bases, as long as there is no carrying, this pattern will continue!
... nice animation. equally nice music. thanks for 'gem'
7:07 for the second puzzle, I noticed in my head that (n-2)^2 + (n-1)^2 + n + (n+1)^2 + (n+2)^2 simplifies to 5.n^2 +10. 12^2=144, so the sum must be equal to 730
Formula: sum(2n^2+k)^2 LHS k=n=>2n. RHS: k=2n+1=>3n
Great Presentation; Great Music!!
Glad you liked it!
It's VERY beautiful