Seahawk10490 Posted September 8, 2006 Share Posted September 8, 2006 My Calculus AP teacher challenged us with what is to be known as the hardest equaiton ever. Me and 5 others spent hours at the library trying to solve this one. We have over 20 pages of work, and about 500 papers in the trash :x . Heres the equation: X^n+Y^n=Z^n (^ that is to the power). Weve come up with this. It has no solution. weve researched this and eventually found the complete answer. Im not going to post it but pm me for the answer. The eqaution is known as Fermats equations. Its baffled mathematicians for 350 years. It was solved by Andrew Wiles, and i must admit, if it wasnt for the internet we would not have been able to come up with the No solution answer. Enjoy this brainstrain! Its better to be judged by twelve, than to be carried by six. Link to comment Share on other sites More sharing options...
Lionheart_0 Posted September 8, 2006 Share Posted September 8, 2006 :? I dotn see a question, all i see is an equation made of all variables. What are you trying to do with it? Sig by IkuraiYour Guide to Posting! Behave or I will send my Moose mounted Beaver launchers at you! Link to comment Share on other sites More sharing options...
unknownmasterofnothing Posted September 8, 2006 Share Posted September 8, 2006 Uhh try to get Y alone? or the derivative? Link to comment Share on other sites More sharing options...
No_OnE Posted September 8, 2006 Share Posted September 8, 2006 Nothing to solve unless you state what we're solving for... Link to comment Share on other sites More sharing options...
Lionheart_0 Posted September 8, 2006 Share Posted September 8, 2006 Uhh try to get Y alone? or the derivative? This would explain it then. I only have grade 10 math, so i only kno whow to solve equations with numbers in them... Sig by IkuraiYour Guide to Posting! Behave or I will send my Moose mounted Beaver launchers at you! Link to comment Share on other sites More sharing options...
knives669 Posted September 8, 2006 Share Posted September 8, 2006 Gah! I don't even want to think about solving that! #-o Link to comment Share on other sites More sharing options...
deloriagod Posted September 8, 2006 Share Posted September 8, 2006 Gah! I don't even want to think about solving that! #-o Hah, I was thinking the same thing. I'm doing my geometry homework and it's proving to be a pain in the a*.. Seahawk, I haven't seen many of your posts but you seem to be quite the intellectual :thumbsup: Personally I would have said screw it when I saw that problem. Internet Marketing For Newbies Link to comment Share on other sites More sharing options...
logic-is-overrated Posted September 8, 2006 Share Posted September 8, 2006 There's nothing to prove. Fermat said a^n + b^n = c^n has no real solution when n is 3 or greater. In 1995 Andrew Wiles published his 150 something page corrected proof that proved Fermat's thereom to be correct. I think that's right. This is the way the world ends. Look at this [bleep]ing shit we're in man. Not with a bang, but with a whimper. And with a whimper, I'm splitting, Jack. Link to comment Share on other sites More sharing options...
No_OnE Posted September 8, 2006 Share Posted September 8, 2006 Like the poster above me said, there shouldn't be an answer if n is anything greater than 2. We know that: x^1+y^1=z^1 because a+b=c x^2+y^2=z^2 because a^2+b^2=c^2 which is also known as the Pythagorean Theorem. Now, I'm only in Algebra 2 right now so I don't know too much about this. I'm not sure whether anything greater than 2 would work or not... Link to comment Share on other sites More sharing options...
knives669 Posted September 8, 2006 Share Posted September 8, 2006 Gah! I don't even want to think about solving that! #-o Hah, I was thinking the same thing. I'm doing my geometry homework and it's proving to be a pain in the a*.. Seahawk, I haven't seen many of your posts but you seem to be quite the intellectual :thumbsup: Personally I would have said screw it when I saw that problem. Lol. My Geometry class was easy. I don't think I learned anything in there though. :( Math is soooo confusing. Link to comment Share on other sites More sharing options...
