P.S. please put all answers in spoiler tags.
Math Problem #2 (Number Theory)
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pat777
United States356 Posts
P.S. please put all answers in spoiler tags. | ||
Solinren
United States2653 Posts
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QuickStriker
United States3694 Posts
That being said, give me several hours and I'll try to see if I have an answer... | ||
Archaic
United States4024 Posts
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mikeymoo
Canada7170 Posts
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Avidkeystamper
United States8551 Posts
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Cheeseball
Australia208 Posts
The only answer is when x=1 y=1 and x=-1 y=1 If x=0 then y=(1/4)^(1/3) If x is negative, its the same as being positive because its squared If x=2 y=(13/4)^(1/3) ---- not integer If x=3 y=7^(1/3) ---- not integer Because y is cubed, it cannot have any other integer values that match the equation. You would need (3x^2+1)/4 to be either 0,1,8,27,64,125, etc. None of those values lead to an integer x value | ||
Muirhead
United States556 Posts
It's easy to see that x must be odd. Set x=2m+1 and simplify to get 3m^2+3m+1=y^3. We recognize that 3m^2+3m+1=(m+1)^3-m^3, which is the tricky step. The equation is equivalent to y^3+m^3=(m+1)^3. By the special case of Fermat's Last Theorem when the exponent is 3, we see that m=0 and y=1 or m=-1 and y=1. Thus the only solutions to the original question are x=1,y=1 and x=-1,y=1 | ||
oshibori_probe
United States2932 Posts
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B1nary
Canada1267 Posts
I guess it's sort of interesting that the solution involve's Fermat's last theorem. So before they started proving the special cases with computers, we, technically, wouldn't have a proof. Or is there another solution that doesn't involve FLT? | ||
Muirhead
United States556 Posts
+ Show Spoiler + The special case of FLT with exponent 3 is quite easy compared to the general result: It was first proven by Euler centuries ago | ||
evanthebouncy!
United States12796 Posts
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numLoCK
Canada1416 Posts
Just a question, what level math is this? | ||
illu
Canada2531 Posts
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moriya
United States54 Posts
+ Show Spoiler + The only solution is (-1,1) and (1,1). Easy to get x is odd,so let x=2z+1 and do some algebra: y^3=3z^2+3z+1=(z+1)^3-z^3. Look at the above Eq. again and recall FLT, the only solution is y=0//impossible or z=0//(1,1) solution or (z+1)=0//(-1,1) solution. Hope it is clear. | ||
moriya
United States54 Posts
He is the first solver. | ||
Self Help
45 Posts
On July 06 2009 11:36 Archaic wrote: Also, don't try to pretend you're quizzing us when it is probably just a question straight out of your Summer HW. I know the guy personally I doubt he's asking you to do his homework for him, he's more of a math enthusiast as given by his major and school he's attending. | ||
doghunter
United States23 Posts
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Raithed
China7077 Posts
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Jonoman92
United States9091 Posts
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pat777
United States356 Posts
On July 06 2009 11:36 Archaic wrote: Also, don't try to pretend you're quizzing us when it is probably just a question straight out of your Summer HW. Next time I'll put the answer in spoiler tags just to prove that I am not trying to get HW help. Also, the answer is already online here (don't click unless you want to be spoiled): http://www.math.washington.edu/~challenge/archive/20090505/flt.pdf btw, thanks to all who defended me. btw, congrats to Muirhead and moriya for solving the problem. Cheeseball had an incomplete proof. | ||
pat777
United States356 Posts
On July 06 2009 16:14 Jonoman92 wrote: Why is it obvious that x is odd? Sorry i'm a bit slow. I see that you set x equal to what its exponent was so I guess you can just disregard the coefficient for some reason. + Show Spoiler + if x is even, then 3x^2+1 will be odd and 4y^3 is always even. | ||
FirstBorn
Romania3955 Posts
This is not an exception. | ||
pat777
United States356 Posts
On July 06 2009 17:17 FirstBorn wrote: Most of these math problem threads make me feel quite stupid. This is not an exception. Don't feel that way, not every smart person loves math. I just post these threads to give math enthusiasts fun. | ||
d3_crescentia
United States4053 Posts
On July 06 2009 16:16 pat777 wrote: Next time I'll put the answer in spoiler tags just to prove that I am not trying to get HW help. Also, the answer is already online here (don't click unless you want to be spoiled): http://www.math.washington.edu/~challenge/archive/20090505/flt.pdf btw, thanks to all who defended me. btw, congrats to Muirhead and moriya for solving the problem. Cheeseball had an incomplete proof. You know, you could avoid such situations by explaining that you're posting math problems for people to solve for fun at the beginning of each post. The other thing is that non-mathematicians will glance at these problems and be reminded of high-school algebra. | ||
pat777
United States356 Posts
On July 06 2009 18:04 d3_crescentia wrote: You know, you could avoid such situations by explaining that you're posting math problems for people to solve for fun at the beginning of each post. The other thing is that non-mathematicians will glance at these problems and be reminded of high-school algebra. Yeah sorry, I just don't have much experience with making threads. | ||
Freyr
United States500 Posts
On July 06 2009 18:04 d3_crescentia wrote: You know, you could avoid such situations by explaining that you're posting math problems for people to solve for fun at the beginning of each post. The other thing is that non-mathematicians will glance at these problems and be reminded of high-school algebra. His critics did not misunderstand his reasons for posting, they just did not believe them. They are obviously idiots who too often reach for their keyboards before their brains. Yes it might look a bit like high school algebra but mathematician or no, trying to solve it will clearly show that it is not. The fact that it says "number theory" in the title is also a bit of a clue. Thanks for these threads pat, I'm really enjoying them. | ||
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