Proving Fermat' s Last Theorem (almost) in just 2 minutes !
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- Опубліковано 17 жов 2024
- 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.
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@@mrgreenskypiano Haha nice.
Following this demo my probability of solving LFT decreases to 0.
For any exponent n (better only p = prime) it is easy to find limitations to the values of x 30
p = 37 x > 1919190
or xyz > 1.19*10^13
But this doesn't solve FLT.
I have a better idea, instead modify them just like how you did in this video, so you can increase the percentage to 100%!
WOAH. Best (partial) proof of the theorem that I have ever seen.
Could this solution be what Fermat himself DIDN'T scribble in the margin?
Very likely !
I proved on 09/14/2016 the ONLY POSSIBLE proof of the Great Fermat's Theorem (Fermata!).
I can pronounce the formula for the proof of Fermath's great theorem:
1 - Fermath's great theorem NEVER! and nobody! NOT! HAS BEEN PROVEN !!!
2 - proven! THE ONLY POSSIBLE proof of Fermat's theorem
3 - Fermath's great theorem is proved universally-proven for all numbers
4 - Fermath's great theorem is proven in the requirements of himself! Fermata 1637 y.
5 - Fermath's great theorem proved in 2 pages of a notebook
6 - Fermath's great theorem is proved in the apparatus of Diophantus arithmetic
7 - the proof of the great Fermath theorem, as well as the formulation, is easy for a student of the 5th grade of the school to understand !!!
8 - Me! opened the GREAT! A GREAT Mystery! Fermath's theorem! (not "simple" - "mechanical" proof)
@@Ал-тайДіЙ what?
@@alephcomputer - Me opened : - EXIST THE ONLY ONE!!!-POSSIBLE proof of Fermat's Great Theorem
Me! opened : - the GREAT! Mystery! of the Fermat's Last theorem! (- !!! not "simple" - "mechanical" proof)
Me! opened : - Pierre de Fermat - was proved! the Fermat's Last theorem!
Me! opened : - my formula of my Proof is completely and absolutely identical with the words of Pierre de Fermat !
Me! opened : - the proof of the theorem - The REAL! Proof! - worth a BILLION !!! !!! !!! , - but do not! a one little-smalest-million
@@Ал-тайДіЙ can you provide me it? i would be interested to read.
"x^n+y^n=z^n is now left as an exercise to to reader"🤣🤣🤣
I have found the solution to that statement, however there isn't enough space in this comment for the proof !
@@Osirion16LOL
@@Osirion16 _Hanc marginis exiguitas non caperet_ :D
for n=2 it is pythagoras and infinite solutions exist with integers.
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.
yea I looked at that and thought it was painfully obvious
Interestingly, the case of n=z are both countable infinite and of the same cardinality, so it's actually incorrect to say that one condition contains "more" possibilities than the other.
@@justinbrentwood1299 Depends what you mean. Both cases contain exactly zero possibilties, such is the nature of the theorem. But in considering every possible value of z and n, yes, there are countably many choices with either case. I mostly meant that n > z is wildly restrictive, and thus unfathomably easier to prove. It's a very specific case. So much so, that I'm not too sure it even qualifies as a relevant contribution, as opposed to a brief trivial observation, but that might be too harsh.
@@fahrenheit2101 Well, I would rate its significance partially by how long it took to figure out. It's progress, and although it's very easy to follow, maybe it took mathematicians a long time to figure it out. Idk the history though, so ....
@@justinbrentwood1299 Idk, mathematicians have an alarming tendency to consider things normal folk might spend ages to figure out as trivial. And given that the actual proof of FLT would probably take a good 2 years of dedicated study to understand from scratch, at least, I think 2 minutes probably isn't enough to make a real dent in the problem
But I admit, I'm not too familiar with FLT myself.
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
Isn't that just (A + B) ^ n = A^n + B ^ n + A ^ (N - 1) B ^ 1 + A ^ (n - 2) B ^ 2 + ........ A ^ 1 B ^ (n- 1)
Expanding Polynomials
And for his next trick, Pancake Man will prove the abc conjecture
@@mathphysicsnerd i get it
@@daijones5558LoL
On the subject of 'Near Misses', are there any examples of x^n + y^n missing z^n by only plus or minus one?
Google might help
Yes, cubic quadruple of which there are infinitely any of both types. 9^3=8^3+6^3+1^3 & 12^3=10^3+9^3-1^3.
Je n'ai jamais écrit ça !???? !
