4 ms·
_DON'T DOWN VOTE JUST BECAUSE YOU CAN'T DO MATH_ The proof Fermat hinted to was about the difference between squares. All whole numbers taken to a power greate
by ox_n 8y ago
_DON'T DOWN VOTE JUST BECAUSE YOU CAN'T DO MATH_
The proof Fermat hinted to was about the difference between squares. All whole numbers taken to a power greater than two (n^3) can be represented as the difference between two whole squares (x^2 - y^2). These differences can then be shown as the sum of consecutive odd numbers:
2^3 = 3^2 - 1^2 = (1+3+5) - (1) = 8,
3^3 = 6^2 - 3^2 = (1+3+5+7+9+11) - (1+3+5) = 27,
4^3 = 10^2 - 6^2 = (1+3+5+7+9+11+13+15+17+19) - (1+3+5+7+9+11) = 64
5^3 = 15^2 - 10^2 = (21+23+25+27+29) = 125
When you examine the odd number series that results from each base, you'll discover that there will always be a gap if you try and combine two odd number series together, which explains Fermat's little joke about margins. The same trick works for higher powers.
It's not that hard people. Stop believing everything you're told about how "hard" something is.
HINT:
The number of odd numbers in the series exactly matches the starting square base number
- x3tm 8y ago> It's not that hard people. Stop believing everything you're told about how "hard" something is. There are still many problems in physics and mathematics which are considered "hard" (e.g., dark energy, Riemann hypothesis, etc). Can we crack them by simply adopting your positive mindset?
- ox_n 8y agoWhat other mindset do you see working better?
- x3tm 8y agoI don't think the "you can do anything" mindset works in real life. It helps self-help book authors sell their stuff, but it's not a good strategy to live by. (Incidentally, this reminds me of Key & Peele's "You can fly" sketch). What does work though is this: advanced formal education in a topic. Once you have that you can start thinking on how to solve some simple open problems. And if you are lucky and turn out to be extremely smart, you may be able to tackle more challenging problems. Some amount of self confidence may also you to keep going but doesn't make you a genius overnight. Simply going to a mindset where things are 'not hard' is closer to delusion than it is to anything else. In academia we get often emails from people who solved quantum gravity (e.g. using fire), show us how einstein is wrong (e.g. using a pendelum), etc. I'm pretty sure they also convinced themselves to "Stop believing everything they're told about how "hard" something is"
- syn0byte 8y agoSo what your saying is... We should ignore letters from patent clerks?
- ColinWright 8y agoNo, but even those from "left field", if genuine, tend to take the time and care to write things up properly, to use the nomenclature of the field, to address obvious potential concerns up front. If you're asserting something that's likely to encounter resistance, it's worth being clear and careful.
- VRay 8y agoOh man, that reminds me of an experience I had in college. I was working with the aerospace department on their fusion reactor (I was just writing software to help them process data from it, not involved in the science itself). My boss kept getting calls from crackpots who'd go on and on and on about their bogus theories, and how they were being shut out of the mainstream by small minded fools, etc etc. It was pretty frustrating. He was too nice a guy to tell them off or even cut them off quickly. My advice to any crackpots who are really sure they're actually geniuses: Get into the stock market (with a SMALL investment). If you're as smart as you think you are, you can find an angle and turn $100 into $1,000,000 or more, and then if anything it'll be GOOD that nobody ever believed in you. I've run across arbitrage opportunities that would have made me fiendishly rich if I'd noticed them sooner myself, believe it or not. Just be careful and don't mess with box spreads.
- deleted 8y ago[deleted]
- yters 8y agoSounds interesting, but not sure what you mean by: > you'll discover that there will always be a gap if you try and combine two odd number series together Can you elaborate?
- aaaaaardvark 8y agoox_n's last theorem
- ox_n 8y agoConsecutive base numbers will necessarily alternate between even and odd. So even the closest base numbers still have a gap between their resulting odd number series, which only increases as the distance between base numbers increases.
- andrepd 8y agoStill don't get it. What do you mean by "base numbers" and what do you mean by "alternate between even and odd"?
- gowld 8y ago'ox_n appears to be trying to prove (by a pigeonhole agument) that the equation x^n + y^n = z^n in unsolvable for _some z_. That's much weaker than proving that it is unsatisfiable at _every_ z.
- HNLurker2 8y agoOff topic: why doesn't HN support LaTeX?
- nappy-doo 8y agoLikely because MathML isn't widely supported, and has even been removed from Chrome.
- ColinWright 8y agoMathJax works an absolute treat.
- gowld 8y agoExcept on Android where the math takes up more horizontal space than the layout engine thinks it should, causing overlap with text after the math.
