The One Sentence Proof (in multiple sentences) - Numberphile

Featuring Professor Matthias Kreck from the University of Bonn.

NUMBERPHILE

Videos by Brady Haran

Hey, why did you make this two videos and not just one?
Because two videos would have been very long and deterred casual viewers from even learning about the problem.
Makes more sense to make a video which describes the problem and the solution, plus its history and the fact a one sentence proof exists... then have a more technical second video for people who want to see the proof itself. I have made links to the second video as clear as possible and I suspect more technical fans subscribe to Numberphile2 anyway.

[cpsof]

1) Let us proof Q.
2) Let us assume that P is true.
3) Let us show that P => Q.
4) Profit.

[ricsixd]

I really like the way Professor Kreck explains things, makes everything easy to follow and still interesting.

[Jose-pqow]

The volume of this video is higher than the other, i want my money back

[deamon]

All the soljutions... Greetings from germany :)

[PersonaRandomNumbers]

I was a bit curious about the assumption that there is an odd number of solutions, so I looked at the image of the proof at 8:06 to find this:

"The verifications of the implicitly made assertions -- that S is finite and that the map is well-defined and involutory (i.e., equal to its own inverse) and has exactly one fixed point -- are immediate and have been left to the reader. "

... Oh boy.

[davidturner]

Proving that |S| is odd IS the whole point, the rest is obvious. Unfortunately the ingenious one-sentence proof is reduced in the video to the mysterious assertion that “there are an odd number of solutions” :-(

[felsenhower]

When he said "Huch" at 2:18, I really had to chuckle.

[Tfin]

LEVELS!

I had to turn up the sound on part one, then came here straight after and woke the house.

[SmileyMPV]

Wonderful!
It took half an hour to understand one sentence.
That first involution is a masterpiece, I mean how does one even begin looking for such a function?

[joakimekern]

Let's all appreciate the perfectly drawn circle at 5:35

[Demki]

First time I see someone spell Q.E.D as "Hurray"

[OnlyPenguian]

The paper itself, shown at 8:00, is clearer than the video, and leaves out nothing important. The key point is that the equation

x^2+4*x*z = p

has exactly one solution when p is a prime of the form 4*k+1.

Why? The left hand side is divisible by x. Since the left hand side equals the prime p, x must be 1 or p. Since z is a positive integer, x cannot equal p. So x=1 and therefore z=k.

The involution shown in the box can only have a fixed point when x=y. Why? The upper and lower cases would have a fixed pont only for z=0 or y=0 (respectively), but these values are not allowed. The middle case must have x=2 * y - x, so x=y.

[koenth]

Consider the involution as defined in the paper (8:02).
We are assuming that |N is defined not to contain 0.

We see that if (x, y, z) is in case 1, so x<y-z, the involution brings it into (x1, y1, z1) = (x+2z, z, y-x-z). Solving this system for x, y, z we get x=x1-2y1, y=x1-y1+z1, z=y1 and from the fact that x>0 we see that x1>2y1. Doing the involution once more thus brings us back to (x, y, z) via case 3.

Similarly, starting with case 3 we go back via case 1.

Here we find no fixed points, because setting x=x1, y=y1, z=z1 results in (x, y, z)=(0, 0, 0), which is outside the domain.

Case 2 (not sure why Zagier calls this 'the last'), when y-z<x<2y, is its own inverse, and this is the route that contain our fixed point.
Fixed points are found when x=2y-x, y=y, and z=x-y+z, which condenses into 0<z<x=y.
That seems to define more than one fixed point... but wait, note that
p = 1+4k = x^2+4yz = x(x+4z) using x=y
And now because p is prime, we see that x must be 1, resulting in the solution (x, y, z)=(1, 1, k).

Check: for (1, 1, k ) y-z<x<2y, is satisfied whenever z>0, and the result is (2-1, 1, 1-1+k)=(1, 1, k)

[heruilin]

Serge Lang used call proofs like these "A let" as in "Let f = blah, expand and voila QED". . are amazingly concise, but give absolutely no motivation for how "the let" was formulated.

[jensraab]

My feeling is that the true purpose of this video is to serve as a clandestine drinking game: Take a shot every time the professor says "solyootiun"!

[garrick]

It amazes me with a proof that seems so simple that no-one came up with it before. It is not as though it was an obscure problem, given that several famous mathematicians came up with proofs, which happened to be more complicated.

[nivolord]

Ok, here we go:
If x is even and y is even, x = 2a, y = 2a, then x^2 + y^2 = 4a^2+4b^2 = 4(a^2+b^2) = 4k for some k.
If x is even and y is odd, x = 2a, y = 2a+1, then x^2 + y^2 = 4a^2+4b^2+4b+1 = 4(a^2+b^2+b) + 1 = 4k+1 for some k
If x is odd and y is even, look above and switch x and y
If x is odd and y is odd, x = 2a+1, y = 2a+1, then x^2+y^2 = 4a^2+4a+1 +4b^2+4b+1 = 4(a^2+a+2b^2+b)+2 = 4k+2 for some k
So the sum of two squares can only be of form 4k, 4k+1 or 4k+2
If a prime p is not 2 (hence not even, and not of form 4k or 4k+2) and not of form 4k+1, then p is of form 4k+3, so it cannot be written as the sum of two squares.

[Harsimran_Singh_]

I just love the accent of the professor. It's so sweet.

[adityarachmat]

Great solution! And now we already knew that the square of prime number more than 3 minus 1 are divisible by 24.
such as : 5^2 - 1 = 24. 7^2 -1 = 24(2), etc.

[Rififi]

Did I miss the _only if_ part of the proof or is that actually omitted?