Je n'ai jamais écrit ça ! ! !
Trivially setting x = 1 gives
1^n + y^n = z^n + 1
Works for x = -1 for odd n as well.
This makes zero sense to me
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!
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.
1:25 - shouldn't this be greater or equal instead of strictly greater?
n >= z, therefore n * y^(n-1) >= z * y^(n-1)
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.
It's not just seeing them, if you don't know the reason for each thing, then obviously you're not going to understand anything
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
He most likely didn't solve it
Or it could be Fermats own theory was flawed, but the theory was proposed still. It could be that it wasn't until 200 years later we had the mathematical research to create an actual proof that supported what Fermat proposed and (inaccurately) proved.
Just saying.
It's generally believed by all mathematicians that Fermat didn't have a proof for it. Or that he did, but it was flawed and innacurate. He probably believed his "proof" was accurate, but yeah he didn't actually quite have it.
The centuries of trying to solve Fermat's last theorum have created whole entire new fields of mathematics. The search for a solution has advanced all of mathematics considerably. In Fermat's time, all that mathematics hadn't been discovered yet. It's almost impossible that he could have actually had a proof.
We don't know that fermat could solve it
Yeah he, uh, didn't... actually... prove it......
"And this is why Math questions cannot be concluded."
Me explaining to my underpaid Math Teacher after asking me what I know about fractions.
Je n'ai jamais écrit ça ! !
Do you have a patreon? When I see content of this quality and at this frequency I want to contribute :)
Couldn't the supposed solution potentially have y > z if x < 0 with an odd n?
Yes, you are completely correct. So at most this proof proves it for n is even case.
x, y, z and n are all natural numbers, meaning x cannot be a negative number
Same concept will apply if you properly switch the signs
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.
I think we missed the case where x = y but it's trivial of course.
I spotted that too but I thought maybe 2x^n = z^n having or having no solutions is trivial to find or prove.
I spotted that too and I actually like that he leaves some cases away. Makes my brain more engaged :-)
@@ujwh2498 that isn't how it goes everytime, sometimes there are equations which we solve to get the result that two (or more) variables have the same value
@@p07aits similar to showing that 2^(1/n) is always irrational
I see the points of everything except for one part, what is the reasoning that (z-y)ny^n-1 is less than the expression before
Because z-y is greater than 1, just looking at the original equation
@MetaMaths, very good video, congratulations. Man, I found your proof very close to the idea of the German mathematician Johann August Grunert or I may have misunderstood. How can you explain the difference?.
Never heard of it, but I will have a look. Thank you !
@@MetaMaths and there, he managed to find Grunert's work. Is your test similar or different from his?.
what are trivial solutions? How did you factored z^n - y^n ?
Trivial solution is when x,y,z are equal to 0. The factoring is possible because x^2 - y^2 = (x-y)(x+y) and it can be generalised for higher powers
@@MetaMaths how do you generalise it for higher powers? Like i Know that you used (x-y)(x+y)
@@danishsamir8807 proofwiki.org/wiki/Difference_of_Two_Powers
@@MetaMaths thank you!
May be this is what Fermat's enigmatic message suggested.
What is the background sound
the background sound is music
Little bit careless by saying all the inequalities are strict, but lovely video regardless!
I was actually glad that he left out some of the gritty details. Made it more concise to me and I like to pause the video sometimes and contemplate, what details are left out and how I would approach them. :-)
@@harriehausenman8623 The proof is ultimately correct, and no small details have been left out. All I was pointing out is that MetaMaths said some of the inequalities were strict, when they are not. Namely at time 1:28, if n>=z, then ny^n-1 >= zy^n-1, ie a weak inequality. In the video a strict inequality is used. However, due to strict inequalities in the earlier steps of this proof, the overall strict inequality still holds and the proof is correct.
The same method works with a slightly weaker condition, n > x.
Can you prove it ?
@@MetaMaths Trivial: x < y < z so n > x is a weaker condition than n > z.
We can assume that x is less than y.
Why?
This but a special case
that's how you solve Fermat on your own: special cases and try to narrow it down.
Please explain me why you can take n greater then z ?
It is not that we can, but simply what happens IF we assume n > z
That is good, but it is only the surface or not even that close.
what was the music you used in this video?
