A Unique Proof Without Induction

We prove Nicomachus' theorem, that (1^3 + 2^3 + ... + n^3) = (1 + 2 + ... + n)^2. The proof relies on manipulation of summations, rather than induction or geometric reasoning.

00:00 Intro
00:45 Setup
01:36 Double summation
04:43 Evaluating the double summation

[nathanisbored]

i like how some geometry still snuck in there with the multiplication table

[rob]

This is somewhat more satisfying than the induction proof.

[eauna]

What a neat proof! Inserting the sum of the first n integers was such an unexpected but great trick over there too; I completely though you'd abuse that "telescopation" in a different way, or you'd exploit reversing the summation order (to go column by column instead of row by row), but this was a very nice one!

[joseivan]

By the Fubini's theorem when both limits in the first final summation are independent the double sum is the product of the simple sums, which gives sum of i from 1 to n, the whole summation squared

[sr]

Your videos inspire me. I guess my level of maths is a lot lower than your viewers. In the past few months I have started running puzzle evenings in a local bar. A vastly simplified version of this could make it into the next evening! Thanks.

[elrichardo]

nice one! I’ve also seen ways to prove this using finite differences

[anastasissfyrides]

By using the closed form for Σk arent you silently using induction in the proof?

[mauisstepsis]

That serious look at the end makes me giggle.

[jackkalver]

But can you prove the inner sum without induction?

[cmilkau]

(n + 1)² - (n - 1)² = 4n. Now multiply by n²/4

[merwan.houiralami]

i wanna die after seeing how you do your sum sign 😂😂

[dalibormaksimovic]

There is a proof that goes like this: If a1 < a2 < a3 < ... < an, where ai is the natural number, and (a1 + a2 +...+ an)^2 = a1^3 + ... + an^3, then ai = i.

[cassianocaro]

this question apered in a similar way in a braziliam olympiad, all tho the exponents were indetermined, and u had determin them

[MyOneFiftiethOfADollar]

Michael Penn sometimes dons a shirt with this particular integer identity.
A suitable definition for a nerd among the nerds.

Nice to see non induction way of showing this.

[georgecswong]

My proof is that (1+2+3+4………….+n)=n(n+1)/2





(Sum of 1 to n)^2-(sum of 1 to n-1)^2=n^3
(Sum of 1 to n-1)^2- (sum of 1 to n-2)^2=(n-1)^3
(Sum of 1 to n)^2=n^3

[MichaelRothwell]

This is a delightful proof! I think it could be nice to run it backwards. Start with the the (sum of the integers from 1 to n)² and use your diagram to split this into two equal triangular double sums and the sum of the squares of the integers from 1 to n, then convert the double triangular sum to the sum of k³-k² for k from 2 to n, by again running your argument backwards, and we are done.

Here are the details. 

We want to prove that 
∑(k=1 to n)k³=[∑(k=1 to n)k]²

For this proof we'll manipulate the RHS until it becomes the LHS. 

[∑(k=1 to n)k]²
=[∑(k=1 to n)k][∑(i=1 to n)i]
=∑(k=1 to n)∑(i=1 to n)ki
=∑(k=1 to n)[∑(1≤i<k)ki + k² + ∑(k<i≤n)ki]
=∑(k=1 to n)∑(1≤i<k)ki + ∑(k=1 to n)k² 
+ ∑(k=1 to n)∑(k<i≤n)ki

Note that the first term ∑(k=1 to n)∑(1≤i<k)ki
=∑(1≤i<k≤n)ki
and the third term ∑(k=1 to n)∑(k<i≤n)ki
=∑(1≤k<i≤n)ki=∑(1≤i<k≤n)ik=∑(1≤i<k≤n)ki,  
so these terms are equal. 

As we saw in the video, they can be thought of as the sums of the entries in the triangular regions above and below the diagonal k=i in the n×n square matrix with entries i×k (i-axis horizontal, k-axis vertical)

[∑(k=1 to n)k]²
=2∑(k=1 to n)∑(1≤i<k)ki + ∑(k=1 to n)k²
=2∑(k=1 to n)k∑(1≤i<k)i + ∑(k=1 to n)k²
=2∑(k=2 to n)k∑(i=1 to k-1)i + ∑(k=1 to n)k²

(where the lower limit for k in the first sum changed from 1 to 2 as for k=1 we have an empty sum)

=2∑(k=2 to n)k[½(k-1)k] + ∑(k=1 to n)k²
=∑(k=2 to n)k²(k-1) + ∑(k=1 to n)k²
=∑(k=1 to n)(k³-k²) + ∑(k=1 to n)k²

(where we can change the lower limit for k in the first sum from 2 to 1 as it is zero for k=1)

=∑(k=1 to n)k³ QED 

Of course, this proof is made from exactly the same ingredients as those in video, but in the opposite order.

[ezvac]

4:00 you say the sum is not defined but it is and is equal to 0

[KSignalEingang]

I first stumbled on this equivalence one night while trying to go to sleep... Not only did it cost me that night's rest, I wasted most of the next couple days trying to figure out why it was true. It seems like the sort of thing you should be able to do with basic grade school algebra, but I quickly got frustrated and wound up looking up some proofs online. Most of these - at least, the ones I understood - were of the visual/geometric variety. Which was fine, but a bit unsatisfying... Like they did not so much show the "why" of it as just restate it in building blocks instead of numbers.

So, this proof was exactly the sort of thing I was looking for! Tbh I still find the equivalence itself fairly incredible (in several senses of the word), but it's nice seeing it proved with calculations I can actually follow.

[JohnBerry-qh]

Imagine a world in which, instead of firing missiles at each other, countries simply sent Olympiad Math Challenges at each other. Or angry Dance Off challenges. Take your pick.

[CutleryChips]

Answers to Questions that no one asked