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!

:? I dotn see a question, all i see is an equation made of all variables. What are you trying to do with it?

Uhh try to get Y alone? or the derivative?

Nothing to solve unless you state what we're solving for...

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...

Gah! I don't even want to think about solving that! #-o

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.

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.

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...

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.

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...

Geometry has proofs... :x :x :x :XD: :XD: :XD:

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.

My I ask. What is the point of such a equation.

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:

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

how about

 

x=2

 

y=3

 

z=5

 

n=1

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=1

The thread starter forgot to mention that Fermat's thereom refers to when the n is 3 or greater.

ah, so close....

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

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.

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)

yup well spotted ^^ :)

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:)

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