Just a random beat from music library
@@MetaMaths Good _random_ *choice* :-)
To Long, Didn't Watch! (inside Fermat joke :) )
Sir , when you replace z with y , the value of the smaller bracket becomes zero , thereby rendering your argument meaningless. Please explain
the replacement only happened on the right bracket, no changes were made to the left bracket.
example:
(8 - 5)(8 ^ 2 + 8 * 5 + 5 ^ 2) > (8 - 5)(5 ^ 2 + 5 * 5 + 5 ^ 2)
Imagine if the proof was torn apart from the margin page
Great. So now prove the theorem when z < n.
What about for n
That was solved by Andrew Wiles !
I proved on 09/14/2016 the ONLY POSSIBLE proof of the Great Fermat's Theorem (Fermata!).
I can pronounce the formula for the proof of Fermath's great theorem:
1 - Fermath's great theorem NEVER! and nobody! NOT! HAS BEEN PROVEN !!!
2 - proven! THE ONLY POSSIBLE proof of Fermat's theorem
3 - Fermath's great theorem is proved universally-proven for all numbers
4 - Fermath's great theorem is proven in the requirements of himself! Fermata 1637 y.
5 - Fermath's great theorem proved in 2 pages of a notebook
6 - Fermath's great theorem is proved in the apparatus of Diophantus arithmetic
7 - the proof of the great Fermath theorem, as well as the formulation, is easy for a student of the 5th grade of the school to understand !!!
8 - Me! opened the GREAT! A GREAT Mystery! Fermath's theorem! -!!!!not a "simple" - "mechanical" proof!
@@MetaMaths I developed an idea that I believe can be used to prove the entire FLT and to prove that idea I would have to check between 10 million and 10 billion equations. My idea is as follows: It is possible to show and prove that in the subtraction z^n - (x^n + y^n) there is a finite and repeatable limit of numbers in their last digits. For example: the last digits of the positive subtraction value (100p + 5) ^n - (100q + 2) ^n - (100r + 1) ^n, with n = 100s + k, with p, q, r, s and k belonging to the natural numbers, and k equal to 10, 20 and 30, are always and respectively 600, 48 and 800. The last digits of the positive value of the subtraction (100p + 5) ^ n - (100q + 2) ^ n - (100r + 1) ^ n, with n = 20s + k, with p, q, r, s and k belonging to the natural numbers, and k equal to 3, 4 and 5, are always and respectively 48, 08 and 92. I believe there must be one standard in this type of analysis and that this same idea can be used for all other equations. Could someone help me prove this idea?.
@@Ал-тайДіЙ You sound crazy af.
@@Ал-тайДіЙ what on earth are you talking about? You sound unhinged, like you're having an episode of psychosis, copying and pasting this exact comment all over the comments section of this video. I realise English is probably not your first language. But what you're saying makes no sense. It's not clear in the slightest what it is you're even trying to say. Talk to a friend who knows English more fluently than you do, and ask them to translate what you're trying to say in a more legible way
If you make the exponent exceed 2 you start to get into "uneven and prime numbers too (and prime numbers are only divisible by one and themselfs and are similar to moebiusstrips which are the non-logic object of a 50% true and 50% false statement)", which inside of the exponent create such a high divergence, that congruency breaks, the homeomorphism of the term isn't given anymore and thus equality of x^2... -> x^n.. doesn't exist.
To visualize this you can imagine Platons analogy of the cave in reverse.
You can ezly figure out that the (90 degree) shadow of a cube (3d) is a rectangular shape (2d), but going the other way is like confusing sub sets with sets they are contained inside of, if you don't know how far away the cube is, you can never know it's real dimensions just from the shadow.
Really fine video!
ONLY! I proved on 09/14/2016 the ONLY POSSIBLE proof of the Great Fermat's Theorem (Fermata!).
I can pronounce the formula for the proof of Fermath's great theorem:
1 - Fermath's great theorem NEVER! and nobody! NOT! HAS BEEN PROVEN !!!
2 - proven! THE ONLY POSSIBLE proof of Fermat's theorem
3 - Fermath's great theorem is proved universally-proven for all numbers
4 - Fermath's great theorem is proven in the requirements of himself! Fermata 1637 y.
5 - Fermath's great theorem proved in 2 pages of a notebook
6 - Fermath's great theorem is proved in the apparatus of Diophantus arithmetic
7 - the proof of the great Fermath theorem, as well as the formulation, is easy for a student of the 5th grade of the school to understand !!!
8 - Me! opened the GREAT! A GREAT Mystery! Fermath's theorem! (not "simple" - "mechanical" proof)
New to math, can some explain how he concluded zx^n-1 > x^n, and how he turned the bracket into ny^n-1?
