I believe I've stumbled upon a Tenth Wonder. Try this- take any number and add it to the one adjacent. 4 + 5 or 6+7 or 566+567 Take the sum of that and to it add the square of the smaller number. It goes something like this: 4+5=9+16= 25 6+7=13+36= 49 566+567=1133+320356=321489 Notice how the number in bold is always the square of the larger number in the addition. Has anyone else discovered this?
Come on Gendy... Please Register or Log in to view the hidden image! ...watch... Real number N N + (N +1) Ok? That's 2N +1 Keep goin' N<sup>2</sup> + 2N + 1 There we are. Of course... (N+1)<sup>2</sup> = (N+1) (N+1) = N<sup>2</sup> + N + N + 1 = N<sup>2</sup> + 2N + 1 Simple, ja, mea amica?
*sniff* *sniff* You smell like Redoubtable, haven't seen that schoolboy in ages- wonder why. Am I right or am I right? Making N=4, that would be 16+8+1= 25 Well hot damn, I believe you are telling me this 'tenth little wonder' I happend to come across is only a quadriatic mutation? I have no fucking clue how useful what "I've" discovered would be.
Rap you've managed to explain why the trick works, but it is still pretty usefull at times. Thanks for the explanation. Gendanken 'you' managed to stumble across a neat little trick, even if that is all it is. It's a bit like those other multiplication tricks I remember learning to spead up mental math. Even if it isn't earth shattering, it's still useful (at times) and interesting.
I found this one in a bit of a different way. I didn't know algebra at the time, so I'll spell it out verbosely, the way I originally thought it through, with the algebraic equivalent in italics after each line: Any number squared is that number multiplied by itself, or that same number of lots of the number. n<sup>2</sup> = n.n So, when a number is added to its square, the result is one extra lot of the number, or the number multiplied by one more than the number. n.n + 1.n = (n+1)n This is the same as that number of lots of one more than the number. (n+1)n = n(n+1) Now, adding an extra lot of one more than the number makes one more than the number lots of one more than the number, which is the square of one more than the number! n(n+1) + 1(n+1) = (n+1)(n+1) = (n+1)<sup>2</sup> I thought I was pretty brilliant at the time Please Register or Log in to view the hidden image! Later on, I related this to pythagorean triads: If 2n+1 is a perfect square for some n, then n, (n+1), √(2n+1) is a triad. Examples are 3,4,5 ; 5,12,13 ; 7,24,25 ; 9,40,41 It's always a little disappointing to find you've rediscovered something that's been know for 2000 years Please Register or Log in to view the hidden image!
Well, have you ever been wrong? Again, ja, mea amica. I've been back awhile, and I'm better than ever. That's a fitting way to put it, considering that quadratic equations were discovered by the Babylonians thousands and thousands of years ago, I believe.
Please Register or Log in to view the hidden image! I've done better then 2000 years, but still not enough to become famous... Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image!
Rap: Knew it, knew it, knew it. Genda's pineal gland knows all, sees all. Better than ever, huh? Are you now.....still haven't found a reference to "bombast" being used as a noun, have you? Pete: No.Shit. I once spent a whole lunch/study fitting out the Pythagorean theorem with exponents up to the hundreds, somehow proving that the theorem only worked for squares. Then lo behold this Wiles shows up saying he has 'proved' Fermat's last theorem, what I in my stupid nublity actually did sans discipline. Mephura: You're...back.
I'm afraid I'm in the dark, as I have frequently found myself when conversing with you. I myself did, once, make reference to both the use of "fustian" as a noun for a person and to your input on this "issue". I was trolling for a response, really, but no one ever noticed. I was beginning to feel sorry for myself when no one surmised my writing style, either. I may've even accidently written "Redoubt" at the end of a few of my posts. I never checked to correct them though. I'm getting remarkably lazy these days.
Rap: Durst you imply that one gets lost in Gendanken's "babble"? Fye! Seriously, as an artist/writer clarity is priority, always. I'll be damned to hear that people do not understand me on account of gassy rhetoric. That's Tessie's realm. Firsts for everything!- a man has actually bested me on my memory. You're right, the bone of contention concerned 'fustian' and not 'bombast'. Look, there is not a soul alive on this planet that can derail a thread as I can. I've done it with elephants and Greek veils, blindspots, hate, and I can do my own thread the same disservice with a battle on 'fustian' I'm right, monsiuer. You were only too narrow or ill-read or advised to see why. No, actually I sniffed a small hint in some thread of mine where you posted. Gibsonitis. I accused you of grand-theft-wordery; in other words, using a thesaurus.