No_OnE Posted September 8, 2006 Share Posted September 8, 2006 Gah! I don't even want to think about solving that! #-o Hah, I was thinking the same thing. I'm doing my geometry homework and it's proving to be a pain in the a*.. Seahawk, I haven't seen many of your posts but you seem to be quite the intellectual :thumbsup: Personally I would have said screw it when I saw that problem. Lol. My Geometry class was easy. I don't think I learned anything in there though. :( Math is soooo confusing. My Geometry teacher was crazy... She talked to herself and she also talked to the wall... She was a good teacher though... Link to comment Share on other sites More sharing options...
Viktorkrum77 Posted September 8, 2006 Share Posted September 8, 2006 Geometry has proofs... :x :x :x :XD: :XD: :XD: Me doing staff. Link to comment Share on other sites More sharing options...
Freshwacka Posted September 8, 2006 Share Posted September 8, 2006 Like the poster above me said, there shouldn't be an answer if n is anything greater than 2. We know that: x^1+y^1=z^1 because a+b=c x^2+y^2=z^2 because a^2+b^2=c^2 which is also known as the Pythagorean Theorem. Now, I'm only in Algebra 2 right now so I don't know too much about this. I'm not sure whether anything greater than 2 would work or not... Technically, you can't say that it can't be anything greater than 2, because it could be 2.1. The correct way to say it would be "3 or greater" Unless you're right. Then the other person is wrong. 99 Crafting Blog Link to comment Share on other sites More sharing options...
Militaris Posted September 8, 2006 Share Posted September 8, 2006 My I ask. What is the point of such a equation. Link to comment Share on other sites More sharing options...
Purecheese Posted September 8, 2006 Share Posted September 8, 2006 x^n+y^n=z^n Assume y=0, then x^n+y^n=x^n=z^n, solution x=z or x=-z if n=2*c. The same can be done when you assume y (or x or z) has a different value. One equation with three variable can't be solved so yea. So if this thing is going to have a solution at all you'd want to assume that you assign atleast one variable a value. Honestly though, this is far from a difficult equation Some googling gave me the following: Fermat claimed to have discovered a proof that the Diophantine equation (x^n+y^n=z^n has no integer solutions (integer is a whole number 0,1,2,3,4,...) for n>2 and x,y,z!=0 (not equal to 0) That's a whole lot harder to prove I think and I know that I'm not going to try :D :P Proof can be found with google though :wink: Link to comment Share on other sites More sharing options...
mad4u689 Posted September 8, 2006 Share Posted September 8, 2006 Quick explanation: If an integer n is greater than 2, then a^n + b^n = c^n has no solutions in non-zero integers a, b, and c. This is Fermat's Last Theorem. Note that it was a theorem - it hadn't been proved, and it hadn't been disproved. However, Fermat wrote in the margin of his notes that he'd found a relatively simple proof for it!! But he died before publishing it, and it had never been found out (for CENTURIES!!) until in 1995 someone finally proved it. Woooo, mathematicians! :D Everybody hug and spread the love :D Link to comment Share on other sites More sharing options...
sheepdean Posted September 8, 2006 Share Posted September 8, 2006 how about x=2 y=3 z=5 n=1 Back by popular demand!And I guess I just wanted to tell you, as the light starts to fade, that you aree the reason, that I am not afraid, and I guess I just wanted to mention, as the heavens will fall, that we will be together soon if we will be anything at all. Link to comment Share on other sites More sharing options...
logic-is-overrated Posted September 8, 2006 Share Posted September 8, 2006 Technically, you can't say that it can't be anything greater than 2, because it could be 2.1. The correct way to say it would be "3 or greater" Unless you're right. Then the other person is wrong.As Mad said, integers. how about x=2 y=3 z=5 n=1The thread starter forgot to mention that Fermat's thereom refers to when the n is 3 or greater. This is the way the world ends. Look at this [bleep]ing shit we're in man. Not with a bang, but with a whimper. And with a whimper, I'm splitting, Jack. Link to comment Share on other sites More sharing options...