Your view is correct ; In fact what you have mentioned is an false inequality .
Because z>x
Substitute z as x and get inequality
Replace zx^(n-1) with (z/x) x^n, since z>x, (z/x) would be greater than 1 thus greater than 1x^n
This is an instance of a youtube video which says something in the title and shows another thing in the video. What was presented, is far far far far away from the actual proof of the theorem. So far that the title loses its meaning and becomes deception. What you did was changing the problem, rather than answering it.
Mathematicians don't study maths they study the language of universe...😋
If math is the language if the universe then you could argue that English or French are also the language of the universe. Math is just a language, invented by humans like all languages.
@@StarsManny maths already existed, we just discovered it same goes with physics, they are all embedded in the fundamentals of our universe and we Humans are just unscarping the fields and trying to find them, Unlikely English and French are INVENTED by us
@@extreme4180 agree with your point regarding physics, but math? It is an entirely man-made construct. I assure you that if aliens were to “discover” math it would be entirely different from ours.
@@OMRIGREEN-xw3he well the way they'd solve will be so similar to us, like even I agree that math was made by us but the axioms are self evident and need no proof (proved by godel too). Also the alien's math can be similar to us except for the language and symbols
This prove is good but cannot prove the main point of the theorem, didn't have integer solution...
Excellent presentation. wow !!
Amazing video!!!
It is not a proof but a new diagnostic of numbers .
No solution, I immediately knew it, but proving could be funny, but why proving, when knowing and nobody understands it or could use it.
How did Fermat managed to be this careless with his stuff? 🤐
He could have saved people from all of the sufferings that went till now 😢
If n=2 then this proof is correct?
If n is 2 , z can only be 1.
Because of the perfect squares trinomial.
Proof of Fermat's Last Theorem for Village Idiots
(works for the case of n=2 as well)
To show: c^n a^n + b^n for all natural numbers, a,b,c,n, n >1
c = a + b
c^n = (a + b)^n = [a^n + b^n] + f(a,b,n) Binomial Expansion
c^n = [a^n + b^n] iff f(a,b,n) = 0
f(a,b,n) 0
c^n [a^n + b^n] QED
n=2
"rectangular coordinates"
c^2 = a^2 + b^2 + 2ab
Note that 2ab = 4[(1/2)ab] represents the areas of four right triangles)
"radial coordinates"
Lete p:= pi, n= 2
multiply by pi
pc^2 = pa^2 + pb^2 + p2ab
Note that pc^2, pa^2, and pb^2 represent areas of circles, wile p2ab = a(2pb) is the product of a radius (a) and a circumference (2pb).
This proof also works for multi-nomial functions.
Note: every number is prime relative to its own base: a = a(a/a) = a(1_a)
a + a = 2a (Godbach's Conjecture (now Theorem.... :)
(Wiles' proof) used modular functions defined on the upper half of the complex plane. Trying to equate the two models is trying to square the circle.
c = a + ib
c* - a - ib
cc* = a^2 + b^2 #^2
But #^2 = [cc*] +[2ab] = [a^2 + b^2] + [2ab] so complex numbers are irrelevant.
Note: there are no positive numbers: - c = a-b, b>a iff b-c = a, a + 0 = a, a-a=0, a+a =2a
Every number is prime relative to its own base: n = n(n/n), n + n = 2n (Goldbach)
1^2 1 (Russell's Paradox)
In particular the group operation of multiplication requires the existence of both elements as a precondition, meaning there is no such multiplication as a group operation)
(Clifford Algebras are much ado about nothing)
Remember, you read it here first)
There is much more to this story, but I don't have the spacetime to write it here.
Bravo!
a high school student could not come up with this
2£ -$32 = 55% is the correct answer. 👍
Name the background music pleased. It's very soothing like the proof!! ❤
it's a standard library sound, it has no name, sorry
Why we assume n< z
IF we assume, then the proof is trivial. It is just a fun experiment.
To create a clickbait content.
How can you assume that z-y > 1?
Because z > y and they are both natural numbers !
@@MetaMaths z can be equal y+1, so the difference can be equal to 1.
@@MerDo138 If that were the case, we would have (z - y)ny^(n-1) = ny^(n-1) > nx^(n-1), which is still greater than x^n since n >= z.
@@OMRIGREEN-xw3he I know friend :) just pointed out that he forgot to put the equal sign and that's the reason for the confustion
@@MerDo138 oh, my bad. Misunderstood your comment
x can equal y ...
