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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...
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.
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.
Originally posted by inverslyproportional
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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)
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.