Billionaire Offers $1 Million to Solve Math Problem

Surely you can't be serious about this equation. I solved it, using six different whole numbers in 3 minutes.

How could such a simple maths problem have lasted for more than 350 years. I think the joke is on me.....

however the confusion from some of the entries seem genuine, and I will feel like a fun-killer if I provide you with the answer...

Originally posted by Trubeeleever

Surely you can't be serious about this equation. I solved it, using six different whole numbers in 3 minutes.

How could such a simple maths problem have lasted for more than 350 years. I think the joke is on me.....

however the confusion from some of the entries seem genuine, and I will feel like a fun-killer if I provide you with the answer...

STOP using numbers. That's not the answer.

HINT: solve it as a relation of X, Y and Z without using numbers at all.. Meaning, say how space IS.


Very basic X^2 + 2 = 4... Already we have 2 answers.. 2 and -2.. The equation in question has nothing to do with numbers.. umm, sort of like how a + b = c, can be rewritten as a = c - b.. re write the equation and use proven properties of math (without numbers) to prove the equation, which I don't see how to do...

Basically prove that 2D eternally effects 3D. And that's a bad description of it. It's the best I can do.

reply to post by Dustytoad


DustyToad, now you've confused me....I thought the answer was simple maths like

A = 1
x = 2

B = 4
y = 10

C = 3
z = 14

1 x 2 = 2
4 x 10 = 40
40 + 2 = 42

3 x 14 = 42

reply to post by Trubeeleever


ok. let's get more simple. Prove A = B..

OR

Someone post the proof of "Pythagorean theorem" So others can get a gist of the idea.

reply to post by Dustytoad


Thank you DustyToad
You might have to be a bit more descriptive, I haven't gone to school for over 25 years, I'm a little rusty.

But the way I read it
A doesn't need to equal B

the equation is (A x X) plus (B x Y) equals (C x Z)

reply to post by Trubeeleever


None of the numbers have to be anything.

The concept, the letters might have meaning, but it goes way beyond my comprehension.

This isn't fill in the blanks.

you have to prove a RANDOM number raised to another random number + a RANDOM number raised to another random number (in 2 dimensions X and Y) = a RANDOM number raised to a random number..

The only way I can see it is proving that dimensionality is relative through each dimension.. Not something I could ever prove, but I do believe it.

You got a star for pointing out what I have been saying.. NONE of the letters have to equal ANYTHING.

Here is something: Mathematical proofs

Must be a problem with the presentation of it because, for 6 variables and no constants, I can see an infinite range of valid solutions no matter how you mix up the operators.

The proof of A^n+B^n=C^n, n>2 (Fermat´s last theorem) being impossible, is over 100 pages long and took 7 years of work,

The claimed (not accepted yet) proof of abc conjecture is 500 pages long.

abc conjecture: For every ε > 0, there are only finitely many triples of coprime positive integers a + b = c such that c > d^(1+ε), where d denotes the product of the distinct prime factors of abc.

Pythagorean Theorem Proof

Some examples of mathemathical proofs

I doubt proving it true is easy (although some proofs are quite easy to make)

What could be done by people not very strong in math is only finding a contradictive option, which disapproves the theory. Basically finding a matching set A^x+B^y=C^z,( x,y,z>2) where A,B and C do not have common prime factor. Although this requires the initial theory to be wrong in the first place, which is unlikely due to it being out there and studied for years already. Making an algorithm searching for such sets is not very hard, so I doubt experimentation would work, but well who knows, some luck and you earn a million

An easier way which does not require much knowledge in the field, just luck, is running the Mersenne prime finding script in your computer. 50k price and just program does not take much resources, runs in the background. Like lottery, but with actually giving something to science, excluding different numbers which are not mersenne primes or finding new ones

Originally posted by Pilgrum
Must be a problem with the presentation of it because, for 6 variables and no constants, I can see an infinite range of valid solutions no matter how you mix up the operators.

The solution isn't numbers so you don't need constants. You can see an infinite range of numbers that can fit as can I. This is not the question being asked.

reply to post by Cabin


that "proof" of pythagorean theorem is not proof..

