[Dismas]

Code:

Dallas banker D. Andrew “Andy” Beal, who in 1997 established a $5,000 prize for solving his namesake equation, the Beal Conjecture number theory problem, has upped the ante in hopes of inspiring young people to pursue math, the Associated Press reports.

“Increasing the prize is a good way to draw attention to mathematics generally and the Beal Conjecture specifically,"  he said in a statement. "I hope many more young people will find themselves drawn into the wonderful world of mathematics."

For just a math problem, that's an amazing reward. I wonder who will end up taking that million dollars (implying someone will). It's insane he'd give away so much, but at least he's putting the money towards something constructive.

Source: International Business Times

[ねこまっしぐら]

I could give it a go. Now where is said problem?

[w00t]

There's a whole bunch of millennium math problems. It isn't the type of math you can do with a calculator.

http://en.wikipedia.org/wiki/Millennium_Prize_Problems

As for this instance:
http://en.wikipedia.org/wiki/Beal's_conjecture

The easier thing to do, unless you have studied number theory extensively, is try to find a counterexample to his conjecture.

[Pyro]

I could give it a go. Now where is said problem?

I believe the equation is a^(x)+b^(y)=c^(z)

[Dismas]

There's a whole bunch of millennium math problems. It isn't the type of math you can do with a calculator.

http://en.wikipedia.org/wiki/Millennium_Prize_Problems

As for this instance:
http://en.wikipedia.org/wiki/Beal's_conjecture

The easier thing to do, unless you have studied number theory extensively, is try to find a counterexample to his conjecture.

It's interesting, because I can only see a professor being able to solve those. Seems somewhat discouraging in my opinion, haha.

[ねこまっしぐら]

I believe the equation is a^(x)+b^(y)=c^(z)

I solved for z, now what?

[w00t]

I solved for z, now what?

The conjecture has nothing to do with solving the equation. To get the award, you must prove or disprove that:


if ( a^x + b^y = c^z ) and (a,b,c,x,y,z > 0) and (x,y,z > 2) then a,b, and c have a common prime factor.

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[ねこまっしぐら]

The conjecture has nothing to do with solving the equation. To get the award, you must prove or disprove that:


if ( a^x + b^y = c^z ) and (a,b,c,x,y,z > 0) and (x,y,z > 2) then a,b, and c have a common prime factor.

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Sounds easy enough. I will try to do this later.

[w00t]

Disproving it would be by far simpler( if it's in fact not always true ) as you could just program an algorithm to try every possible combination until it breaks.

Proving it isn't as simple as it seems. There's no mathematical law that states any correlation between powers of separate numbers.

[Customer]

I have scrambled eggs for brains.

I could never figure that out.