But the theorem doesnt have this condition n≥z
26/7/2023. Vous avez fait une erreur en ""tour de Pise"". Contrairement en PYTHAGORE, en FERMAT on peut avoir X = Y avec solutions .Et votre théorie tombe à l'eau !
L'équation générale UNIVERSELLE de FERMAT, son équation cachée fut sûrement :
Zpuissance(N)) = Xpuissance(N-1) + Ypuissance(N-1) avec +1
Did this take 350 years to solve?
the general proof ? yep
Wiles took Taniyama-Shimura conjecture for the proof, because T-S solves Fermat as byproduct.
the peer review of Wiles's work took a long time because most mathematicians were unable to understand the proof; it is hardcore math still :D
Add another condition huh. Nice one. 😂
Why is z lower limit for n
Because we make this our assumption
@@MetaMaths so that does not prove the fermer theorem in entirety
@@abdurrahmanwahid7794 it does not, he stated that at the beginning of the video.
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/ UA-cam. I did prove 11/2/2022.
If I add the condition that x, y and z are all odd, then I can show there is no solution in 20 seconds 😂.
Do you imply that "n > z" is just as big of a simplification ?)
@@MetaMaths
No, I grant that proof of the theorem for n ≥ z is a worthwhile result, but there is a reason why it required two papers with 129 pages and many years to complete the proof.
If the condition n < z is not included, then there are infinitely many cases not covered by your proof.
@@davidbrisbane7206 True, but I think the video does not make an attempt to say that "it is almost FLT". I just wanted to show how a seemingly harmless twist simplifies the problem greatly.
@@MetaMaths
Not what the title implies.
great! but why not X less n
?
Homer proved this in Simpsons 😂😂😂
But proof by contradiction is like cutting of your own legs in order to go on crutches 😊
sure thought you didn't wish to hear from me. let's acknowlge Fermst never claimed to have to a proof for it. I've yet to see his original marginal notation but what I recall it wasn't in English. Might not even been in French or Latin or even Coney Greek but a recall it reading that he had a "demonstration" that there were no solutions high to adding two square numbers together that would always result in their sum yielding a number that has a positive square rooted number ahead for a solution, excepting of course the squares shown by the Greeks. Let's demonstrate this by showing if we take the square of 3, which is nine and add to sixteen which is the square of 4 or 16 and it to 9 we will get to 25 which does have a square rooted solution of five time five. Right or wrong. Now try that with the squares of 2x2=4
and 3x3=9 and which sums to 13. Unless I'm mistaken. Fermat knew what I've just demonstrated. But he knew more. Just try that with take the cube of two which is what...2x2x2 which yields eight 8 now take any other cubic number 3x3x3 which summed gives us 27 and then add 27 and 8 together and they sum to 35, which even a fool knows isn't or doesn't have as a solution any two cubic numbers that can be added together to yield what we needed. now take ten years like that guy who won the prize for solving the pulze I believe Fermat never claimed to have reached a proof for. Now I'll share what I believe Fermat and I came quickly to understand why there are NO soulitions to the cubes and because of that there can never be any solutions for and positive integers when taken to powers higher than cubes it's easy wait for me to point out why...don't we all know any time we take zero times any positive number the answer is...you know it xerro. Now pause again since Muller that there was NO solutions in to adding any two cubics together to reach a number that will have a solution in any sum arrived at that have be reached by adding two sums that were reached by adding powers including those of three because all powers higher than some squares known as Pyragorean Squares are also equal to zero. Now does that demonstration stand as a Proof. If so please send me the prize money and take it from that Cambridge guy who only thought he'd offered a proof. He didn't.
A rigious proof can not be offered. Such a demonstration would allow zero times empty sets of positive itergers not to be zero.
I can be reached at baker43john gmail.com By phone 1 360 8805276 522 E Main Apt 2 Centralia WA 98531
Why we need proof...
We have pythogaras theorem..
Poor deduction
the equation has solutions contrary to what is believed even for n> 2
Elaborate.
I'm going to pretend I understood that.
Before I tap in this video, I believe you want to show us the famous fail attempt (but enlight the development of classical Algebraic number theory) using cyclotomic extension and prime decomposition. But this, this is ridiculous, (almost) tells us nothing.
The thumbnail says 'for n>z', so learn to read please
What is the answer then with proof i wanna roast my math teacher thanks
Where the hell did you get the inequality "n" is greater or equal "z"? The evidence is totally wrong. I am sorry 🙂
I did not “get it”. I simply demonstrate what happens when you make an assumption that n is greater or equal to z
@@MetaMaths But that's wrong assumption tho. You're cutting out too much n.
afaik fermat's actual theorem didn't cut it out to n>=z, it was n>=3
And even n can't be desimplified like that. (z-y)(z^n...) only applies to odd n.
