Proving Fermat' s Last Theorem (almost) in just 2 minutes !

Andrew Wiles spent almost a decade proving a theorem nobody else could do before him. If you add a small extra condition to the statement of Fermat's Last Theorem , it becomes almost trivial ! See my video to learn more.

#fermat
#manim
#Wiles

[MetaMaths]

Subscription increases your chances of solving one of Millennium problems by 20 % !

[justinernest]

"x^n+y^n=z^n is now left as an exercise to to reader"🤣🤣🤣

[jonathanl]

Could this solution be what Fermat himself DIDN'T scribble in the margin?

[akselai]

WOAH. Best (partial) proof of the theorem that I have ever seen.

[iwersonsch]

I think this proves something stronger than just "x^n+y^n is not exactly z^n if n>=z"
Specifically, I think it proves "x^n+y^n is *smaller than* z^n if *n>=x* ",
because we can substitute n for x and we still get x^n, and the strict inequality between the second and third expression is still guaranteed by y<z.
Examples: 2^2+2^2<3^2
3^3+3^3<4^3
4^4+4^4<5^4
etc.

[fahrenheit]

This is a pretty specific case isn't it. N < z is inconceivably more difficult, so its a little dishonest to claim that this is almost a proof of fermat's last theorem, when it doesn't even scrape the surface of the actual proof of the full theorem. Still a good argument.

[gedstrom]

On the subject of 'Near Misses', are there any examples of x^n + y^n missing z^n by only plus or minus one?

[sytalsbtw]

I don't even know what I'm looking at. I thought by looking at some super hard math problems would automatically make me smarter so I can finish my geometry homework. Apparently that's not the case.

[williamchurcher]

Little bit careless by saying all the inequalities are strict, but lovely video regardless!

[miguelrezende]

its incredible that Fermat could solve it with the math of his time, but we had to use the pinacle of the modern math to get it. It should be a very tricky proof we weren't prepared for

[davidbrisbane]

Great. So now prove the theorem when z < n.

[mohameda.]

I love the attempt as I am a big fan of this theorem, the conditions are specific but there is one more condition missing, with y > x there is a chance that z-y=1 which disqualifies the inequality mentioned @ 1:16. So you have to state from the start that z-y > 1… this makes your case more and more specific. Yet, I still salute the partial proof, good job!

[mkwarlock]

1:25 — shouldn't this be greater or equal instead of strictly greater?

n >= z, therefore n * y^(n-1) >= z * y^(n-1)

[legendarybelt]

"And this is why Math questions cannot be concluded."

Me explaining to my underpaid Math Teacher after asking me what I know about fractions.

[regularsense]

Let us assume that x=1, y=2, z=3, n=3. (z-y)*n*y^(n-1) = (3-2)*3*2^2 and n*y^(n-1) = 3*2^2 so in that case (z-y)*n*y^(n-1) is equal to n*y^(n-1) and it means that the inequality is not strict.

[tahamuhammad]

You added a condition, then you are out of Fermat's Last Theorem. Your solution and Sir Andrew Wiles solution are not acceptable. My Solution is acceptable and you can look at it (5 Fermat's videos) and 5 for General Fermat's Last Theorem. My solution is the normal and best. Taha M. Muhammad/ YouTube. I did prove 11/2/2022.

[pablocopello]

n>=z is no "small" condition, can be "loosened" to n>=min(x, y, z) (still no "small" condition), and should be more specific with "trivial": min(x, y, z)>1 and n>2.

[BRORIGIN]

Do you have a patreon? When I see content of this quality and at this frequency I want to contribute :)

[primephoenix.]

Mathematicians don't study maths they study the language of universe...😋

[mychannel-teke]

I think we missed the case where x = y but it's trivial of course.