- ColinWright 8y agoI'll pass that report on to them - have you reported it already? In which browser is that happening?
- ColinWright 8y agoDo you really believe that: (a) This constitutes a proof; (b) This is the "proof" that Fermat had; (c) Mathematicians missed this for over 350 year? I'm not quite sure exactly what you are claiming.
- ox_n 8y agoIt's completely arrogant to assume that because it hasn't been solved by "better" people that I couldn't solve it.
- ColinWright 8y agoOK, so I'm taking it from your comment that you really do believe this constitutes a proof. Thanks for the reply.
- ox_n 8y agoI see it more like walking away knowing that I have checkmate whether the other person can see it or not. If you pay attention to the conditions that must necessarily exist, like the that the difference between two whole squares can be represented as EITHER an integer base taken to a power higher than two, or the sum of a series of consecutive odd numbers, the number of terms equaling the original base number of the higher power equation.
- andrepd 8y ago>about: Fuck you, hater. Oh, you're that guy.
- ox_n 8y agoI got to call them like I see them. Thanks for announcing that you stepped on that land mine.
- ColinWright 8y agoIt looks like you edited this comment, but I'm serious. I'm trying to understand your proof, but I'm having trouble seeing what the steps are for higher powers than 3 or 4. I already know the proofs for then cases n=3 and n=4, but I can't see how what you say works in the case, say, n=5, or n=13. Seriously, can you walk us through the steps of why x^5+y^5=z^5 has no (non-trivial) solutions? And to be fair, there are cases where, say, undergrads have proven significant results that had been outstanding for a long time. Proving that prime recognition is in P is one such case. but in that case they published a complete, clear paper. In this case I can't really see what you're saying you've done, or why it's true, which is why a walk-through of the case n=5 would be so helpful. Thanks. ======== For anyone interested, this is what the comment used to say ... The series of odd numbers must be consecutive and they simply are not when you add two different series together for all powers greater than two. It's okay. You'll get it. 1) a whole number, n, taken to a power greater than two 2) can be represented as a consecutive series of odd numbers 3) where there will always be a gap in the series between consecutive base numbers, for all p and p+1 4) therefore there will be a gap for all p and p+n, n>=1 combinations
- keymone 8y ago> The same trick works for higher powers. can you demonstrate?
- ColinWright 8y agoI'm working to understand this, but I can't seem to fit it together. Following Feynman's lead in this sort of thing, can you give me an explicit example of why the equation x^13+y^13=z^13 has no solutions? Or even just use your technique to explain why x^5+y^5=z^5 has no solutions? Thanks.
- dang 8y agoCan you please read the site guidelines and follow them? https://news.ycombinator.com/newsguidelines.html https://news.ycombinator.com/newsguidelines.html
- zorga 8y agoFuck you ya sjw cunt. Take your pussy ass guidelines and shove them up your ass.
- throwawayznxnyn 8y agoThis argument puzzled me, but unfortunately this is as far as I got: ~~Lemma 1~~: z^n can be written as a difference of squares: Proof: z^n = x^2 - y^2 = (x+y)(x-y), leading to the system of equations x + y = z^(n-1) x - y = z which may be solved for x and y: x = (z^(n-1) + z)/2 y = x - z QED. ~~Lemma 2~~: Any number squared can be written as a sum of sequantial odd numbers starting at 1. By induction: (n)^2 = (n-1)^2 + (2(n-1)+1) = \sum_{k=0}^{n-1}(2k + 1) QED. ~~~Fermat's Last Theorem~~~: z^n = x^n + y^n has no solutions for n>2, and positive z, x, y. Proof: Without loss of generality, assume a > b, then rewrite z^n as a sum of sequential odd numbers starting at b: z^n = a^2 - b^2 = \sum_{k=b}^{a-1}(2k+1) x^n and y^n can similarly be written as a sum of sequenatal odd numbers: x^n = c^2 - d^2 = \sum_{k=d}^{c-1}(2k+1) y^n = e^2 - f^2 = \sum{k=f}^{e-1}(2k+1) By substitution to the Theorem's equation: \sum_{k=b}^{a-1}(2k+1) = \sum_{k=d}^{c-1}(2k+1) + \sum_{f}^{e-1}(2k+1) Which is true if and only if there are no gaps in the bounds of summation on the right side, so d = b, c = f, and e = a. But then a^2 - b^2 = (f^2 + (-a^2))^n + (b^2 + (-f^2))^n and this is certainly not true by the binomial theorem. We've reached a contradiction so QED. ^^ This is where I'm stuck. I don't actually know why that would not be true by the binomial theorem? It seems like simple expansion should do it.