So no. It's just very partial proof. Not whole proof.
Disliked because i was searching full proof.
@@bxyhxyh you didn't get the point of the video then
@@lietpi Maybe I got word "almost" wrong.
It seems he meant "almost proving".
But I read it as "almost in 2 minutes"
For those unfamiliar with LaTex.
The key to the proof is that 161 can be written as the difference of the squares where
161=(15^2)-(8^2).
The Negation of Fermat's last Theorem: Proof IV}
begin{theorem}
Fermat's Conjecture, also known as Fermat's Last Theorem is false and therefore negated.
end{theorem}
begin{proof IV.}
1. The difference of the squares is given as a^2-b^2=(a+b)(a-b).
2. Given are the two Pythagorean triples:(8, 15, 17) and (161, 240, 289).
3. Therefore, via the Pythagorean theorem
(17^2=8^2+ 15^2) and (17^2=289=161+240).
4.Therefore
(8^2+ 15^2)= 17^2= 289=(161+240) =289
=[(15^2- 8^2)] +[ 2(8\times 15)]
=289=[(15-8)(15+8)]+[2(8\times 15)]
5. Squaring 289 yields
(8^4+ 15^4 = \quad17^4
ight)= (289^2=161^2+240^2)=\left([(15^2-8^2)]^2
=161^2
ight)+\left([2(8\times 15)]^2=240^2=\left([(15-8)(15+8)]^2
=161^2
ight)+\left([2(8\times 15)]^2=240^2
ight)=17^4
=289^2=(161^2+240^2)=83521.
6. Given the difference of the squares then
[(15^2-8^2)]^2=\left[15^4+\left[-(8\times 15)^2 -(8\times 15)^2
ight] + 8^4
ight]
=[(15^2-8^2)]^2=\left[15^4+(-\left[2(8\times 15)
ight]^2) + 8^4
ight]
7. Therefore
[(15^2-8^2)]^2+[2(8\times 15)]^2
=\left[15^4+8^4
ight]+\left[\left(\left[2(\left 8\times 15
ight)
ight]^2-\left[2\left(8\times 15
ight)
ight]^2
ight)= 0
ight]
=[15^4+8^4]= 17^4 =289^2=(161^2+240^2)=83521.
8. Fermat's Last Theorem states that no three positive integers, a,b, and c can satisfy the equation a^n+b^n=c^n for any positive integer value of n>2.
9. Given step #7 if a=15, b=8, c=17 and n=4 then Fermat's Last Theorem is proven false and thus negated.
\end{proof}
15^4+8^4=54721
Por causa do trinomio quadrados perfeitos.
Clickbait
no
@@MetaMaths Replybait.
@@ozan4702 Dickheadbait
pro
You can not put > in first video you should check this and put any thing so l will check
Xⁿ ? (Z-y)nyⁿ-¹
When n >=z this lead to
Xⁿ < (z-y)nyⁿ-¹
So not > and that make your proof is not efficient
sorry .
Fermat couldn't prove it in his lifetime and you just did it in 2 minutes 🤯
That is false scientist.
LOL
Ah oui metamatiques
It is the formula for turning lead into gold fill in a and B with atomic numbers
start page is bullshit: n > z
Didn't understand anything. Kinda a bad video lol, very poorly explained
EASY!!!!!!!!
💚💚"Great Video!"💚💚 (Jesimiel Millar Fernåndez)💙💙 1M2K101 [1 Chr. 4×36 = Matt. 2(4)×(36)]
I wonder how the solution above has eluded the great mathematicians for 350 years, namely Hilbert, Gauss, Euler. Hmmmmm
@shaharudinhamidun2117 You must be unbelievably stupid, if you actually think that none of them knew such a simple proof.
If smart people are watching your videos, do you really need to hold their attention with annoying background music?
Sorry, got no music education. Can you suggest a better melody for my vids ?
No music, Welsh word for carrot.
I have the real solution
I found the equation as n>z
And n can be anything
So x^n+y^n=z
So x=0
Y=0
Z=0
proof was required to yield nontrivial results
What if n less than z?
This is the essence of FLT
As a kid i only understand why x is y🥲👹
x^n+y^n=z^n =∈3 commentary design
L^3 space