So are you saying that you managed to prove Fermat's Last Theorem for the first few hundred exponents during your lunch hour, without any high powered techniques? If this is true, and you'll forgive my skepticism, it would be incredibly interesting and potentially quite important.
I believe I've stumbled upon a Tenth Wonder. Whats the 9th wonder? I thought there were eight? (and then chyna but that was just a gimmick).
Really? Evan a simple proof that a<sup>7</sup> + b<sup>7</sup> = c<sup>7</sup> has no integer solutions for a, b, and c would be astounding. Can you recall how you did it?
He wanted to write it as a footnote, but he didn't have the time, so he'll leave it to the reader, it's a simple proof. Please Register or Log in to view the hidden image!
Schmoe: Well, that I 'proved' his theorem is just something I like telling myself and yes I got up to around 100, quite simple. Using a simple calculator and cutting the multiplication down to conveninet halves- for example, if you know that 42 times 42 is so and so, 84 times 84 becomes easier if you acquaint yourself with a pattern. It does not matter which numbers you decided to use as halves, its all arbitrary. Not really, I only looked down one day bored as fuck and wondered if any other numbers would fit as exponents. Of course, none did- only squares work. Logially Sound: The 9th wonder would be the patchy mold on fat woman's undergarments. Kidding. I believe in the natural world we have 9 wonders- Victoria Falls in Africa, Angel falls in Venezuela (?) and the others I'm too lazy to look up now. Pete: Pause. Dude- I know that. *That* is what Fermat wrote on his margin, that no other integer save 2 has a solution to the formula. We all know this, but only Wiles had the tedious discipline stretcehd over years to show why. 1100f: Ha, ha.
I'm sorry but I don't understand what you're talking about. What does this halving business accomplish in realation to the problem? Could you do as Pete requested and show us why the exponent 7 will have no integer solutions using your method? If you were able to get up to 100 quickly, it should be no problem showing us your work for one exponent. If you don't like 7, I'll let you choose your own exponent.
gendanken: I think you're a little confused. Suppose you want to show that there are no integer values of x,y and z which satisfy: x<sup>4</sup> + y<sup>4</sup> = z<sup>4</sup> There is a huge difference between trying a few combinations of numbers, only to discover that none of them satisfy the equation, and actually coming up with a PROOF that NO combination of integers will satisfy the equation. If you actually have a simple proof that NO combination of integers can satisfy the above equation, mathematicians across the world will be very interested indeed. And that's only for 4-th power exponents. The next step, after you've done that one, is to extend your proof to all the other powers (except for 2).
Look, people. This is not nuclear science- I did not do anything of much skill nor did I 'prove' anything. Fermat states in his theory that there are no natural number solutions for this formula with an exponeent >/= 3. This means that its impossible for a cube to equal the sum of two cubes or for a fourth power to equal the sum of two fourth powers, yes? And if we are speaking natural numbers (not integers) then its the common 1,2,3,4,5...and and so on. Therefore: That is exactly what I am saying- there is no way that someone of my caliber, skill, or patience would undergo what Wiles was competent enough to do. All we have here is a little girl fiddling with the numbers in boredom. Given: 10^2= 8^2 + 6^2 20^2= 16^2 + 12^2 40^2= 32^2 + 24^2 80^2= 64^2 + 48^2 160^2= 128^2 + 96^2 are perfect all the way up in the hundreds and hundreds, the pattern among them being a factor of 4. as opposed to (for Schmoe): 10^7= 8^7 +6^7 10000000= 577088 is simply not true. For that matter x^3=3^3 + 4^3 gives an irrational number, try any cube with any number combination and the answer is always an rational answer. Try this for 3, 4, 5, 5, 7, 100 powers and you'll find no other natural number as solution for that power, it only applies if the exponent is 2. Logs you can use to figure out the larger exponents. I did not prove anything- I only found that for all those hundreds of attempts all other solutions were either integers or rational numbers, not natural numbers.
