None of this nursery school stuff - a proper maths problem. 25$ reward - Page 2

jason_t · #11 07-22-2005, 04:10 PM

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x^2+68-y^5
x=+/- (y^5-68)^1/2
y has to be >= 68^1/5
x can be any real number

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No, x can not be. I assure you that the OP meant, as he clarified later, for x and y to be natural numbers. Otherwise the problem is incredibly uninteresting and not related to any entire body of mathematics called Diophantine equations which is clearly what he intended.

ihardlyknowher · #12 07-22-2005, 04:11 PM

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the problem is too difficult for me.

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FYP

GFunk911 · #13 07-22-2005, 04:13 PM

I'm assuming X and Y are integers?

Edit: nm, clicked reply when there were no replies

ThaSaltCracka · #14 07-22-2005, 04:17 PM

[censored], if I had my TI-92 or whatever its called, I could figure this out.

TheIrishThug · #15 07-22-2005, 04:19 PM

yeah, i just googled Diophantine equations. this is way over my head. u guys have fun.

durron597 · #16 07-22-2005, 04:32 PM

Why do I think this is one of those "unanswered" problems that gets a million bucks if the correct answer is published.

hotsauce615 · #17 07-22-2005, 04:35 PM

We need matt damon

jakethebake · #18 07-22-2005, 04:36 PM

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We need matt damon

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I expect his gimmick account any moment.

citanul · #19 07-22-2005, 04:38 PM

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Why do I think this is one of those "unanswered" problems that gets a million bucks if the correct answer is published.

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it's not. it's just some random diophantine equation.

citanul

Arnfinn Madsen · #20 07-22-2005, 04:39 PM

I am a born mathematical genius (solved high school problems at age of 7) but haven't kept up with it lately. My intuition though tells me there are no solutions without quite finding the proof in my head.

So I will offer my 2 cents to lay the foundation for somebody else. When you ^2 a number you basically just double up the number of factors it consist of (i.e. 30=2*3*5 while 900=2*2*3*3*5*5). When you add 68 you add 2*2*17. The number you get after adding 68 has to contain a number of each factor dividable by 5. I think the assymetric nature of 68 containing of 1 factor of 1 kind and 2 of the other messes that up.

No proof, so I don't claim the $25.

EDIT: Just for clarification I know that addition does not mean that you add the factors (that is multiplying), so somebody need to look into the effect of adding the assymetric number of 68.