sheepdean Posted September 8, 2006 Share Posted September 8, 2006 ah, so close.... Back by popular demand!And I guess I just wanted to tell you, as the light starts to fade, that you aree the reason, that I am not afraid, and I guess I just wanted to mention, as the heavens will fall, that we will be together soon if we will be anything at all. Link to comment Share on other sites More sharing options...
rft Posted September 8, 2006 Share Posted September 8, 2006 looks like your calculus teacher was having a bit of a laugh with you there. Shame that you spent a long time trying to prove it. To answer the question above: how about x=2 y=3 z=5 n=1 this is a triangle with zero area! Here's one to get your revenge on your calculus teacher with: Prove that 1=2!! (there is a a flaw in this, see if you can figure it out) A=B AB=BB AB-AA=BB-AA A(B-A)=(B-A)(B+A) A=B+A as A=B, then A=2A, therefore 1=2! :D Best Barrows Chest: guth skirt, dh helm, dh axe, racks, gp (23/08/06); best tt reward: robin, guthix helm, guthix legs, nats, rune legs Link to comment Share on other sites More sharing options...
DragonFirenze Posted September 8, 2006 Share Posted September 8, 2006 Erm... Not quite the hardest equation. And if you really did spend 5 hours on it i suggest you get out more :-k . My maths teacher was telling me about an equation that took the entire maths department (10 or more teachers) 12 hours to figure out. Lvl 80 construction. Dragon Drops: 11 (4 Chains, 2 Axes, 1 Med, 2 Skirts, 1 Legs, 1 2h) God Wars Drops: 4 Zamorakian Spears, 1 Godsword Shard 1 Link to comment Share on other sites More sharing options...
runesmithie Posted September 8, 2006 Share Posted September 8, 2006 looks like your calculus teacher was having a bit of a laugh with you there. Shame that you spent a long time trying to prove it. To answer the question above: how about x=2 y=3 z=5 n=1 this is a triangle with zero area! Here's one to get your revenge on your calculus teacher with: Prove that 1=2!! (there is a a flaw in this, see if you can figure it out) A=B AB=BB AB-AA=BB-AA A(B-A)=(B-A)(B+A) A=B+A as A=B, then A=2A, therefore 1=2! :D No, silly rabbit, you're dividing by zero! (At the bolded part) I just posted something! ^_^ to the terrorist...er... kirbybeam. Link to comment Share on other sites More sharing options...
rft Posted September 8, 2006 Share Posted September 8, 2006 yup well spotted ^^ :) Best Barrows Chest: guth skirt, dh helm, dh axe, racks, gp (23/08/06); best tt reward: robin, guthix helm, guthix legs, nats, rune legs Link to comment Share on other sites More sharing options...
dogfever Posted September 8, 2006 Share Posted September 8, 2006 Are you suggesting we all try and solve Fermat's Last Theorum? This was reputedly one of the most impressive feats of mathematics in recent times. Although you haven't actually given the full specification of the problem which is: For the system: a^n + b^n = c^n it is impossible to find a,b,c > 0 which solve the equation for any power of n>2. This was proven in 1994. If you had proven it during your five hours in the library I woudl have been very impressed:) Link to comment Share on other sites More sharing options...
Black_Flat Posted September 8, 2006 Share Posted September 8, 2006 Your kidding me right, what is so difficult about this awnser? X^n+Y^n=Z^n By looking at this problem by means of subtraction, we get. X+Y=Z and so Z-X = Y and Z-Y = X Thats my look on it, i may be way off but, to me that looks like a pritty simplified awnser. please feel free to correct me if im wrong but there is no awnser (Or ignore my post as its been 2 years since ive done any form of maths) Black flat Link to comment Share on other sites More sharing options...
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