Also, stop adding [greater than 2] to the equation..

If someone could prove this equation the meaning of it would be astounding.

Then I can see an application in the field of cryptography

Some people have a little too much time on their hands

reply to post by Dustytoad


en.wikipedia.org...

There are actually over 100 ways of proving pythagorean theorem, but the one I posted can also be considered a proof


Greater than 2 is important piece of information. Without that the conjecture/theorem would not be the same. It makes a huge difference.

Originally posted by Cabin
reply to post by Dustytoad

 

en.wikipedia.org...

There are actually over 100 ways of proving pythagorean theorem, but this can also be considered a proof

Greater than 2 is important piece of information. Without that the conjecture/theorem would not be the same. It makes a huge difference.

Where is that in the OP (greater than 2)? I don't see it as a condition at all.

And yes that shows proof of pythagorean theorem. The other link skipped steps so..

I doubt the proof would be in algebra or calculus...

reply to post by Dustytoad


OP is wrong, it just copy-pasted from an article where mistakes where I also posted it before.

The million prize is for proving or disapproving the Beal Conjecture

Million Dollar Prize (set by the Billionaire) for proving or disapproving the Beal Conjecture


Beal Conjecture:
If A^x + B^y =C^z ,
where A, B, C, x, y, and z are positive integers with x, y, z > 2,
then A, B, and C have a common prime factor.

reply to post by Cabin


sorry. I see it now. The article (/OP) is way off then.

It's an if then statement. Completely provable if I was smarter.

Thanks for the link. WOW.. I still have no clue how to prove anything about it. But a million bucks is worth my spare time.

Star for respect.

Originally posted by inverslyproportional
[

In math there is only ever one correct answer to the problem, I am simply trying to find it for the fun of stimulation of my brain before I go to bed.



I am pretty certain that is an Incorrect statement.

Have you ever seen a Rubik's Cube?
It has an Ungodly number of Possible Solutions.

Originally posted by Dustytoad
reply to post by TeflonBear

 

Are you guys silly? 3 variables times 3 variables arranged in an equation..
There is no answer..
It's a lot like life, actually..
He'd bet more than 1/8 thousandth of his worth if he expected something.
It's only there to make you think in 3 dimensions, which is the future of rational thought. Not that I can do that coherently either..

edit on 6/7/2013 by Dustytoad because: (no reason given)

On the surface, it sounds as if you are correct. However, after a bit of thought, I see there is an answer. Now all I need to do is a proof.

reply to post by Cabin


Excellent! You are very clear. I appreciate your adding something useful to this discussion.

reply to post by Cabin


Thank you for your addition(s) to the thread. This is my first and I should have done a little more research.

Originally posted by Cabin
reply to post by boymonkey74

 

Ok.

I´ll explain the problem then

Beal Conjecture:

If A^x+B^y=C^z,where A, B, C, x, y, and z are positive integers with x, y, z > 2,
then A, B, and C have a common prime factor.

Basically:

There are 2 numbers A and B. Both of them are powered of some number larger than 2. For example,

A^x=3^3=27
B^y=6^3=216

If you sum these up, you get A^x+B^y= 216+27=243

243 is 3 powered of 5. 3^5

This means the equation is true. 3^3+6^3=3^5
A=3, B=6, C=3, x=3, y=3 and z=5

If we take a look at A,B and C, all these are divisible by 3. 3 is a prime number, so it is the prime factor of all these numbers, this shows that on this case the equation is correct.

What this reward requires is proving that such equation is correct in all situations or finding a case, where this equation is not true, (so that when you add two powered numbers (larger power than 2) up and get another powered number (larger power than 2), then these three numbers A,B and C do not have a common prime factor)

I am not very good at explaining math in english, although hopefully it helps at least somebody to understand this problem.

You highlighted that the OP wrote the equation incorrectly. He didn't include the '^' indicating an exponent.
The equation is a^x+b^y=c^z or, a to the x power + b to the y power = c to the z power. Another example is 2^2=4, 2 squared.